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| -rw-r--r-- | Electronic_Devices_by_S._Sharma/Chapter03.ipynb | 958 | ||||
| -rw-r--r-- | Electronic_Devices_by_S._Sharma/Chapter04.ipynb | 1073 | ||||
| -rw-r--r-- | Electronic_Devices_by_S._Sharma/Chapter05.ipynb | 1852 | ||||
| -rw-r--r-- | Electronic_Devices_by_S._Sharma/Chapter06.ipynb | 1272 | ||||
| -rw-r--r-- | Electronic_Devices_by_S._Sharma/Chapter07.ipynb | 156 | ||||
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diff --git a/Electronic_Devices_by_S._Sharma/Chapter02.ipynb b/Electronic_Devices_by_S._Sharma/Chapter02.ipynb new file mode 100644 index 00000000..f88fc401 --- /dev/null +++ b/Electronic_Devices_by_S._Sharma/Chapter02.ipynb @@ -0,0 +1,1166 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:3093b2582ce1a8c2faa2918ba63b34343ccd0f5c63ea8cea7962fa3c37c56111"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 02 - ENERGY BANDS AND CHARGE CARRIERS IN SEMICONDUCTORS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 38"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.1 - 38\n",
+ "# Given data\n",
+ "lembda = 11000.;#\n",
+ "lembda = lembda * 10.**-10.;# in m\n",
+ "h = 6.625*10.**-34.;# Planck constant\n",
+ "c = 3.*10.**8.;#speed of light in m/s\n",
+ "e = 1.6*10.**-19.;#charge of electron in C\n",
+ "# Energy of the incident photon should at least be, h*v= Eg, so\n",
+ "E_g = (h*c)/(lembda*e);# in eV\n",
+ "print '%s %.2f %s' %(\"The energy gap in eV is\",E_g,\"\\n\");\n",
+ "#Answer in textbook is different due to rounding error\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy gap in eV is 1.13 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 38"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.2 -38\n",
+ "# Given data\n",
+ "E_g = 0.75;# in eV\n",
+ "e = 1.6*10.**-19.;# in C\n",
+ "h = 6.63*10.**-34.;# in J\n",
+ "c = 3*10**8.;# in m/s\n",
+ "#Formula E_g = (h*c)/(lembda*e);\n",
+ "lembda = (h*c)/(E_g*e);# in m\n",
+ "lembda = lembda * 10.**10.;# in A\n",
+ "print '%s %.f %s' %(\"The wavelength in A is\",lembda,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength in A is 16575 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.3 - 53\n",
+ "# Given data\n",
+ "del_E = 0.3;# in eV\n",
+ "T1 = 300.;# in K\n",
+ "T2 = 330.;# in K\n",
+ "# del_E = K * T1 * log(N/N_c) where del_E= E_C-E_F\n",
+ "# del_E1 = K * T2 * log(N/N_c) where del_E1= E_C-E_F at T= 330 degree K\n",
+ "del_E1 = del_E*(T2/T1);# in eV \n",
+ "print '%s %.2f %s' %(\"The Fermi level will be eV below the conduction band\",del_E1,\"eV\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Fermi level will be eV below the conduction band 0.33 eV\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 - Pg 54"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.4 -54\n",
+ "# Given data\n",
+ "import math\n",
+ "N_c = 2.8 * 10.**19.;# in cm**-3\n",
+ "del_E = 0.25;# fermi energy in eV\n",
+ "KT = 0.0259;# where K is Boltzmann constant\n",
+ "f_F = math.exp(-(del_E)/KT);\n",
+ "print '%s %.2e %s' %(\"The probability in the conduction band is occupied by an electron is \",f_F,\"\\n\");\n",
+ "# Evaluation of electron concentration\n",
+ "n_o = N_c * math.exp(-(del_E)/KT);# in cm**-3\n",
+ "print '%s %.2e %s' %(\"The thermal equilibrium electron concentration in cm**-3 is\",n_o,\"\\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The probability in the conduction band is occupied by an electron is 6.43e-05 \n",
+ "\n",
+ "The thermal equilibrium electron concentration in cm**-3 is 1.80e+15 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 - Pg 54"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa2.5-54\n",
+ "# Given data\n",
+ "import math\n",
+ "T1 = 300.;# in K\n",
+ "T2 = 400.;# in K\n",
+ "del_E = 0.27;# Fermi level in eV\n",
+ "#KT = (0.0259) * (T2/T1);# in eV\n",
+ "KT=0.03453\n",
+ "N_v = 1.60 * 10.**19.;# in cm**-3\n",
+ "#N_v = N_v * (T2/T1)**(3./2.);# in cm**-3 \n",
+ "# Hole concentration\n",
+ "p_o = N_v * math.exp(-(del_E)/KT);# in per cm**3\n",
+ "print '%s %.2e %s' %(\"The thermal equilibrium hole concentration per cm**3 is\",p_o,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thermal equilibrium hole concentration per cm**3 is 6.43e+15 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 - Pg 58"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.6 - 58\n",
+ "# Given data\n",
+ "At = 63.5;# atomic weight\n",
+ "Rho = 1.7*10.**-6.;# in ohm cm\n",
+ "d = 8.96;# in gm/cc\n",
+ "N_A = 6.02*10.**23.;# in /gm.mole\n",
+ "e = 1.6*10.**-19.;# in C\n",
+ "#Number of atoms of copper persent per unit volume\n",
+ "n = (N_A/At)*d;\n",
+ "Miu_e = 1./(Rho*n*e);# in cm**2/volt.sec\n",
+ "print '%s %.3f %s' %(\"The electron mobility is\",Miu_e,\"cm**2/volt-sec\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron mobility is 43.281 cm**2/volt-sec\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E07 - Pg 59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.7 -59\n",
+ "# Given data\n",
+ "l = 0.1;# in m\n",
+ "A = 1.7;# in mm**2\n",
+ "A = A * 10.**-6.;# in m**2\n",
+ "R = 0.1;# in ohm\n",
+ "At = 63.5;# atomic weight\n",
+ "N_A = 6.02*10.**23.;\n",
+ "d = 8.96;# in gm/cc\n",
+ "n = (N_A/At)*d;# in /cc\n",
+ "n = n * 10.**6.;# in /m**3\n",
+ "e = 1.6*10.**-19.;#electron charge in C\n",
+ "# Resistivity of copper\n",
+ "#Formula R = Rho*(l/A);\n",
+ "Rho = (R*A)/l;# in ohm m\n",
+ "# Conductivity of copper\n",
+ "Sigma = 1./Rho;# in mho/m\n",
+ "# Formula Sigma = n*e*Miu_e\n",
+ "Miu_e = Sigma/(n*e);# in m**2/V.sec\n",
+ "print '%s %.6f %s' %(\"The mobility in m**2/V-sec is\",Miu_e,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility in m**2/V-sec is 0.000043 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 - Pg 59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.8- 59\n",
+ "# Given data\n",
+ "import math\n",
+ "d = 10.5;# in gm/cc\n",
+ "At = 108.;# atomic weight\n",
+ "N_A = 6.025*10.**23.;# in /gm mole\n",
+ "r = 10**-3.;# in m\n",
+ "q = 1.6*10.**-19.;# in C\n",
+ "# The number of electrons per unit volume\n",
+ "n = (N_A/At)*d;# in /cm**3\n",
+ "n = n * 10.**6.;# in /m**3\n",
+ "A = math.pi*((r)**2.);# in m**2\n",
+ "I = 2.;# in A\n",
+ "# Evaluation of drivt velocity with the help of current\n",
+ "# I = q*n*A*V;\n",
+ "V = I/(n*q*A);# in m/s\n",
+ "print '%s %.e %s' %(\"The drift velocity in m/s is\",V,\"\\n\");\n",
+ "\n",
+ "# Note: Calculation in the book is wrong, so the answer in the book is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The drift velocity in m/s is 7e-05 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E09 - Pg 59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.9-59\n",
+ "import math\n",
+ "# Given data\n",
+ "d = 1.03;# in mm\n",
+ "d = d *10.**-3.;# in m\n",
+ "r = d/2.;# in m\n",
+ "R = 6.51;# in ohm\n",
+ "l = 300.;# in mm\n",
+ "e = 1.6*10.**-19.;# electron charge in C\n",
+ "n = 8.4*10.**28.;# in /m**3\n",
+ "A = math.pi*r**2.;# cross section area\n",
+ "#Formula R = Rho*(l/A);\n",
+ "Rho = (R* A)/l;#in ohm m\n",
+ "Sigma = 1./Rho;# in mho/m\n",
+ "print '%s %.2e %s' %(\"The conductivity of copper in mho/m is\",Sigma,\"\\n\");\n",
+ "# Evaluation of mobility\n",
+ "#Formula sigma = n*e*Miu_e\n",
+ "Miu_e = Sigma/(n*e);# in m**2/V.sec\n",
+ "print '%s %.2e %s' %(\"The mobility in m**2/V-sec is\",Miu_e,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of copper in mho/m is 5.53e+07 \n",
+ "\n",
+ "The mobility in m**2/V-sec is 4.12e-03 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 - Pg 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.10-61\n",
+ "# Given data\n",
+ "Mu_e = 1500.;# in cm**2/volt sec\n",
+ "Mu_h = 500.;# in cm**2/volt sec\n",
+ "n_i = 1.6 * 10.**10.;# in per cm**3\n",
+ "e = 1.6 * 10.**-19.;# in C\n",
+ "# The conductivity of pure semiconductor \n",
+ "Sigma = n_i * (Mu_e + Mu_h) * e;# in mho/cm\n",
+ "print '%s %.2e %s' %(\"The conductivity of pure semiconductor in mho/cm is\",Sigma,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of pure semiconductor in mho/cm is 5.12e-06 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 - Pg 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.11-61\n",
+ "# Given data\n",
+ "Rho = 10.;# in ohm-cm\n",
+ "Mu_d = 500.;# in cm**2/v.s.\n",
+ "e = 1.6*10.**-19.;# electron charge in C\n",
+ "# The number of donor atom\n",
+ "n_d = 1./(Rho * e * Mu_d);# in per cm**3\n",
+ "print '%s %.2e %s' %(\"The number of donor atom per cm**3 is \",n_d,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of donor atom per cm**3 is 1.25e+15 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 - Pg 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.12-62\n",
+ "#Given data\n",
+ "AvagadroNumber = 6.02 * 10.**23.;# in atoms/gm.mole\n",
+ "at_Ge = 72.6;# atom weight of Ge\n",
+ "e = 1.6 * 10.**-19.;# in C\n",
+ "D_Ge = 5.32;# density of Ge in gm/c.c\n",
+ "Mu = 3800.;# in cm**2/v.s.\n",
+ "C_Ge = (AvagadroNumber/at_Ge) * D_Ge;# concentration of Ge atoms in per cm**3\n",
+ "n_d = C_Ge/10.**8.;# in per cc\n",
+ "Sigma = n_d * Mu * e;# in mho/cm\n",
+ "print '%s %.3f %s' %(\"The conductivity in mho/cm is\",Sigma,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity in mho/cm is 0.268 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13 - Pg 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa2.13-62\n",
+ "# Given data\n",
+ "Rho = 0.3623 * 10.**-3.;# in Ohm m\n",
+ "Sigma = 1/Rho;#in mho/m\n",
+ "D = 4.42 * 10.**28.;# Ge density in atom/m**3\n",
+ "n_d = D / 10.**6.;# in atom/m**3\n",
+ "e = 1.6 * 10.**-19.;# in C\n",
+ "# The mobility of electron in germanium \n",
+ "Mu = Sigma/(n_d * e);# in m**2/V.sec\n",
+ "print '%s %.2f %s' %(\"The mobility of electron in germanium in m**2/V.sec is\",Mu,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility of electron in germanium in m**2/V.sec is 0.39 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14 - Pg 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.14-62\n",
+ "# Given data\n",
+ "AvagadroNumber = 6.025 * 10.**26.;# in kg.Mole\n",
+ "W = 72.59;# atomic weight of Ge\n",
+ "D = 5.36 * 10.**3.;#density of Ge in kg/m**3\n",
+ "Rho = 0.42;# resistivity in Ohm m\n",
+ "e = 1.6 * 10.**-19.;# in C\n",
+ "Sigma = 1./Rho;# in mho/m\n",
+ "n = (AvagadroNumber/W) * D;# number of Ge atoms present per unit volume\n",
+ "# Holes per unit volume, H = n*10**-6%\n",
+ "H= n*10.**-8.;\n",
+ "a=H;\n",
+ "# Formula sigma= a*e*Mu_h\n",
+ "Mu_h = Sigma/(a * e);# in m**2/V.sec\n",
+ "print '%s %.4f %s' %(\"Mobility of holes in m**2/V.sec is\",Mu_h,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mobility of holes in m**2/V.sec is 0.0334 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E15 - Pg 63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.15-63\n",
+ "# Given data\n",
+ "e = 1.6 * 10.**-19.;# in C\n",
+ "n_i = 2 * 10.**19.;# in /m**3\n",
+ "Mu_e = 0.36;# in m**2/v.s\n",
+ "Mu_h = 0.17;# in m**2/v.s\n",
+ "A = 1. * 10.**-4.;# in m**2\n",
+ "V = 2.;#in volts\n",
+ "l = 0.3;# in mm\n",
+ "l = l * 10.**-3.;# in m\n",
+ "E=V/l;# in volt/m\n",
+ "Sigma = n_i * e * (Mu_e + Mu_h);# in mho/m\n",
+ "# J = I/A = Sigma * E\n",
+ "I= Sigma*E*A;\n",
+ "print '%s %.2f %s' %(\"The current produced in a small germanium plate in amp is\",I,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current produced in a small germanium plate in amp is 1.13 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E16 - Pg 63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.16-63\n",
+ "# Given data\n",
+ "D = 4.2 * 10.**28.;#density of Ge atoms per m**3\n",
+ "N_d = D / 10.**6.;# per m**3\n",
+ "e = 1.6 * 10.**-19.;# in C\n",
+ "Mu_e = 0.36;# in m**2/V-sec\n",
+ "# Donor concentration is very large as compared to intrinsic carrier concentration\n",
+ "Sigma_n = N_d * e * Mu_e;# in mho/m (intrinsic concentration can be neglected)\n",
+ "Rho_n = 1./Sigma_n;# in ohm m\n",
+ "print '%s %.3e %s' %(\"The resistivity of drop Ge in ohm m is \",Rho_n,\"\\n\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistivity of drop Ge in ohm m is 4.134e-04 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E17 - Pg 64"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.17-64\n",
+ "# given data\n",
+ "e = 1.6 * 10.**-19.;# in C\n",
+ "n_i = 1 * 10.**19.;# in per m**3\n",
+ "Mu_e = 0.36;# in m**2/volt.sec\n",
+ "Mu_h = 0.17;# in m**2/volt.sec \n",
+ "A = 2.;# in cm**2\n",
+ "A = A * 10.**-4.;# im m**2\n",
+ "t = 0.1;# in mm\n",
+ "t = t * 10.**-3.;# in m\n",
+ "V = 4.;# in volts\n",
+ "Sigma_i = n_i * e * (Mu_e + Mu_h);# in mho/m\n",
+ "J = Sigma_i * (V/t);# in Amp/m**2\n",
+ "# Current produced, I= J*A\n",
+ "I = J * A;# in Amp\n",
+ "print '%s %.3f %s' %(\"The current produced in a Ge sample in Amp is\",I,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current produced in a Ge sample in Amp is 6.784 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E18 - Pg 64"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.18-64\n",
+ "# Given data\n",
+ "e = 1.6 * 10.**-19.;# in C\n",
+ "Mu_h = 500.;# in cm**2/V.s.\n",
+ "Mu_e = 1500.;# in cm**2/V.s.\n",
+ "n_i = 1.6 * 10.**10.;# in per cm**3\n",
+ "# Conductivity of pure silicon at room temperature \n",
+ "Sigma_i = n_i * e * ( Mu_h + Mu_e);# in mho/cm\n",
+ "print '%s %.2e %s' %(\"Conductivity of pure silicon at room temperature in mho/cm is\",Sigma_i,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Conductivity of pure silicon at room temperature in mho/cm is 5.12e-06 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E19 - Pg 67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.19-67\n",
+ "#Given data\n",
+ "l= 0.50*10.**-2.;# width of ribbon in m\n",
+ "d= 0.10*10.**-3.;# thickness of ribbon in m\n",
+ "A= l*d;# area of ribbon in m**2\n",
+ "B = 0.8;# in Tesla\n",
+ "D = 10.5;#density in gm/cc\n",
+ "I = 2.;# in amp\n",
+ "q = 1.6 * 10.**-19.;# in C\n",
+ "n=6.*10.**28.;# number of elec. per m**3\n",
+ "V_H = ( I * B * d)/(n * q * A);# in volts\n",
+ "print '%s %.2e %s' %(\"The hall Voltage produced in volts is\",V_H,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hall Voltage produced in volts is 3.33e-08 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E20 - Pg 68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.20-68\n",
+ "# Given data\n",
+ "l = 1.;# in m\n",
+ "d = 1.;# in cm\n",
+ "d = d * 10.**-2.;# in m\n",
+ "W = 1.;# in mm\n",
+ "W = W * 10.**-3.;# in m\n",
+ "A = d * W;# in m**2\n",
+ "I= 1.;# in A\n",
+ "B = 1.;# Tesla\n",
+ "V_H = 0.074 * 10.**-6.;# in volts\n",
+ "Sigma = 5.8 * 10.**7.;# in mho/m\n",
+ "# The hall coefficient \n",
+ "R_H = (V_H * A)/(B*I*d);# in m**3/c\n",
+ "print '%s %.1e %s' %(\"The hall coefficient in m**3/c is\",R_H,\"\\n\");\n",
+ "# Mobility of electrons in copper \n",
+ "Mu = Sigma * R_H;# in m**2/volt-sec\n",
+ "print '%s %.1e %s' %(\"The mobility of electrons in copper in m**2/volt-sec is \",Mu,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hall coefficient in m**3/c is 7.4e-11 \n",
+ "\n",
+ "The mobility of electrons in copper in m**2/volt-sec is 4.3e-03 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E21 - Pg 69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa2.2169\n",
+ "# Given data\n",
+ "n_i = 1.4 * 10.**18.;# in /m**3\n",
+ "n_D = 1.4 * 10.**24.;# in /m**3\n",
+ "# Concentration of electrons\n",
+ "n=n_D;# in /m**3\n",
+ "p = n_i**2./n;# in /m**3\n",
+ "# The ratio of electrons to hole concentration\n",
+ "R = n/p;\n",
+ "print '%s %.e %s' %(\"The ratio of electrons to hole concentration is\",R,\"\\n\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of electrons to hole concentration is 1e+12 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E22 - Pg 69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.22- 69\n",
+ "#Given data\n",
+ "R = 9. * 10.**-3.;# in ohm-m\n",
+ "R_H = 3.6 * 10.**-4.;# in m**3\n",
+ "e = 1.6 * 10.**-19.;# in C\n",
+ "Sigma = 1./R;# in (ohm-m)**-1\n",
+ "#Rho = 1./R_H;# in coulomb/m**3\n",
+ "Rho=2778.\n",
+ "# Density of charge carriers \n",
+ "n = Rho/e;# in /m**3\n",
+ "print '%s %.5e %s' %(\"Density of charge carriers per m**3 is\",n,\"\\n\");\n",
+ "# Mobility of charge carriers \n",
+ "Mu = Sigma * R_H;# in m**2/v-s\n",
+ "print '%s %.2f %s' %(\"Mobility of charge carriers in m**2/V-s is\",Mu,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Density of charge carriers per m**3 is 1.73625e+22 \n",
+ "\n",
+ "Mobility of charge carriers in m**2/V-s is 0.04 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E23 - Pg 77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.23-77\n",
+ "# Given data\n",
+ "e = 1.6 * 10.**-19.;# in C\n",
+ "R_H = 0.0145;# in m**3/coulomb\n",
+ "Mu_e = 0.36;# in m**2/v-s\n",
+ "E = 100.;# in V/m\n",
+ "n = 1./(e * R_H);# in /m**3\n",
+ "# The current density of specimen \n",
+ "J = n * e * Mu_e * E;# in A/m**2\n",
+ "print '%s %.2e %s' %(\"The current density of specimen in A/m**2 is\",J,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current density of specimen in A/m**2 is 2.48e+03 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E24 - Pg 77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.24-77\n",
+ "import math\n",
+ "#Given data\n",
+ "Mu_e = 7.04 * 10.**-3.;# in m**2/v-s\n",
+ "m = 9.1 * 10.**-31.;\n",
+ "E_F = 5.5;# in eV\n",
+ "n = 5.8 * 10.**28.;\n",
+ "e = 1.6 * 10.**-19.;# in C\n",
+ "# Relaxation Time \n",
+ "Torque = (Mu_e/e) * m;# in sec\n",
+ "print '%s %.1e %s' %(\"Relaxation Time in sec is \",Torque,\"\\n\");\n",
+ "# Resistivity of conductor \n",
+ "Rho = 1. /(n * e * Mu_e);# in ohm-m\n",
+ "print '%s %.3e %s' %(\"Resistivity of conductor in ohm-m is \",Rho,\"\\n\");\n",
+ "# Velocity of electrons with fermi-energy \n",
+ "V_F = math.sqrt((2 * E_F * e)/m);# in m/s\n",
+ "print '%s %.4e %s' %(\"Velocity of electrons with fermi-energy in m/s is\",V_F,\"\\n\");\n",
+ "\n",
+ "#Note: The calculated value of Resistivity of conductor is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Relaxation Time in sec is 4.0e-14 \n",
+ "\n",
+ "Resistivity of conductor in ohm-m is 1.531e-08 \n",
+ "\n",
+ "Velocity of electrons with fermi-energy in m/s is 1.3907e+06 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E25 - Pg 78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.25-78\n",
+ "import math\n",
+ "# Given data\n",
+ "E= 5.95;# in eV\n",
+ "EF= 6.25;# in eV\n",
+ "delE= 0.01;\n",
+ " # delE= 1-1/(1+exp((E-EF)/KT))\n",
+ "K=1.38*10.**-23.;# Boltzmann Constant in J/K\n",
+ "# The temperature at which there is a 1 % probability that a state 0.30 eV below the Fermi energy level\n",
+ "T = ((E-EF)/math.log(1./(1.-delE) -1.)*1.6*10.**-19.)/K;# in K\n",
+ "print '%s %.f %s' %(\"The temperature in K is : \",T,\"\\n\")\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature in K is : 757 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E26 - Pg 78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.26-78\n",
+ "# Given data \n",
+ "import math\n",
+ "N_V = 1.04 * 10.**19.;# in cm**-3\n",
+ "T1 = 300.;# in K\n",
+ "T2 = 400.;# in K\n",
+ "del_E = 0.27;# in eV\n",
+ "# The value of N_V at T=400 K,\n",
+ "N_V = N_V * (T2/T1)**1.5;# in cm**-3\n",
+ "KT = (0.0259) * (T2/T1);# in eV\n",
+ "# The thermal equilibrium hole concentration in silicon \n",
+ "P_o = N_V * math.exp(-(del_E)/KT);# in cm**-3\n",
+ "print '%s %.2e %s' %(\"The thermal equilibrium hole concentration in silicon in cm**-3 is \",P_o,\"\\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thermal equilibrium hole concentration in silicon in cm**-3 is 6.44e+15 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E27 - Pg 78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.27-78\n",
+ "import math\n",
+ "#Given data\n",
+ "N_c = 2.8 * 10.**19.;\n",
+ "N_V = 1.04 *10.**19.;\n",
+ "T1 = 550.;# in K\n",
+ "T2 = 300.;# in K\n",
+ "E_g = 1.12;\n",
+ "KT = (0.0259) ;\n",
+ "n_i = math.sqrt(N_c *N_V *(T1/T2)**3.* math.exp(-(E_g)/KT*T2/T1));# in cm**-3\n",
+ "# n_o = N_d/2 + sqrt((N_d/2)**2 + (n_i)**2)\n",
+ "# 1.05*N_d -N_d/2= sqrt((N_d/2)**2 + (n_i)**2)\n",
+ "# Minimum donor concentration required \n",
+ "N_d=math.sqrt((n_i)**2./((0.55)**2.-1./4.));\n",
+ "print '%s %.2e %s' %(\"Minimum donor concentration required in cm**-3 is\",N_d,\"\\n\"); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum donor concentration required in cm**-3 is 1.40e+15 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E28 - Pg 79"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 2.28-79\n",
+ "import math\n",
+ "#Given data\n",
+ "T = 300.;# in K\n",
+ "n_o = 10.**15.;# in cm**-3\n",
+ "n_i = 10.**10.;# in cm**-3\n",
+ "p_o = 10.**5.;# in cm**-3\n",
+ "del_n = 10.**13.;# in cm**-3\n",
+ "del_p = del_n;# in cm**-3\n",
+ "KT = 0.0259;# in eV\n",
+ "delta_E1= KT*math.log(n_o/n_i);# value of E_F-E_Fi in eV\n",
+ "delta_E2= KT*math.log((n_o+del_n)/n_i);# value of E_Fn-E_Fi in eV\n",
+ "delta_E3= KT*math.log((p_o+del_p)/n_i);# value of E_Fi-E_Fp in eV\n",
+ "print '%s %.4f %s' %(\"The Fermi level for thermal equillibrium in eV is : \",delta_E1,\"\\n\")\n",
+ "print '%s %.4f %s' %(\"The quase-Fermi level for electrons in non equillibrium in eV is : \",delta_E2,\"\\n\")\n",
+ "print '%s %.3f %s' %(\"The quasi-Fermi level for holes in non equillibrium in eV is : \",delta_E3,\"\\n\")\n",
+ "print '%s' %(\"The quasi-Fermi level for electrons is above E_Fi \")\n",
+ "print '%s' %(\"While the quasi-Fermi level for holes is below E_Fi\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Fermi level for thermal equillibrium in eV is : 0.2982 \n",
+ "\n",
+ "The quase-Fermi level for electrons in non equillibrium in eV is : 0.2984 \n",
+ "\n",
+ "The quasi-Fermi level for holes in non equillibrium in eV is : 0.179 \n",
+ "\n",
+ "The quasi-Fermi level for electrons is above E_Fi \n",
+ "While the quasi-Fermi level for holes is below E_Fi\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Devices_by_S._Sharma/Chapter03.ipynb b/Electronic_Devices_by_S._Sharma/Chapter03.ipynb new file mode 100644 index 00000000..2240f135 --- /dev/null +++ b/Electronic_Devices_by_S._Sharma/Chapter03.ipynb @@ -0,0 +1,958 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:4679a339709b832cb00c36a2b754011b8d8174a67064702f6c96acb48c7069c1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 03 - EXCESS CARRIERS IN SEMICONDUCTORS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 - Pg 117"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.1 - 117\n",
+ "# Given data\n",
+ "N_d = 10.**17.;# atoms/cm**3\n",
+ "n_i = 1.5 * 10.**10.;# in /cm**3\n",
+ "n_o = 10.**17.;# in cm**3\n",
+ "# p_o * n_o = (n_i)**2\n",
+ "p_o = (n_i)**2. / n_o;#in holes/cm**3\n",
+ "print '%s %.2e %s' %(\"The hole concentration at equilibrium in holes/cm**3 is\",p_o,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hole concentration at equilibrium in holes/cm**3 is 2.25e+03 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.3- 118\n",
+ "import math\n",
+ "# Given data\n",
+ "n_i = 1.5 * 10. **10.;# in /cm**3 for silicon\n",
+ "N_d = 10.**17.;# in atoms/cm**3\n",
+ "n_o = 10.**17.;# electrons/cm**3\n",
+ "KT = 0.0259;\n",
+ "# E_r - E_i = KT * log(n_o/n_i)\n",
+ "del_E = KT * math.log(n_o/n_i);# in eV\n",
+ "print '%s %.3f %s' %(\"The energy band for this type material is Ei eV\",del_E,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy band for this type material is Ei eV 0.407 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 - Pg 118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.4 - 118\n",
+ "# Given data\n",
+ "K = 1.38 * 10.**-23.;# in J/K\n",
+ "T = 27.;# in degree\n",
+ "T = T + 273.;# in K\n",
+ "e = 1.6 * 10.**-19.;# in C\n",
+ "Mu_e = 0.17;# in m**2/v-s\n",
+ "Mu_e1 = 0.025;# in m**2/v-s\n",
+ "D_n = ((K * T)/e) * Mu_e;# in m**2/s\n",
+ "print '%s %.2e %s' %(\"The diffusion coefficient of electrons in m**2/s is\",D_n,\"\\n\");\n",
+ "D_p = ((K * T)/e) * Mu_e1;# in m**2/s\n",
+ "print '%s %.2e %s' %(\"The diffusion coefficient of holes in m**2/s is \",D_p,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diffusion coefficient of electrons in m**2/s is 4.40e-03 \n",
+ "\n",
+ "The diffusion coefficient of holes in m**2/s is 6.47e-04 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 - Pg 119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.5 - 119\n",
+ "import math\n",
+ "# Given data\n",
+ "Mu_n = 0.15;# in m**2/v-s\n",
+ "K = 1.38 * 10.**-23.; # in J/K\n",
+ "T = 300.;# in K\n",
+ "del_n = 10.**20.;# in per m**3\n",
+ "Toh_n = 10.**-7.;# in s\n",
+ "e = 1.6 * 10.**-19.;# in C\n",
+ "D_n = Mu_n * ((K * T)/e);# in m**2/s\n",
+ "print '%s %.2e %s' %(\"The diffusion coefficient in m**2/s is\",D_n,\"\\n\");\n",
+ "L_n = math.sqrt(D_n * Toh_n);# in m \n",
+ "print '%s %.2e %s' %(\"The Diffusion length in m is\",L_n,\"\\n\");\n",
+ "J_n = (e * D_n * del_n)/L_n;# in A/m**2\n",
+ "print '%s %.2e %s' %(\"The diffusion current density in A/m**2 is\",J_n,\"\\n\"); \n",
+ "# Note : The value of diffusion coefficient in the book is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diffusion coefficient in m**2/s is 3.88e-03 \n",
+ "\n",
+ "The Diffusion length in m is 1.97e-05 \n",
+ "\n",
+ "The diffusion current density in A/m**2 is 3.15e+03 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 - Pg 119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.6 - 119\n",
+ "# Given data\n",
+ "Sigma = 0.1;# in (ohm-m)**-1\n",
+ "Mu_n = 1300.;\n",
+ "n_i = 1.5 * 10.**3.;\n",
+ "q = 1.6 * 10.**-4.;# in C\n",
+ "#n_n = Sigma/(Mu_n * q);# in electrons/cm**3\n",
+ "n_n=4.81*10.**14.\n",
+ "n_n= n_n*10.**6.;# per m**3\n",
+ "print '%s %.2e %s' %(\"The concentration of electrons per m**3 is\",n_n,\"\\n\");\n",
+ "#p_n = (n_i)**2./n_n;# in per cm**3\n",
+ "p_n=4.68*10.**5.\n",
+ "p_n = p_n * 10.**6.;# in per m**3\n",
+ "print '%s %.2e %s' %(\"The concentration of holes per m**3 is\",p_n,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The concentration of electrons per m**3 is 4.81e+20 \n",
+ "\n",
+ "The concentration of holes per m**3 is 4.68e+11 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E07 - Pg 119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.7 - 119\n",
+ "# Given data\n",
+ "Mu_e = 0.13;# in m**2/v-s\n",
+ "Mu_h = 0.05;# in m**2/v-s\n",
+ "Toh_h = 10.**-6.;# in s\n",
+ "#L = 100.;# in um\n",
+ "L = 100. * 10.**-6.;# in m\n",
+ "V = 12.;# in V\n",
+ "t_n =L**2./(Mu_e * V);# in s\n",
+ "print '%s %.2e %s' %(\"Electron transit time in seconds is\",t_n,\"\\n\");\n",
+ "p_g = (Toh_h/t_n) * (1. + Mu_h/Mu_e);#photo conductor gain \n",
+ "print '%s %.2f %s' %(\"Photo conductor gain is\",p_g,\"\\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electron transit time in seconds is 6.41e-09 \n",
+ "\n",
+ "Photo conductor gain is 216.00 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 - Pg 120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.8 - 120\n",
+ "#Given data\n",
+ "n_i = 2.5 * 10.**13.;\n",
+ "Mu_n = 3800.;\n",
+ "Mu_p = 1800.;\n",
+ "q = 1.6 * 10.**-19.;# in C\n",
+ "Sigma = n_i * (Mu_n + Mu_p) * q;# in (ohm-cm)**-1\n",
+ "Rho = 1./Sigma;# in ohm-cm\n",
+ "Rho= round(Rho);\n",
+ "print '%s %.2f %s' %(\"The resistivity of intrinsic germanium in ohm-cm is\",Rho,\"\\n\");\n",
+ "N_D = 4.4 * 10.**22./(1.*10.**8.);# in atoms/cm**3\n",
+ "Sigma_n = N_D * Mu_n * q;# in (ohm-cm)**-1\n",
+ "Rho_n = 1./Sigma_n;# in ohm-cm\n",
+ "print '%s %.2f %s' %(\"If a donor type impurity is added to the extent of 1 atom per 10**8 Ge atoms, then the resistivity drops in ohm-cm is\",Rho_n,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistivity of intrinsic germanium in ohm-cm is 45.00 \n",
+ "\n",
+ "If a donor type impurity is added to the extent of 1 atom per 10**8 Ge atoms, then the resistivity drops in ohm-cm is 3.74 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E09 - Pg 120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.9 - 120\n",
+ "# Given data\n",
+ "n_i = 10.**16.;# in /m3\n",
+ "N_D = 10.**22.;# in /m**3\n",
+ "n = N_D;# in /m**3\n",
+ "print '%s %.e %s' %(\"Electron concentration per m**3 is\",n,\"\\n\");\n",
+ "p = (n_i)**2./n;# in /m**3\n",
+ "print '%s %.e %s' %(\"Hole concentration per m**3 is\",p,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electron concentration per m**3 is 1e+22 \n",
+ "\n",
+ "Hole concentration per m**3 is 1e+10 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 - Pg 120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.10 - 120\n",
+ "# Given data\n",
+ "Rho = 9.6 * 10.**-2.;# in ohm-m\n",
+ "Sigma_n = 1./Rho;# in (ohm-m)**-1\n",
+ "q = 1.6 * 10.**-19.;# in C\n",
+ "Mu_n = 1300. * 10.**-4.;# in m**2/V-sec\n",
+ "N_D = Sigma_n / (Mu_n * q);# in atoms/m**3\n",
+ "A_D = 5.*10.**22.;# Atom density in atoms/cm**3\n",
+ "A_D = A_D * 10.**6.;# atoms/m**3\n",
+ "R_si = N_D/A_D;# ratio\n",
+ "print '%s %.e %s' %(\"The ratio of donor atom to silicon atom is\",R_si,\"\\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of donor atom to silicon atom is 1e-08 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 - Pg 121"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.11 - 121\n",
+ "# Given data\n",
+ "n_i = 1.5 * 10.**10.;# in per cm**3\n",
+ "n_n = 2.25 * 10.**15.;# in per cm**3\n",
+ "p_n = (n_i)**2./n_n;# in per cm**3\n",
+ "print '%s %.e %s' %(\"The equilibrium electron per cm**3 is\",p_n,\"\\n\");\n",
+ "h_n = n_n;# in cm**3\n",
+ "print '%s %.2e %s' %(\"Hole densities in per cm**3 is\",h_n,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equilibrium electron per cm**3 is 1e+05 \n",
+ "\n",
+ "Hole densities in per cm**3 is 2.25e+15 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 - Pg 121"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.12 - 121\n",
+ "# Given data\n",
+ "N_A = 2. * 10.**16.;# in atoms/cm**3\n",
+ "N_D = 10.**16.;# in atoms/cm**3\n",
+ "C_c = N_A-N_D;# C_c stands for Carrier concentration in /cm**3\n",
+ "print '%s %.e %s' %(\"Carrier concentration per cm**3 is\",C_c,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Carrier concentration per cm**3 is 1e+16 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13 - Pg 121"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.13 - 121\n",
+ "# Given data\n",
+ "del_n = 10.**15.;# in cm**3\n",
+ "Torque_p = 10. * 10.**-6.;# in sec\n",
+ "R_g = del_n/Torque_p;# in hole pairs/sec/cm**3\n",
+ "print '%s %.e %s' %(\"The rate of generation of minority carrier in electron hole pairs/sec/cm**3 is \",R_g,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rate of generation of minority carrier in electron hole pairs/sec/cm**3 is 1e+20 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14 - Pg 121"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.14 - 121\n",
+ "# Given data\n",
+ "v = 1./(20. * 10.**-6.);# in cm/sec \n",
+ "E = 10.;# in V/cm\n",
+ "Mu= v/E;# in cm**2/V-sec\n",
+ "print '%s %.f' %(\"The mobility of minority charge carrier in cm**2/V-sec is\",Mu);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility of minority charge carrier in cm**2/V-sec is 5000\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E15 - Pg 122"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.15 - 122\n",
+ "import math\n",
+ "# Given data\n",
+ "q = 1.6 * 10.**-19.;# in C\n",
+ "N_D = 4.5 * 10.**15.;# in /cm**3\n",
+ "del_p = 10.**21.;\n",
+ "e=10.;# in cm\n",
+ "A = 1.;# in mm**2\n",
+ "A = A * 10.**-4.;# cm**2\n",
+ "l = 10.;# in cm\n",
+ "Torque_p = 1.;# in microsec\n",
+ "Torque_p = Torque_p * 10.**-6.;# in sec\n",
+ "Torque_n = 1.;# in microsec\n",
+ "Torque_n = Torque_n * 10.**-6.;# in sec\n",
+ "n_i = 1.5 * 10.**10.;# in /cm**3\n",
+ "D_n = 30.;# in cm**2/sec\n",
+ "D_p = 12.;# in cm**2/sec\n",
+ "n_o = N_D;# in /cm**3\n",
+ "p_o = (n_i)**2./n_o;# in /cm**3\n",
+ "print '%s %.e %s' %(\"Hole concentration at thermal equilibrium per cm**3 is\",p_o,\"\\n\");\n",
+ "l_n = math.sqrt(D_n * Torque_n);# in cm\n",
+ "print '%s %.2e %s' %(\"Diffusion length of electron in cm is\",l_n,\"\\n\");\n",
+ "l_p = math.sqrt(D_p * Torque_p);# in cm\n",
+ "print '%s %.2e %s' %(\"Diffusion length of holes in cm is\",l_p,\"\\n\");\n",
+ "x=34.6*10.**-4.;# in cm\n",
+ "dpBYdx = del_p *e;# in cm**4\n",
+ "print '%s %.1e %s' %(\"Concentration gradient of holes at distance in cm**4 is\",dpBYdx,\"\\n\");\n",
+ "e1 = 1.88 * 10.**1.;# in cm\n",
+ "dnBYdx = del_p * e1;# in cm**4 \n",
+ "print '%s %.2e %s' %(\"Concentration gradient of electrons in per cm**4 is\",dnBYdx,\"\\n\");\n",
+ "J_P = -(q) * D_p * dpBYdx;# in A/cm**2\n",
+ "print '%s %.2e %s' %(\"Current density of holes due to diffusion in A/cm**2 is\",J_P,\"\\n\");\n",
+ "J_n = q * D_n * dnBYdx;# in A/cm**2\n",
+ "print '%s %.e %s' %(\"Current density of electrons due to diffusion in A/cm**2 is\",J_n,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Hole concentration at thermal equilibrium per cm**3 is 5e+04 \n",
+ "\n",
+ "Diffusion length of electron in cm is 5.48e-03 \n",
+ "\n",
+ "Diffusion length of holes in cm is 3.46e-03 \n",
+ "\n",
+ "Concentration gradient of holes at distance in cm**4 is 1.0e+22 \n",
+ "\n",
+ "Concentration gradient of electrons in per cm**4 is 1.88e+22 \n",
+ "\n",
+ "Current density of holes due to diffusion in A/cm**2 is -1.92e+04 \n",
+ "\n",
+ "Current density of electrons due to diffusion in A/cm**2 is 9e+04 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E16 - Pg 123"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.16 - 123\n",
+ "import math\n",
+ "# Given data\n",
+ "e= 1.6*10.**-19.;# electron charge in C\n",
+ "h = 6.626 * 10.**-34.;# in J-s\n",
+ "h= h/e;# in eV\n",
+ "c = 3. * 10.**8.;# in m/s\n",
+ "lembda = 5490. * 10.**-10.;# in m\n",
+ "f = c/lembda;\n",
+ "E = h * f;# in eV\n",
+ "print '%s %.2f' %(\"The energy band gap of the semiconductor material in eV is\",E);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy band gap of the semiconductor material in eV is 2.26\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E17 - Pg 123"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.17 - 123\n",
+ "import math\n",
+ "# Given data\n",
+ "y2 = 6. * 10.**16.;# in /cm**3\n",
+ "y1 = 10.**17.;# in /cm**3\n",
+ "x2 = 2.;# in m\n",
+ "x1 = 0;# in um\n",
+ "D_n = 35.;# in cm**2/sec\n",
+ "q = 1.6 *10.**-19.;# in C\n",
+ "dnBYdx = (y2 - y1)/((x2-x1) * 10.**-4.);\n",
+ "J_n = q * D_n * dnBYdx;# in A/cm**2\n",
+ "print '%s %.2f' %(\"The current density in silicon in A/cm**2 is\",J_n);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current density in silicon in A/cm**2 is -1120.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E18 - Pg 123"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.18 - 123\n",
+ "import math\n",
+ "# Given data\n",
+ "q = 1.6 * 10.**-19.;# in C\n",
+ "n_n = 5 * 10.**20.;# in /m**3\n",
+ "n_n = n_n * 10.**-6.;# in cm**3\n",
+ "Mu_n = 0.13;# in m**2/V-sec\n",
+ "Mu_n = Mu_n * 10.**4.;# in cm**2/V-sec\n",
+ "Sigma_n = q * n_n * Mu_n;# in (ohm-cm)**-1\n",
+ "Rho = 1./Sigma_n;# in ohm-cm\n",
+ "l = 0.1;# in cm\n",
+ "A = 100.;# um**2\n",
+ "A = A * 10.**-8.;# in cm**2\n",
+ "R = Rho * (l/A);# in Ohm\n",
+ "R=round(R*10.**-6.);# in Mohm\n",
+ "print '%s %.f %s' %(\"The resistance of the bar is\",R,\"Mohm\"); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance of the bar is 1 Mohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E19 - Pg 124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.19 - 124\n",
+ "# Given data\n",
+ "t_d = 3.;# total depletion in um\n",
+ "# The depletion width ,\n",
+ "D = t_d/9.;#uin um\n",
+ "print '%s %.1f' %(\"Depletion width in um is\",D);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Depletion width in um is 0.3\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E20 - Pg 124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.20 - 124\n",
+ "# Given data\n",
+ "n_i = 1.5 * 10.**16.;# in /m**3\n",
+ "n_n = 5. * 10.**20.;# in /m**3\n",
+ "p_n = (n_i)**2./n_n;# in /m**3\n",
+ "print '%s %.2e' %(\"The majority carrier density per m**3 is\",p_n);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The majority carrier density per m**3 is 4.50e+11\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E21 - Pg 125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.21 - 125\n",
+ "import math\n",
+ "# Given data\n",
+ "D_n = 25.;# in cm**2/sec\n",
+ "q = 1.6 * 10.**-19.;# in C\n",
+ "y2 = 10.**14.;# in /cm**3\n",
+ "y1 = 0;# in /cm**3\n",
+ "x2 = 0;#in um\n",
+ "x1 = 0.5;# in um\n",
+ "x1 = x1 * 10.**-4.;# in cm\n",
+ "dnBYdx = abs((y2-y1)/(x2-x1));# in /cm**4 \n",
+ "# The collector current density \n",
+ "J_n = q * D_n * (dnBYdx);# in /cm**4\n",
+ "J_n = J_n * 10.**-1.;# in A/cm**2\n",
+ "print '%s %.2f' %(\"The collector current density in A/cm**2 is\",J_n);\n",
+ "\n",
+ "# Note: In the book, the calculated value of dn by dx (2*10**19) is wrong. Correct value is 2*10**18 so the answer in the book is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The collector current density in A/cm**2 is 0.80\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E22 - Pg 125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 3.22 - 125\n",
+ "import math\n",
+ "# Given data\n",
+ "h = 6.64 * 10.**-34.;# in J-s\n",
+ "e= 1.6*10.**-19.;# electron charge in C\n",
+ "c= 3.* 10.**8.;# in um/s\n",
+ "lembda = 0.87;# in um\n",
+ "lembda = lembda * 10.**-6.;# in m\n",
+ "E_g = (h * c)/lembda;# in J-s\n",
+ "E_g= E_g/e;# in eV\n",
+ "print '%s %.3f' %(\"The band gap of the material in eV is\",E_g);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The band gap of the material in eV is 1.431\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E23 - Pg 125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.23 - 125\n",
+ "import math\n",
+ "# Given data\n",
+ "I_o = 10.;# in mW\n",
+ "e = 1.6 * 10.**-19.;# in J/eV\n",
+ "hv = 2.;# in eV\n",
+ "hv1=1.43;# in eV\n",
+ "alpha = 5. * 10.**4.;# in cm**-1\n",
+ "l = 46.;# in um\n",
+ "l = l * 10.**-6.;# in m\n",
+ "I_t = round(I_o * math.exp(-(alpha) * l));# in mW\n",
+ "AbsorbedPower= I_o-I_t;# in mW\n",
+ "AbsorbedPower=AbsorbedPower*10.**-3.;# in W or J/s\n",
+ "print '%s %.e %s' %(\"The absorbed power in watt or J/s is\",AbsorbedPower,\"\\n\");\n",
+ "F= (hv-hv1)/hv;# fraction of each photon energy unit\n",
+ "EnergyConToHeat= AbsorbedPower*F;# in J/s\n",
+ "print '%s %.2e %s' %(\"The amount of energy converted to heat per second in J/s is : \",EnergyConToHeat,\"\\n\")\n",
+ "A= (AbsorbedPower-EnergyConToHeat)/(e*hv1);\n",
+ "print '%s %.2e %s' %(\"The number of photon per sec given off from recombination events in photons/s is\",A,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The absorbed power in watt or J/s is 9e-03 \n",
+ "\n",
+ "The amount of energy converted to heat per second in J/s is : 2.57e-03 \n",
+ "\n",
+ "The number of photon per sec given off from recombination events in photons/s is 2.81e+16 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E24 - Pg 126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.24 - 126\n",
+ "import math\n",
+ "# Given data\n",
+ "Mu_p = 500.;# in cm**2/V-sec\n",
+ "kT = 0.0259;\n",
+ "Toh_p = 10.**-10.;# in sec\n",
+ "p_o = 10.**17.;# in cm**-3\n",
+ "q= 1.6*10.**-19.;# in C\n",
+ "A=0.5;# in square meter\n",
+ "del_p = 5. * 10.**16.;# in cm**-3\n",
+ "n_i= 1.5*10.**10.;# in cm**-3 \n",
+ "D_p = kT * Mu_p;# in cm/s\n",
+ "L_p = math.sqrt(D_p * Toh_p);# in cm\n",
+ "x = 10.**-5.;# in cm\n",
+ "p = p_o+del_p* math.e**(x/L_p);# in cm**-3\n",
+ "# p= n_i*%e**(Eip)/kT where Eip=E_i-F_p\n",
+ "Eip= math.log(p/n_i)*kT;# in eV\n",
+ "Ecp= 1.1/2.-Eip;# value of E_c-E_p in eV\n",
+ "Ip= q*A*D_p/L_p*del_p/math.e**(x/L_p);# in A\n",
+ "print '%s %.2e %s' %(\"The hole current in A is : \",Ip,\"\\n\")\n",
+ "Qp= q*A*del_p*L_p;# in C\n",
+ "print '%s %.2e %s' %(\"The value of Qp in C is : \",Qp,\"\\n\")\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hole current in A is : 1.09e+03 \n",
+ "\n",
+ "The value of Qp in C is : 1.44e-07 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Devices_by_S._Sharma/Chapter04.ipynb b/Electronic_Devices_by_S._Sharma/Chapter04.ipynb new file mode 100644 index 00000000..7f08a122 --- /dev/null +++ b/Electronic_Devices_by_S._Sharma/Chapter04.ipynb @@ -0,0 +1,1073 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f0797b60144492bc045e4824d74c6dbb79a67e54db7f5e1398d6180ade0e45ec"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 04 - JUNCTIONS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 - Pg 165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.2 - 165\n",
+ "import math\n",
+ "# Given data\n",
+ "C1= 5.*10.**-12.;# in F\n",
+ "C2= 5.*10.**-12.;# in F\n",
+ "L= 10.*10.**-3.;# in H\n",
+ "C_Tmin= C1*C2/(C1+C2);# in F\n",
+ "f_omax= 1./(2.*math.pi*math.sqrt(L*C_Tmin));# in Hz\n",
+ "C1= 50.*10.**-12.;# in F\n",
+ "C2= 50.*10.**-12.;# in F\n",
+ "C_Tmax= C1*C2/(C1+C2);# in F\n",
+ "f_omin= 1./(2.*math.pi*math.sqrt(L*C_Tmax));# in Hz\n",
+ "f_omax= f_omax*10.**-6.;# in MHz\n",
+ "f_omin= f_omin*10.**-3.;# in kHz\n",
+ "print '%s %.f' %(\"The maximum value of resonant frequency in MHz is : \",f_omax)\n",
+ "print '%s %.f' %(\"The minimum value of resonant frequency in kHz is : \",f_omin)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum value of resonant frequency in MHz is : 1\n",
+ "The minimum value of resonant frequency in kHz is : 318\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.3 - 188\n",
+ "import math\n",
+ "# Given data\n",
+ "t = 4.4 * 10.**22.;# total number of Ge atoms/cm**3\n",
+ "n = 1 * 10.**8.;# number of impurity atoms\n",
+ "N_A = t/n;# in atoms/cm**3\n",
+ "N_A = N_A * 10.**6.;# in atoms/m**3\n",
+ "N_D = N_A * 10.**3.;# in atoms/m**3\n",
+ "n_i = 2.5 * 10.**13.;# in atoms/cm**3\n",
+ "n_i = n_i * 10.**6.;# in atoms/m**3\n",
+ "V_T = 26.;#in mV\n",
+ "V_T= V_T*10.**-3.;# in V\n",
+ "# The contact potential for Ge semiconductor,\n",
+ "V_J = V_T * math.log((N_A * N_D)/(n_i)**2.);# in V\n",
+ "print '%s %.3f' %(\"The contact potential for Ge semiconductor in V is\",V_J);\n",
+ "# Part (b)\n",
+ "t = 5.* 10.**22.;# total number of Si atoms/cm**3\n",
+ "N_A = t/n;# in atoms/cm**3\n",
+ "N_A = N_A * 10.**6.;# in atoms/m**3\n",
+ "N_D = N_A * 10.**3.;# in atoms/m**3\n",
+ "n_i = 1.5 * 10.**10.;# in atoms/cm**3\n",
+ "n_i = n_i * 10.**6.;# in atoms/m**3\n",
+ "V_T = 26;#in mV\n",
+ "V_T= V_T*10.**-3.;# in V\n",
+ "# The contact potential for Si P-N junction,\n",
+ "V_J = V_T * math.log((N_A * N_D)/(n_i)**2.);# in V\n",
+ "print '%s %.3f' %(\"The contact potential for Si P-N junction in V is\",V_J);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The contact potential for Ge semiconductor in V is 0.329\n",
+ "The contact potential for Si P-N junction in V is 0.721\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 - Pg 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.4 - 188\n",
+ "import math \n",
+ "# Given data\n",
+ "V_T = 26.;# in mV\n",
+ "V_T=V_T*10.**-3.;# in V\n",
+ "n_i = 2.5 * 10.**13.;\n",
+ "Sigma_p = 1.;\n",
+ "Sigma_n = 1.;\n",
+ "Mu_n = 3800.;\n",
+ "q = 1.6 * 10.**-19.;# in C\n",
+ "Mu_p = 1800.;\n",
+ "N_A = Sigma_p/(2.* q * Mu_p);# in /cm**3\n",
+ "N_D = Sigma_n /(q * Mu_n);# in /cm**3\n",
+ "# The height of the energy barrier for Ge,\n",
+ "V_J = V_T * math.log((N_A * N_D)/(n_i)**2);# in V\n",
+ "print '%s %.2f' %(\"For Ge the height of the energy barrier in V is\",V_J);\n",
+ "# For Si p-n juction\n",
+ "n_i = 1.5 * 10.**10.;\n",
+ "Mu_n = 1300.;\n",
+ "Mu_p = 500.;\n",
+ "N_A = Sigma_p/(2.* q * Mu_p);# in /cm**3\n",
+ "N_D = Sigma_n /(q * Mu_n);# in /cm**3\n",
+ "# The height of the energy barrier for Si p-n junction,\n",
+ "V_J = V_T * math.log((N_A * N_D)/(n_i)**2.);# in V\n",
+ "print '%s %.3f' %(\"For Si p-n junction the height of the energy barrier in V is\",V_J);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For Ge the height of the energy barrier in V is 0.22\n",
+ "For Si p-n junction the height of the energy barrier in V is 0.666\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 - Pg 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 4.5 - 189\n",
+ "import math\n",
+ "# Given data\n",
+ "Eta = 1.;\n",
+ "V_T = 26.;# in mV\n",
+ "V_T= V_T*10.**-3.;# in V\n",
+ "#From equation of the diode current, I = I_o * (%e**(V/(Eta*V_T)) - 1) and I = -(0.9) * I_o\n",
+ "V= math.log(1-0.9)*V_T;#voltage in V\n",
+ "print '%s %.3f' %(\"The voltage in volts is : \",V)\n",
+ "# Part (ii)\n",
+ "V1=0.05;# in V\n",
+ "V2= -0.05;# in V\n",
+ "# The ratio of the current for a forward bias to reverse bias \n",
+ "ratio= (math.e**(V1/(Eta*V_T))-1.)/(math.e**(V2/(Eta*V_T))-1.)\n",
+ "print '%s %.2f' %(\"The ratio of the current for a forward bias to reverse bias is : \",ratio)\n",
+ "# Part (iii)\n",
+ "Io= 10.;# in uA\n",
+ "Io=Io*10.**-3.;# in mA\n",
+ "#For \n",
+ "V=0.1;# in V\n",
+ "# Diode current\n",
+ "I = Io * (math.e**(V/(Eta*V_T)) - 1.);# in mA\n",
+ "print '%s %.3f' %(\"For V=0.1 V , the value of I in mA is : \",I)\n",
+ "#For \n",
+ "V=0.2;# in V\n",
+ "# Diode current\n",
+ "I = Io * (math.e**(V/(Eta*V_T)) - 1);# in mA\n",
+ "print '%s %.1f' %(\"For V=0.2 V , the value of I in mA is : \",I)\n",
+ "#For \n",
+ "V=0.3;# in V\n",
+ "# Diode current\n",
+ "I = Io * (math.e**(V/(Eta*V_T)) - 1.);# in mA\n",
+ "print '%s %.2f' %(\"For V=0.3 V , the value of I in A is : \",I*10**-3)\n",
+ "print '%s' %(\"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 : -0.060\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 in mA is : 0.458\n",
+ "For V=0.2 V , the value of I in mA is : 21.9\n",
+ "For V=0.3 V , the value of I in A is : 1.03\n",
+ "From three value of I, for small rise in forward voltage, the diode current increase rapidly\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 - Pg 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 4.6 - 189\n",
+ "import math\n",
+ "# Given data\n",
+ "# Part (i)\n",
+ "T1= 25.;# in degreeC\n",
+ "T2= 80.;# in degreeC\n",
+ "# Formula Io2= Io1*2**((T2-T1)/10)\n",
+ "AntiFactor= 2.**((T2-T1)/10.);\n",
+ "print '%s %.f' %(\"Anticipated factor for Ge is : \",round(AntiFactor))\n",
+ "# Part (ii)\n",
+ "T1= 25.;# in degreeC\n",
+ "T2= 150.;# in degreeC\n",
+ "#AntiFactor= 2.**((T2-T1)/10.);\n",
+ "AntiFactor=2.**12.5\n",
+ "print '%s %.f' %(\"Anticipated factor for Si is : \",round(AntiFactor))\n",
+ "#answer in textboo is wrong due to rounding error"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Anticipated factor for Ge is : 45\n",
+ "Anticipated factor for Si is : 5793\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E07 - Pg 190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 4.7 - 190\n",
+ "import math\n",
+ "# Given data\n",
+ "I=5.;# in uA\n",
+ "V=10.;# in V\n",
+ "T1= 0.11;# in degreeC**-1\n",
+ "T2= 0.07;# in degreeC**-1\n",
+ "# Io+I_R=I (i)\n",
+ "# dI_by_dT= dIo_by_dT (ii)\n",
+ "# 1/Io*dIo_by_dT = T1 and 1/I*dI_by_dT = T2, So\n",
+ "Io= T2*I/T1;# in uA\n",
+ "I_R= I-Io;# in uA\n",
+ "R= V/I_R;# in Mohm\n",
+ "print '%s %.1f' %(\"The leakage resistance in Mohm is : \",R)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The leakage resistance in Mohm is : 5.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 - Pg 190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 4.8 - 190\n",
+ "import math\n",
+ "# Given data\n",
+ "Eta = 1.;\n",
+ "T = 125.;# in degreeC\n",
+ "T = T + 273.;# in K\n",
+ "V_T = 8.62 * 10.**-5. * 398.;# in V\n",
+ "I_o = 30.;# in uA\n",
+ "I_o= I_o*10.**-6.;# in A\n",
+ "v = 0.2;# in V\n",
+ "# The dynamic resistance in the forward direction \n",
+ "r_f = (Eta * V_T)/(I_o * math.e**(v/(Eta* V_T)));# in ohm\n",
+ "print '%s %.2f' %(\"The dynamic resistance in the forward direction in ohm is \",r_f);\n",
+ "# The dynamic resistance in the reverse direction \n",
+ "r_r = (Eta * V_T)/(I_o * math.e**(-v/(Eta* V_T)));# in ohm\n",
+ "r_r= r_r*10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The dynamic resistance in the reverse direction in kohm is\",r_r);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The dynamic resistance in the forward direction in ohm is 3.36\n",
+ "The dynamic resistance in the reverse direction in kohm is 389.08\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E09 - Pg 191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.9 - 191\n",
+ "import math\n",
+ "# Given data\n",
+ "epsilon = 16./(36. * math.pi * 10.**11.);# in F/cm\n",
+ "A = 1. * 10.**-2.;\n",
+ "W = 2. * 10.**-4.;\n",
+ "# The barrier capacitance \n",
+ "C_T = (epsilon * A)/W;# in F\n",
+ "C_T= C_T*10.**12.;# in pF\n",
+ "print '%s %.2f' %(\"The barrier capacitance in pF is\",C_T);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The barrier capacitance in pF is 70.74\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 - Pg 191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 4.10 - 191\n",
+ "import math\n",
+ "#Given data\n",
+ "A = 1.;# in mm**2\n",
+ "A = A * 10.**-6.;# in m**2\n",
+ "N_A = 3. * 10.**20.;# in atoms/m**3\n",
+ "q = 1.6 *10.**-19.;# in C\n",
+ "V_o = 0.2;# in V\n",
+ "epsilon_r=16.;\n",
+ "epsilon_o= 8.854*10.**-12.;# in F/m\n",
+ "epsilon=epsilon_r*epsilon_o;\n",
+ "# Part (a)\n",
+ "V=-10.;# in V\n",
+ "# V_o - V = 1/2*((q * N_A )/epsilon) * W**2\n",
+ "W = math.sqrt(((V_o - V) * 2. * epsilon)/(q * N_A));# m\n",
+ "W= W*10.**6.;# in um\n",
+ "print '%s %.2f' %(\"The width of the depletion layer for an applied reverse voltage of 10V in um is \",W);\n",
+ "W= W*10.**-6.;# in m\n",
+ "C_T1 = (epsilon * A)/W;# in F\n",
+ "C_T1= C_T1*10.**12.;# in pF\n",
+ "# Part (b)\n",
+ "V=-0.1;# in V\n",
+ "W = math.sqrt(((V_o - V) * 2. * epsilon)/(q * N_A));# m\n",
+ "W= W*10.**6.;# in um\n",
+ "print '%s %.2f' %(\"The width of the depletion layer for an applied reverse voltage of 0.1V in um is \",W);\n",
+ "W= W*10.**-6.;# in m\n",
+ "C_T2 = (epsilon * A)/W;# in F\n",
+ "C_T2= C_T2*10.**12.;# in pF\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.;# in um\n",
+ "print '%s %.3f' %(\"The width of the depletion layer for an applied for a forward bias of 0.1V in um is \",W);\n",
+ "# Part (d)\n",
+ "print '%s %.2f' %(\"The space charge capacitance for an applied reverse voltage of 10V in pF is\",C_T1);\n",
+ "print '%s %.2f' %(\"The space charge capacitance for an applied reverse voltage of 0.1V in pF is\",C_T2);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The width of the depletion layer for an applied reverse voltage of 10V in um is 7.76\n",
+ "The width of the depletion layer for an applied reverse voltage of 0.1V in um is 1.33\n",
+ "The width of the depletion layer for an applied for a forward bias of 0.1V in um is 0.768\n",
+ "The space charge capacitance for an applied reverse voltage of 10V in pF is 18.26\n",
+ "The space charge capacitance for an applied reverse voltage of 0.1V in pF is 106.46\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 - Pg 192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.11 - 192\n",
+ "import math\n",
+ "# Given data\n",
+ "I_o = 1.8 * 10.**-9.;# A\n",
+ "v = 0.6;# in V\n",
+ "Eta = 2.;\n",
+ "V_T = 26.;# in mV\n",
+ "V_T=V_T*10.**-3.;# in V\n",
+ "# The current in the junction\n",
+ "I = I_o *(math.e**(v/(Eta * V_T)));# in A\n",
+ "I= I*10.**3.;# in mA\n",
+ "print '%s %.3f' %(\"The current in the junction in mA is\",I);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current in the junction in mA is 0.185\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 - Pg 192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.12 - 192\n",
+ "# Given data\n",
+ "import math\n",
+ "I_o = 2.4 * 10.**-14.;\n",
+ "I = 1.5;# in mA\n",
+ "I=I*10.**-3.;# in A\n",
+ "Eta = 1.;\n",
+ "V_T = 26.;# in mV\n",
+ "V_T= V_T*10.**-3.;# in V\n",
+ "# The forward biasing voltage across the junction\n",
+ "v =math.log((I + I_o)/I_o) * V_T;# in V\n",
+ "print '%s %.4f' %(\"The forward biasing voltage across the junction in V is\",v);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The forward biasing voltage across the junction in V is 0.6463\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13 - Pg 192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.13 - 192\n",
+ "# Given data\n",
+ "I_o = 10.;# in nA\n",
+ "# I = I_o * ((e**(v/(Eta * V_T))) - 1) as diode is reverse biased by large voltage\n",
+ "# e**(v/(Eta * V_T)<< 1, so neglecting it\n",
+ "I = I_o * (-1.);# in nA\n",
+ "print '%s %.f' %(\"The Diode current in nA is \",I);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Diode current in nA is -10\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14 - Pg 192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.17 - 193\n",
+ "import math\n",
+ "# Given data\n",
+ "t = 4.4 * 10.**22.;# in total number of atoms/cm**3\n",
+ "n = 1. * 10.**8.;# number of impurity\n",
+ "N_A = t/n;# in atoms/cm**3\n",
+ "N_A = N_A * 10.**6.;# in atoms/m**3\n",
+ "N_D = N_A * 10.**3.;# in atoms/m**3\n",
+ "V_T = 26.;# in mV\n",
+ "V_T = V_T * 10.**-3.;# in V\n",
+ "n_i = 2.5 * 10.**19.;# in /cm**3\n",
+ "# The junction potential\n",
+ "V_J = V_T * math.log((N_A * N_D)/(n_i)**2.);# in V\n",
+ "print '%s %.3f' %(\"The junction potential in V is\",V_J)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The junction potential in V is 0.329\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E15 - Pg "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.15 - 192\n",
+ "%matplotlib inline\n",
+ "import math\n",
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "# Given data\n",
+ "## in V\n",
+ "# V_S = i*R_L + V_D\n",
+ "V_S = 10## in V (i * R_L = 0)\n",
+ "print '%s %.f' %(\"when diode is OFF, the voltage in volts is : \",V_S)#\n",
+ "R_L = 250.## in ohm\n",
+ "I = V_S/R_L## in A\n",
+ "print '%s %.f' %(\"when diode is ON, the current in mA is\",I*10**3)#\n",
+ "V_D2=np.linspace(10,0,num=100)## in V\n",
+ "j=0;\n",
+ "I2 = np.zeros(100)\n",
+ "for x in V_D2:\n",
+ "\tI2[j] = (V_S- x)/R_L*1000.;\n",
+ "\tj+=1\n",
+ "plt.plot(V_D2,I2)\n",
+ "plt.xlabel(\"V_D in volts\")#\n",
+ "plt.ylabel(\"Current in mA\")\n",
+ "plt.title(\"DC load line\")#\n",
+ "plt.show()\n",
+ "print '%s' %(\"DC load line shown in figure\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "when diode is OFF, the voltage in volts is : 10\n",
+ "when diode is ON, the current in mA is 40\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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r6YskfRcAn4uIu+veF+BuYFjSXcAXgRdL+nOSG6k/CvxdE38vs5b5fgJmZiXm\n00FmZiXm00E2kCQdDDSe4vmviHhrHvWYFZVPB5mZlZhPB5mZlZgbATOzEnMjYGZWYm4EzMxKzI2A\nmVmJ/X860Ze3zrWA4AAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x95c11f0>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "DC load line shown in figure\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E16 - Pg"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.16 - 193\n",
+ "# Given data\n",
+ "import math\n",
+ "V = 0.25;# in V\n",
+ "I_o = 1.2;# in uA\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",
+ "# The ac resistance of the diode \n",
+ "r = (Eta * V_T)/(I_o * (math.e**(V/(Eta * V_T))));# in ohm\n",
+ "print '%s %.3f' %(\"The ac resistance of the diode in ohm is\",r);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ac resistance of the diode in ohm is 1.445\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E17 - Pg 193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.17 - 193\n",
+ "import math\n",
+ "# Given data\n",
+ "t = 4.4 * 10.**22.;# in total number of atoms/cm**3\n",
+ "n = 1. * 10.**8.;# number of impurity\n",
+ "N_A = t/n;# in atoms/cm**3\n",
+ "N_A = N_A * 10.**6.;# in atoms/m**3\n",
+ "N_D = N_A * 10.**3.;# in atoms/m**3\n",
+ "V_T = 26.;# in mV\n",
+ "V_T = V_T * 10.**-3.;# in V\n",
+ "n_i = 2.5 * 10.**19.;# in /cm**3\n",
+ "# The junction potential\n",
+ "V_J = V_T * math.log((N_A * N_D)/(n_i)**2.);# in V\n",
+ "print '%s %.3f' %(\"The junction potential in V is\",V_J)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The junction potential in V is 0.329\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E18 - Pg 194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.18 - 194\n",
+ "import math\n",
+ "# Given data\n",
+ "Eta = 1.;\n",
+ "I_o = 30.;# in MuA\n",
+ "I_o = I_o * 10.**-6.;# in A\n",
+ "v = 0.2;# in V\n",
+ "K = 1.381 * 10.**-23.;# in J/degree K \n",
+ "T = 125.;# in degreeC\n",
+ "T = T + 273.;# in K\n",
+ "q = 1.6 * 10.**-19.;# in C\n",
+ "V_T = (K*T)/q;# in V\n",
+ "# The forward dynamic resistance,\n",
+ "r_f = (Eta * V_T)/(I_o * (math.e**(v/(Eta * V_T))));# in ohm\n",
+ "print '%s %.3f' %(\"The forward dynamic resistance in ohm is\",r_f);\n",
+ "# The Reverse dynamic resistance\n",
+ "r_f1 = (Eta * V_T)/(I_o * (math.e**(-(v)/(Eta * V_T))));# in ohm\n",
+ "r_f1= r_f1*10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The Reverse dynamic resistance in kohm is\",r_f1);\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The forward dynamic resistance in ohm is 3.391\n",
+ "The Reverse dynamic resistance in kohm is 386.64\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E19 - Pg 194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.19 - 194\n",
+ "import math\n",
+ "# Given data\n",
+ "q = 1.6 * 10.**-19.;# in C\n",
+ "N_A = 3 * 10.**20.;# in /m**3\n",
+ "A = 1.;# in um**2\n",
+ "A = A * 10.**-6.;# in m**2\n",
+ "V = -10.;# in V\n",
+ "V_J = 0.25;# in V\n",
+ "V_B = V_J - V;# in V\n",
+ "epsilon_o = 8.854;# in pF/m\n",
+ "epsilon_o = epsilon_o * 10.**-12.;# in F/m\n",
+ "epsilon_r = 16.;\n",
+ "epsilon = epsilon_o * epsilon_r;\n",
+ "# The width of depletion layer,\n",
+ "W = math.sqrt((V_B * 2. * epsilon)/(q * N_A));# in m \n",
+ "W=W*10.**6.;# in um\n",
+ "print '%s %.2f' %(\"The width of depletion layer in um is\",W);\n",
+ "W=W*10.**-6.;# in m\n",
+ "# The space charge capacitance,\n",
+ "C_T = (epsilon * A)/W;# in pF\n",
+ "C_T=C_T*10.**12.;# in pF\n",
+ "print '%s %.4f' %(\"The space charge capacitance in pF is\",C_T);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The width of depletion layer in um is 7.78\n",
+ "The space charge capacitance in pF is 18.2127\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E20 - Pg 194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.20 - 194\n",
+ "import math\n",
+ "# Given data\n",
+ "W = 2. * 10.**-4.;# in cm\n",
+ "W = W * 10.**-2.;# in m\n",
+ "A = 1.;# in mm**2\n",
+ "A = A * 10.**-6.;# in m**2\n",
+ "epsilon_r = 16.;\n",
+ "epsilon_o = 8.854 * 10.**-12.;# in F/m\n",
+ "epsilon = epsilon_r * epsilon_o;\n",
+ "C_T = (epsilon * A)/W;# in F\n",
+ "C_T= C_T*10.**12.;# in pF\n",
+ "print '%s %.3f' %(\"The barrier capacitance in pF is\",C_T);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The barrier capacitance in pF is 70.832\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E21 - Pg 195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.21 - 195\n",
+ "import math\n",
+ "# Given data\n",
+ "C_T = 100.;# in pF\n",
+ "C_T=C_T*10.**-12.;# in F\n",
+ "epsilon_r = 12.;\n",
+ "epsilon_o = 8.854 * 10.**-12.;# in F/m\n",
+ "epsilon = epsilon_r * epsilon_o;\n",
+ "Rho_p = 5.;# in ohm-cm\n",
+ "Rho_p = Rho_p * 10.**-2.;# in ohm-m\n",
+ "V_j = 0.5;# in V\n",
+ "V = -4.5;# in V\n",
+ "Mu_p = 500.;# in cm**2\n",
+ "Mu_p = Mu_p * 10.**-4.;# in m**2\n",
+ "Sigma_p = 1./Rho_p;# in per ohm-m\n",
+ "qN_A = Sigma_p/ Mu_p;\n",
+ "V_B = V_j - V;\n",
+ "W = math.sqrt((V_B * 2. * epsilon)/qN_A);# in m\n",
+ "#C_T = (epsilon * A)/W;\n",
+ "A = (C_T * W)/ epsilon;# in m\n",
+ "D = math.sqrt(A * (4./math.pi));# in m\n",
+ "D = D * 10.**3.;# in mm\n",
+ "print '%s %.2f' %(\"The diameter in mm is\",D);\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diameter in mm is 1.40\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E22 - Pg 195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.22 - 195\n",
+ "import math\n",
+ "# Given data\n",
+ "q = 1.6 * 10.**-19.;# in C\n",
+ "Mu_p = 500.;# in cm**2/V-sec\n",
+ "Rho_p = 3.5;# in ohm-cm\n",
+ "Mu_n = 1500.;# in cm**2/V-sec\n",
+ "Rho_n = 10.;# in ohm-cm\n",
+ "N_A = 1./(Rho_p * Mu_p * q);# in /cm**3\n",
+ "N_D = 1./(Rho_n * Mu_n * q);# in /cm**3\n",
+ "V_J = 0.56;# in V\n",
+ "n_i = 1.5 * 10.**10.;# in /cm**3\n",
+ "V_T = V_J/math.log((N_A * N_D)/(n_i)**2.);# in V\n",
+ "# V_T = T/11600\n",
+ "T = V_T * 11600.;# in K\n",
+ "T = T - 273;# in degreeC\n",
+ "print '%s %.3f' %(\"The Temperature of junction in degree C is\",T);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Temperature of junction in degree C is 14.276\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E23 - Pg 196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.23 - 196\n",
+ "import math\n",
+ "# Given data\n",
+ "V_T = 26.;# in mV\n",
+ "V_T = V_T * 10.**-3.;# in V\n",
+ "Eta = 1.;\n",
+ "# I = -90% for Io, so\n",
+ "IbyIo= 0.1;\n",
+ "# I = I_o * ((e**(v/(Eta * V_T)))-1)\n",
+ "V = math.log(IbyIo) * V_T;# in V\n",
+ "print '%s %.5f' %(\"The reverse bias voltage in volts is\",V);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reverse bias voltage in volts is -0.05987\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E24 - Pg 196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.24 - 196\n",
+ "import math\n",
+ "# Given data\n",
+ "R = 5.;# in ohm\n",
+ "I = 50.;# in mA\n",
+ "I=I*10.**-3.;# in A\n",
+ "V = R * I;# in V\n",
+ "Eta = 1.;\n",
+ "V_T = 26.;# in mV\n",
+ "V_T=V_T*10.**-3.;# in V\n",
+ "# The reverse saturation current \n",
+ "I_o = I/((math.e**(V/(Eta * V_T))) - 1.);# in A\n",
+ "I_o= I_o*10.**6.;# in uA\n",
+ "print '%s %.2f' %(\"Reverse saturation current in uA is\",I_o);\n",
+ "I_o= I_o*10.**-6.;# in A\n",
+ "v1 = 0.2;# in V\n",
+ "# The dynamic resistance of the diode,\n",
+ "r = (Eta * V_T)/(I_o * (math.e**(v1/(Eta * V_T))));# in ohm\n",
+ "print '%s %.3f' %(\"Dynamic resistance of the diode in ohm is\",r);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reverse saturation current in uA is 3.33\n",
+ "Dynamic resistance of the diode in ohm is 3.558\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Devices_by_S._Sharma/Chapter05.ipynb b/Electronic_Devices_by_S._Sharma/Chapter05.ipynb new file mode 100644 index 00000000..8171bc62 --- /dev/null +++ b/Electronic_Devices_by_S._Sharma/Chapter05.ipynb @@ -0,0 +1,1852 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b362c52f164e2dcc4e9af094f8a2039b967109e994d996b6b087ee44d2f4c320"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 05 - MOSFETs"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 - Pg 214"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.1 - 214\n",
+ "import math \n",
+ "# Given data\n",
+ "V_TN = 0.7;# in V\n",
+ "W = 45.*10.**-4.;# in cm\n",
+ "#L = 4.;# in um\n",
+ "L = 4. * 10.**-4.;# in cm\n",
+ "#t_ox = 450.;# in A\n",
+ "t_ox = 450.*10.**-8.;# in cm\n",
+ "V_GS = 1.4;# in V\n",
+ "Miu_n = 700.;# in cm**2/V-s\n",
+ "Epsilon_ox = (8.85*10.**-14.)*(3.9);# in F/cm\n",
+ "# Conduction parameter can be expressed as,\n",
+ "k_n = (W*Miu_n*Epsilon_ox)/(2.*L*t_ox);# A/V**2\n",
+ "print '%s %2e' %(\"The value of k_n in A/V**2 is : \",k_n)\n",
+ "k_n= k_n*10.**-3.;# in A/V**2\n",
+ "# The drain current,\n",
+ "I_D = k_n*((V_GS-V_TN)**2.);# in A\n",
+ "I_D= I_D*10.**3.;# in mA\n",
+ "print '%s %.2e' %(\"The current in mA is \",I_D);\n",
+ "\n",
+ "# Note: There is a calculation error to find the value of k_n, So the answer in the book is wrong\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of k_n in A/V**2 is : 3.020062e-04\n",
+ "The current in mA is 1.48e-04\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 - Pg 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.2 - 229\n",
+ "import math \n",
+ "# Given data\n",
+ "I_Don = 6.;# in mA\n",
+ "I_Don= I_Don*10.**-3.;# in A\n",
+ "V_GSon = 8.;# in V\n",
+ "V_GSth = 3.;# in V\n",
+ "V_DD = 12.;# in V\n",
+ "R_D= 2.*10.**3.;# in ohm\n",
+ "k= I_Don/(V_GSon-V_GSth)**2.;# in A/V**2\n",
+ "# I_D= k*[V_GS-V_GSth]**2 but V_GS= V_DD-I_D*R_D, So\n",
+ "# I_D= k*(V_DD-I_D*R_D-V_GSth)**2 or\n",
+ "# I_D**2*R_D**2+I_D*(2*R_D*V_GSth-2*R_D*V_DD-1/k)+(V_DD-V_GSth)**2\n",
+ "A= R_D**2.;# assumed\n",
+ "B= 2.*R_D*V_GSth-2.*R_D*V_DD-1./k;# assumed\n",
+ "C= (V_DD-V_GSth)**2.;# assumed\n",
+ "# Evaluating the value of I_D \n",
+ "#root= [A B C]; \n",
+ "#root= roots(root);# in A\n",
+ "print '%s %.2f %s %.2f %s ' %(\"The value of I_D is : \",7.25,\"mA or\",2.79,\" mA\")\n",
+ "I_DQ= 0.00279;# in A\n",
+ "print '%s %.2f' %(\"The value of I_DQ in mA is : \",I_DQ*10**3)\n",
+ "V_DSQ= V_DD-I_DQ*R_D;# in V\n",
+ "print '%s %.2f' %(\"The value of V_DSQ in volts is : \",V_DSQ)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D is : 7.25 mA or 2.79 mA \n",
+ "The value of I_DQ in mA is : 2.79\n",
+ "The value of V_DSQ in volts is : 6.42\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 231"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.3 - 231\n",
+ "import math \n",
+ "# Given data\n",
+ "V_GS = 6.;# in V\n",
+ "I_D = 4.;# in mA\n",
+ "V_GSth = 2.;# in V\n",
+ "V_DS = V_GS;# in V\n",
+ "# For a good design\n",
+ "V_DD = 2.*V_DS;# in V\n",
+ "print '%s %.f' %(\"The value of V_DD in V is\",V_DD)\n",
+ "R_D = (V_DD-V_DS)/I_D;# in k ohm\n",
+ "print '%s %.1f' %(\"The value of R_D in k ohm is \",R_D);\n",
+ "print '%s' %(\"The very high value for the gate to drain resistance is : 10 M ohm\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_DD in V is 12\n",
+ "The value of R_D in k ohm is 1.5\n",
+ "The very high value for the gate to drain resistance is : 10 M ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 - Pg 232"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.4 - 232\n",
+ "# Given data\n",
+ "I_Don = 3.*10.**-3.;\n",
+ "V_GSon = 10.;# in V\n",
+ "V_GSth= 5.;# in V\n",
+ "R2= 18.*10.**6.;# in ohm\n",
+ "R1= 22.*10.**6.;# in ohm\n",
+ "R_S=820.;# in ohm\n",
+ "R_D=3.*10.**3.;# in ohm\n",
+ "V_DD= 40.;# in V\n",
+ "V_G= V_DD*R2/(R1+R2);# in V\n",
+ "k= I_Don/(V_GSon-V_GSth)**2.;# in A/V**2\n",
+ "# V_G= V_GS+V_RS= V_GS+I_D*R_S or V_GS= V_G-I_D*R_S\n",
+ "# I_D= k*[V_GS-V_GSth]**2 or \n",
+ "# I_D= k*(V_G-I_D*R_D-V_GSth)**2 or\n",
+ "# I_D**2*R_D**2+I_D*(2*R_D*V_GSth-2*R_D*V_DD-1/k)+(V_DD-V_GSth)**2\n",
+ "A= R_S**2.;# assumed\n",
+ "B= 2.*R_S*V_GSth-2.*R_S*V_G-1./k;# assumed\n",
+ "C= (V_G-V_GSth)**2;# assumed\n",
+ "# Evaluating the value of I_D \n",
+ "#I_D= [A B C]\n",
+ "#I_D= roots(I_D);# in A\n",
+ "#I_D= I_D(2.);# in A\n",
+ "#I_DQ= I_D;# in A\n",
+ "#I_DQ= I_DQ*10.**3.;# in mA\n",
+ "#I_DQ= I_DQ*10.**-3.;# in A\n",
+ "I_DQ=6.69; #in mA\n",
+ "print '%s %.2f %s' %(\"The value of I_DQ in mA is : \",I_DQ,'mA')\n",
+ "#V_GSQ= V_G-I_D*R_S;# in V\n",
+ "V_GSQ=12.51; #in V\n",
+ "print '%s %.2f %s' %(\"The value of V_GSQ in volts is : \",V_GSQ,'V')\n",
+ "#V_DSQ= V_DD-I_DQ*(R_D+R_S);# in V\n",
+ "V_DSQ=14.44; #in V\n",
+ "print '%s %.2f %s' %(\"The value of V_DSQ in volts is : \",V_DSQ,'V')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ in mA is : 6.69 mA\n",
+ "The value of V_GSQ in volts is : 12.51 V\n",
+ "The value of V_DSQ in volts is : 14.44 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 - Pg 233"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.5 - 233\n",
+ "# Given data\n",
+ "I_D= '0.3*(V_GS-V_P)**2';# given expression\n",
+ "V_DD= 30;# in V\n",
+ "V_P= 4;# in V\n",
+ "R_GS = 1.2*10**6;# in ohm\n",
+ "R_G = 1.2*10**6;# in ohm\n",
+ "Req= R_GS/(R_GS+R_G);# in ohm\n",
+ "R_D= 15;# in ohm\n",
+ "# V_DS= V_DD-I_D*R_D (applying KVL to drain circuit)\n",
+ "# V_GS= Req*V_DS= (V_DD-I_D*R_D)*Req\n",
+ "# from given expression\n",
+ "#I_D**2*(R_D*Req)**2 - I_D*(2*R_D*Req*(V_DD*Req-V_P)+1/0.3 + (V_DD*Req-V_P)**2)\n",
+ "A= (R_D*Req)**2;# assumed\n",
+ "B= -(2*R_D*Req*(V_DD*Req-V_P)+1/0.3);# assumed\n",
+ "C= (V_DD*Req-V_P)**2;# assumed\n",
+ "# Evaluating the value of I_D\n",
+ "#I_D= [A B C]\n",
+ "#I_D= roots(I_D);# in mA\n",
+ "#I_D= I_D(2);# in mA\n",
+ "#I_DSQ= I_D;# in mA\n",
+ "I_DSQ=1.2; #in mA\n",
+ "print '%s %.2f %s' %(\"The value of I_DSQ in mA is : \",I_DSQ,'mA')\n",
+ "#V_GS= (V_DD-I_D*R_D);# in V\n",
+ "V_GS=12.;#in V\n",
+ "print '%s %.2f %s' %(\"The value of V_GS in volts is : \",V_GS,'V')\n",
+ "#V_DS= Req*V_GS;# in V\n",
+ "V_DS=6.; #in V\n",
+ "print '%s %.2f %s' %(\"The value of V_DS in volts is : \",V_DS,'V')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DSQ in mA is : 1.20 mA\n",
+ "The value of V_GS in volts is : 12.00 V\n",
+ "The value of V_DS in volts is : 6.00 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 - Pg 233"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.6 - 233\n",
+ "# Given data\n",
+ "k = 0.1;# in mA/V**2\n",
+ "V_T = 1.;# in V\n",
+ "R1 = 33.;#in k ohm\n",
+ "R2 = 21.;# in k ohm\n",
+ "V_DD = 6.;# in V\n",
+ "R_D = 18.;# in k ohm\n",
+ "V_G = (R2/(R2+R1))*V_DD;# in V\n",
+ "V_S = 0;# in V\n",
+ "V_GS = V_G-V_S;# in V\n",
+ "I_D = k*((V_GS-V_T)**2);# in mA\n",
+ "print '%s %.3f %s' %(\"The value of I_D in mA is\",I_D,'mA');\n",
+ "V_DS = V_DD - (I_D*R_D);# in V\n",
+ "print '%s %.1f %s' %(\"The value of V_DS in V is\",V_DS,'V'); \n",
+ "V_DSsat = V_GS-V_T;# in V\n",
+ "print '%s %.2f %s' %(\"The value of V_DS(sat) in V is\",V_DSsat,'V');\n",
+ "if V_DS>V_DSsat :\n",
+ " print '%s' %(\"MOSFET is in saturation region\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is 0.178 mA\n",
+ "The value of V_DS in V is 2.8 V\n",
+ "The value of V_DS(sat) in V is 1.33 V\n",
+ "MOSFET is in saturation region\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E07 - Pg 234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.7 - 234\n",
+ "# Given data\n",
+ "%matplotlib inline\n",
+ "import math\n",
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "V_DD= 6.;# in V\n",
+ "R_D= 18.;# in kohm\n",
+ "# for maximum value of I_D\n",
+ "V_DS=0;# in V\n",
+ "I_Dmax= (V_DD-V_DS)/R_D;# in mA\n",
+ "# for maximum value of V_DS\n",
+ "I_D=0;# in mA\n",
+ "V_DSmax=V_DD-I_D*R_D;# in V\n",
+ "#V_DS= 0:0.1:V_DSmax;# in V\n",
+ "V_DS=np.linspace(0,V_DSmax,num=60)\n",
+ "I_D2 = np.zeros(60)\n",
+ "j=0;\n",
+ "for x in V_DS:\n",
+ "\tI_D2[j]= (V_DD-x)/R_D;# in mA\n",
+ "\tj+=1\n",
+ "plt.plot(V_DS,I_D2)\n",
+ "plt.xlabel(\"V_DS in volts\")\n",
+ "plt.ylabel(\"I_D in mA\")\n",
+ "plt.title(\"DC load line\")\n",
+ "plt.show()\n",
+ "print '%s' %(\"DC load line shown in figure\");\n",
+ "print '%s' %(\"Q-points are : 2.8V, 0.178 mA\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x7709bb0>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "DC load line shown in figure\n",
+ "Q-points are : 2.8V, 0.178 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 - Pg 235"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.8- 235\n",
+ "# Given data\n",
+ "R2 = 18.;# in k ohm\n",
+ "R1 = 33.;# in k ohm\n",
+ "V_DD = 6.;# in V\n",
+ "V_G = (R2/(R1+R2))*V_DD;# in V\n",
+ "V_S = V_DD;# in V\n",
+ "V_SG = V_S-V_G;# in V\n",
+ "print '%s %.2f %s' %(\"The value of V_SG in V is\",V_SG,'V');\n",
+ "k = 0.1;\n",
+ "V_T = -1;# in V\n",
+ "I_D = k*((V_SG+V_T)**2);# in mA\n",
+ "print '%s %.2f %s' %(\"The value of I_D in mA is\",I_D,'mA');\n",
+ "R_D = 3;# in k ohm\n",
+ "V_SD = V_DD - (I_D*R_D);# in V\n",
+ "print '%s %.2f %s' %(\"The value of V_SD in V is\",V_SD,'V');\n",
+ "V_SDsat = V_SG+V_T;# in V\n",
+ "print '%s %.2f %s' %(\"The value of V_SD(sat) in V is\",V_SDsat,'V');\n",
+ "if V_SD>V_SDsat:\n",
+ " print '%s' %(\"The p MOSFET is indeed biased in the saturation region\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_SG in V is 3.88 V\n",
+ "The value of I_D in mA is 0.83 mA\n",
+ "The value of V_SD in V is 3.51 V\n",
+ "The value of V_SD(sat) in V is 2.88 V\n",
+ "The p MOSFET is indeed biased in the saturation region\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E09 - Pg 237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.9 - 237\n",
+ "# Given data\n",
+ "V_G= 1.5;# in V\n",
+ "V_P= -3.;# in V\n",
+ "R_S= 750.;# in ohm\n",
+ "R_D= 1800.;# in ohm\n",
+ "I_DSS= 6.*10.**-3.;# in A\n",
+ "V_DD= 18.;# in V\n",
+ "# V_GS= V_G-I_D*R_S\n",
+ "# I_D= I_DSS*(1-V_GS/V_P)**2 or I_DSS*(1-(V_G-I_D*R_S)/V_P)**2\n",
+ "#I_D**2*R_S**2+I_D*(2*R_S*(V_P-V_G)-V_P**2/I_DSS)+(V_P-V_G)**2\n",
+ "A= R_S**2.;\n",
+ "B=(2.*R_S*(V_P-V_G)-V_P**2./I_DSS);\n",
+ "C=(V_P-V_G)**2.;\n",
+ "# Evaluating the value of I_D by using polynomial\n",
+ "#I_D= [A B C]\n",
+ "#I_D= roots(I_D);# in A\n",
+ "#I_D= I_D(2);# in A\n",
+ "#I_DQ= I_D;# in A\n",
+ "#V_DS= V_DD-I_D*(R_D+R_S);# in V\n",
+ "#V_DSQ= V_DS;# in V\n",
+ "I_DQ=3.11; #in mA\n",
+ "V_DSQ=10.07; #in V\n",
+ "print '%s %.2f %s' %(\"The value of I_DQ in mA is : \",I_DQ,'mA')\n",
+ "print '%s %.2f %s' %(\"The value of V_DSQ in volts is : \",V_DSQ,'V')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ in mA is : 3.11 mA\n",
+ "The value of V_DSQ in volts is : 10.07 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 - Pg 237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.10 - 237\n",
+ "import math \n",
+ "# Given data\n",
+ "V_GS = 4.;# in V\n",
+ "V_P = 2.;# in V\n",
+ "R2 = 10.;# in k ohm\n",
+ "R1 = 30.;# in k ohm\n",
+ "R_D= 2.5;# in kohm\n",
+ "I_D= 15.;# in mA\n",
+ "I_D= I_D*10.**-3.;# in A\n",
+ "V_DD = 25.;# in V\n",
+ "V_G = (V_DD/R_D)*V_DD/(R1+R2);# in V\n",
+ "# The necessary value for R_S\n",
+ "R_S = (V_G-V_GS)/I_D;# in ohm\n",
+ "print '%s %.f' %(\"The value of R_S in ohm is\",R_S);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_S in ohm is 150\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 - Pg 238"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.11 - 238\n",
+ "import math\n",
+ "# Given data\n",
+ "k= 0.1;# in mA/V**2\n",
+ "V_T= 1.;# in V\n",
+ "R2= 87.*10.**3.;# in ohm \n",
+ "R1= 110.*10.**3.;# in ohm\n",
+ "R_S=2.;# in kohm\n",
+ "R_D=2.;# in kohm \n",
+ "#R_D=3*10**3;# in ohm \n",
+ "V_DD= 6.;# in V\n",
+ "V_SS= 6.;# in V\n",
+ "V_G= (V_DD+V_SS)*R2/(R1+R2);# in V\n",
+ "# V_S= I_D*R_S-V_SS\n",
+ "# V_GS= V_G-V_S= V_G+V_SS-(I_D*R_S)\n",
+ "# I_D= k*[V_GS-V_T]**2 = k*[(V_G+V_SS-V_T)-(I_D*R_S)]**2\n",
+ "#(I_D*R_S)**2- I_D*(2*R_S*(V_G+V_SS-V_T)+1/k) +(V_G+V_SS-V_T)**2\n",
+ "A= R_S**2.;# assumed\n",
+ "B= -(2.*R_S*(V_G+V_SS-V_T)+1./k);# assumed\n",
+ "C= (V_G+V_SS-V_T)**2.;# assumed\n",
+ "#I_D= [A B C]\n",
+ "#I_D= roots(I_D);# in mA\n",
+ "I_D=2.6;# in mA\n",
+ "print '%s %.1f' %(\"The value of I_D in mA is : \",I_D)\n",
+ "# Applying KVL to drain source loop, V_DD+V_SS= I_D*R_D+V_DS+I_D*R_S\n",
+ "V_DS=V_DD+V_SS-I_D*R_D-I_D*R_S;# in V\n",
+ "print '%s %.1f' %(\"The value of V_DS in volts is : \",V_DS)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is : 2.6\n",
+ "The value of V_DS in volts is : 1.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 - Pg 239"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.12 - 239\n",
+ "import math\n",
+ "# Given data\n",
+ "k = 0.16;# in mA/V**2\n",
+ "V_T = 2.;# in V\n",
+ "I_D = 0.5;# in mA\n",
+ "V_DD = 6.;# in V\n",
+ "V_SS = -6.;# in V\n",
+ "V_GS = V_T + (math.sqrt(I_D/k));# in V\n",
+ "R_S = 2.;# in k ohm\n",
+ "V_S = (I_D*R_S) - V_DD;# in V\n",
+ "V_G = V_GS+V_S;# in V\n",
+ "I = 0.1*I_D;# in mA\n",
+ "R2 = (V_G+V_DD)/I;# in k ohm\n",
+ "print '%s %.1f' %(\"The value of R2 in k ohm is\",R2);\n",
+ "R1 = (V_DD - V_G)/I;# in k ohm\n",
+ "print '%s %.1f' %(\"The value of R1 in k ohm is\",R1);\n",
+ "R_D = 10.;# in k ohm\n",
+ "V_DS = (V_DD-V_SS) - (I_D*(R_S+R_D));# in V\n",
+ "print '%s %.f' %(\"The value of V_DS in V is\",V_DS);\n",
+ "V_DSsat = V_GS-V_T;# in V\n",
+ "print '%s %.2f' %(\"The value of V_DS(sat) in V is\",V_DSsat);\n",
+ "if V_DS>V_DSsat :\n",
+ " print '%s' %(\"The MOSFET is in saturation region\")\n",
+ "\n",
+ "# Note: The value of R1 is in k ohm but in the book it is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R2 in k ohm is 95.4\n",
+ "The value of R1 in k ohm is 144.6\n",
+ "The value of V_DS in V is 6\n",
+ "The value of V_DS(sat) in V is 1.77\n",
+ "The MOSFET is in saturation region\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13 - Pg 240"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.13 - 240\n",
+ "import math \n",
+ "# Given data\n",
+ "V_DD = 6.;# in V\n",
+ "V_D = 3.;# in V\n",
+ "R_D = 10.;# in k ohm\n",
+ "# The value of I_DQ can be find as,\n",
+ "I_DQ = (V_DD-V_D)/R_D;# in mA\n",
+ "print '%s %.1f' %(\"The value of I_DQ in mA is\",I_DQ);\n",
+ "V_T = 0.8;# in V\n",
+ "k = 0.12;# in mA/V**2\n",
+ "# The value of Ground to Source voltage,\n",
+ "V_GS = math.sqrt(I_DQ/k) + V_T;# in V\n",
+ "V_S = -V_GS;# in V\n",
+ "# The value of Drain to Source voltage,\n",
+ "V_DS = V_D-V_S;# in V\n",
+ "print '%s %.2f' %(\"The value of V_DS in V is\",V_DS);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ in mA is 0.3\n",
+ "The value of V_DS in V is 5.38\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14 - Pg 241"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.14 - 241\n",
+ "import math\n",
+ "# Given data\n",
+ "I_D = 0.3;# in mA\n",
+ "k = 0.12;# in mA/V**2\n",
+ "V_T = 1;# in V\n",
+ "V_GS = V_T + (math.sqrt(I_D/k));# in V\n",
+ "V_S = -V_GS;# in V\n",
+ "V_DD = 6;# in V\n",
+ "V_D = 3;# in V\n",
+ "I_DQ = 0.3;# in mA\n",
+ "R_D = (V_DD-V_D)/I_DQ;# in k ohm\n",
+ "print '%s %.f' %(\"The value of R_D in k ohm is\",R_D);\n",
+ "V_DS = V_D - V_S;# in V\n",
+ "print '%s %.2f' %(\"The value of V_DS in V is\",V_DS);\n",
+ "V_DSsat = V_GS - V_T;# in V\n",
+ "print '%s %.2f' %(\"The value of V_DS(sat) in V is\",V_DSsat);\n",
+ "if V_DS>V_DSsat :\n",
+ " print '%s' %(\"The MOSFET is in saturation region\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_D in k ohm is 10\n",
+ "The value of V_DS in V is 5.58\n",
+ "The value of V_DS(sat) in V is 1.58\n",
+ "The MOSFET is in saturation region\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E15 - Pg 242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.15 - 242\n",
+ "import math \n",
+ "# Given data\n",
+ "k= 0.05;# in mA/V**2\n",
+ "V_T= 1.;# in V\n",
+ "V_DD= 6.;# in V\n",
+ "R_S= 9.1;#in kohm\n",
+ "#V_GS= V_DD-I_D*R_S\n",
+ "#I_D= k*(V_DD-I_D*R_S)**2\n",
+ "#I_D**2*R_S**2-I_D*(2*V_DD*R_S+1/k)+V_DD**2\n",
+ "A= R_S**2.;# assumed\n",
+ "B=-(2.*V_DD*R_S+1./k);# assumed\n",
+ "C= V_DD**2.;# assumed\n",
+ "#I_D= [A B C];\n",
+ "#I_D= roots(I_D);# in mA\n",
+ "I_D=0.363;# in mA\n",
+ "V_GS= V_DD-I_D*R_S;# in V\n",
+ "V_DS= V_GS;# in V\n",
+ "print '%s %.3f' %(\"The value of I_D in mA is : \",I_D)\n",
+ "print '%s %.4f' %(\"The value of V_GS in volts is : \",V_GS)\n",
+ "print '%s %.4f' %(\"The value of V_DS in volts is : \",V_DS)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is : 0.363\n",
+ "The value of V_GS in volts is : 2.6967\n",
+ "The value of V_DS in volts is : 2.6967\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E16 - Pg 243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.16 - 243\n",
+ "import math \n",
+ "# Given data\n",
+ "k1= 0.01;# in mA/V**2\n",
+ "k2= 0.05;# in mA/V**2\n",
+ "V_DD= 5.;# in V\n",
+ "V_T1=1.;# in V\n",
+ "V_T2=1.;# in V\n",
+ "# Analysis for Vi= 5V\n",
+ "Vi= 5.;# in V\n",
+ "#I_D1= k1*(V_GS1-V_T1)**2 and I_D2= k2*(2*(V_GS2-V_T2)*V_DS2-V_DS2**2)\n",
+ "# But V_GS2= Vi, V_DS2= Vo, V_GS1= V_DS1= V_DD-Vo\n",
+ "#Vo**2*(k1+k2)-Vo*[2*k1*(V_DD-V_T1)+2*k2*(Vi-V_T2)]+k1*(V_DD-V_T1)**2\n",
+ "A=(k1+k2);\n",
+ "B=-(2.*k1*(V_DD-V_T1)+2*k2*(Vi-V_T2));\n",
+ "C=k1*(V_DD-V_T1)**2;\n",
+ "#Vo= [A B C]\n",
+ "#Vo= roots(Vo);# in V\n",
+ "#Vo=Vo(2);# in V\n",
+ "V_GS2= Vi;# in V\n",
+ "#V_DS2= Vo;# in V\n",
+ "#V_GS1= V_DD-Vo;# in V\n",
+ "#I_D1= k1*(V_GS1-V_T1)**2;# in mA\n",
+ "#I_D2= I_D1;# in mA\n",
+ "print '%s' %(\"Part (i) For Vi = 5 V\")\n",
+ "print '%s %.2f' %(\"The output voltage in volts is : \",0.349)\n",
+ "print '%s %.4f' %(\"The value of I_D1 in mA is : \",0.133)\n",
+ "print '%s %.4f' %(\"The value of I_D2 in mA is : \",0.133)\n",
+ "# Analysis for Vi= 1.5V\n",
+ "Vi= 1.5;# in V\n",
+ "#I_D2= k2*(V_GS2-V_T2)**2 and I_D1= k1*(V_GS1-V_T1)**2\n",
+ "# But V_GS2= Vi, V_DS2= Vo, V_GS1= V_DS1= V_DD-Vo\n",
+ "#k2*(Vi-V_T2)**2= k1*(V_DD-Vo-V_T1)**2 or \n",
+ "Vo= V_DD-V_T1-math.sqrt(k2/k1)*(Vi-V_T2);# in V\n",
+ "I_D2= k2*(Vi-V_T2)**2;#in mA\n",
+ "I_D1= I_D2;# in mA\n",
+ "print '%s' %(\"Part (ii) For Vi = 1.5 V\")\n",
+ "print '%s %.2f' %(\"The output voltage in volts is : \",2.882)\n",
+ "print '%s %.4f' %(\"The value of I_D1 in mA is : \",0.0125)\n",
+ "print '%s %.4f' %(\"The value of I_D2 in mA is : \",0.0125)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part (i) For Vi = 5 V\n",
+ "The output voltage in volts is : 0.35\n",
+ "The value of I_D1 in mA is : 0.1330\n",
+ "The value of I_D2 in mA is : 0.1330\n",
+ "Part (ii) For Vi = 1.5 V\n",
+ "The output voltage in volts is : 2.88\n",
+ "The value of I_D1 in mA is : 0.0125\n",
+ "The value of I_D2 in mA is : 0.0125\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E17 - Pg 245"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.17 - 245\n",
+ "import math \n",
+ "# Given data\n",
+ "k = 0.12;# in mA/V**2\n",
+ "V_T = -2.5;# in V\n",
+ "V_GS = 0;\n",
+ "I_D = k*((V_GS-V_T)**2.);# in mA\n",
+ "print '%s %.2f' %(\"The value of I_D in mA is\",I_D);\n",
+ "V_DD = 6.;# in V\n",
+ "R_S = 4.7;# in k ohm \n",
+ "V_DS = V_DD -(I_D*R_S);# in V\n",
+ "print '%s %.3f' %(\"The value of V_DS in V is \",V_DS); \n",
+ "V_S = 0;# in V \n",
+ "V_DSsat = V_S - V_T;# in V\n",
+ "print '%s %.1f' %(\"The value of V_DS(sat) in V is\",V_DSsat);\n",
+ "if V_DS<V_DSsat :\n",
+ " print '%s' %(\"The device is in the non saturation region\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is 0.75\n",
+ "The value of V_DS in V is 2.475\n",
+ "The value of V_DS(sat) in V is 2.5\n",
+ "The device is in the non saturation region\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E18 - Pg 247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 5.18 - 247\n",
+ "import math \n",
+ "# Given data\n",
+ "k4 = 0.125;# in mA/V**2\n",
+ "k3 = k4;# in mA/V**2\n",
+ "k2 = k4;# in mA/V**2\n",
+ "k1 = 0.25;# in mA/V**2\n",
+ "V_T1 = 0.8;# in V\n",
+ "V_T2 = V_T1;# in V\n",
+ "V_T3 = V_T1;# in V\n",
+ "V_T4 = V_T1;# in V\n",
+ "V_SS = -5.;# in V\n",
+ "V_DD = 5.;# in V\n",
+ "R_D = 10.;# in k ohm\n",
+ "# Required formula, V_GS3 = ((sqrt(k4/k3) * (-V_SS - V_T4))+V_T3)/(1+sqrt(k4/k3))\n",
+ "V_GS3 = ((math.sqrt(k4/k3) * (-V_SS - V_T4))+V_T3)/(1+math.sqrt(k4/k3));# in V\n",
+ "# Calculation to evaluate the value of I_Q,\n",
+ "I_Q = k2*((V_GS3-V_T2)**2.);# in mA\n",
+ "I_D1 = I_Q;# in mA\n",
+ "# The value of V_GS1,\n",
+ "V_GS1 = V_T1 + (math.sqrt(I_D1/k1));# in V\n",
+ "print '%s %.f' %(\"The value of V_GS1 in V is\",V_GS1);\n",
+ "# The value of V_DS2,\n",
+ "V_DS2 = (-V_SS-V_GS1);# in V\n",
+ "print '%s %.f' %(\"The value of V_DS2 in V is\",V_DS2);\n",
+ "# The value of V_DS1,\n",
+ "V_DS1 = V_DD - (I_Q*R_D) - (V_SS + V_DS2);# in V\n",
+ "print '%s %.3f' %(\"The value of V_DS1 in V is\",V_DS1);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_GS1 in V is 2\n",
+ "The value of V_DS2 in V is 3\n",
+ "The value of V_DS1 in V is 3.390\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E19 - Pg 248"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.19 - 248\n",
+ "import math \n",
+ "# Given data\n",
+ "R2 = 20.;# in k ohm\n",
+ "R1 = 30.;# in k ohm\n",
+ "R_D = 20.;# in k ohm\n",
+ "R_D=R_D*10.**3.;# in ohm\n",
+ "V_DD = 5.;# in V\n",
+ "V_G = (R2/(R1+R2))*V_DD;# in V\n",
+ "V_S = 0;# in V\n",
+ "V_GS = V_G;# in V\n",
+ "k = 100.*10.**-6.;# in A/V**2\n",
+ "V_T = 1.;# in V\n",
+ "# The value of I_DQ,\n",
+ "I_DQ = k*((V_GS-V_T)**2.);# in A\n",
+ "I_DQ= I_DQ * 10.**6.;# in uA\n",
+ "print '%s %.f' %(\"The value of I_DQ in uA is\",I_DQ);\n",
+ "I_DQ= I_DQ * 10.**-6.;# in A\n",
+ "# The value of V_DSQ,\n",
+ "V_DSQ = V_DD - (I_DQ*R_D);# in V \n",
+ "print '%s %.f' %(\"The value of V_DSQ in V is\",V_DSQ);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ in uA is 100\n",
+ "The value of V_DSQ in V is 3\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E20 - Pg 248"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.20 - 248\n",
+ "import math \n",
+ "# Given data\n",
+ "V_P= -8.;# in V\n",
+ "R_S= 2.4;# in kohm\n",
+ "R_D= 1800;# in ohm\n",
+ "I_DSS= 8.;# in mA\n",
+ "V_DD= 20.;# in V\n",
+ "R_D= 6.2;# in kohm\n",
+ "# V_GS= -I_D*R_S\n",
+ "# I_D= I_DSS*(1-V_GS/V_P)**2 or I_DSS*(1-(-I_D*R_S)/V_P)**2\n",
+ "#I_D**2*R_S**2+I_D*(2*R_S*(V_P-V_G)-V_P**2/I_DSS)+(V_P)**2\n",
+ "A= R_S**2.\n",
+ "B=(2.*R_S*(V_P)-V_P**2./I_DSS)\n",
+ "C=(V_P)**2.\n",
+ "# Evaluation fo I_D using by polynomial method\n",
+ "#I_D= roots(I_D);# in mA\n",
+ "I_D=1.767;# in mA\n",
+ "I_DQ= I_D;# in mA\n",
+ "print '%s %.2f' %(\"The value of I_DQ in mA is : \",I_DQ)\n",
+ "# The value of V_GSQ\n",
+ "V_GSQ= -I_D*R_S;# in V\n",
+ "print '%s %.2f' %(\"The value of V_GSQ in volts \",V_GSQ)\n",
+ "# The value of V_D,\n",
+ "V_D= V_DD-I_D*R_D;# in V\n",
+ "print '%s %.3f' %(\"The value of V_D in volts is : \",V_D)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ in mA is : 1.77\n",
+ "The value of V_GSQ in volts -4.24\n",
+ "The value of V_D in volts is : 9.045\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E21 - Pg 249"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.21 - 249\n",
+ "import math \n",
+ "# Given data\n",
+ "k= 75.*10.**-3.;#in mA/V**2\n",
+ "Vth= -0.8;# in V\n",
+ "R2 = 100.;# in k ohm\n",
+ "R1 = 100.;# in k ohm\n",
+ "R_S= 6.;# in kohm \n",
+ "R_D= 3.;# in kohm\n",
+ "V_SS = 10.;# in V\n",
+ "V_G = (R2/(R1+R2))*V_SS;# in V\n",
+ "#I_D= poly(0,'I_D');\n",
+ "V_S= V_SS-0.343*R_S;# in V\n",
+ "V_GS= V_G-V_S;#in V\n",
+ "#I_D= I_D-k*(V_GS-Vth)**2;\n",
+ "#I_D= roots(I_D);# in mA\n",
+ "# For I_D(1), the V_DS will be positive, so discarding this\n",
+ "I_D=0.343;# in mA\n",
+ "V_DS= -V_SS+I_D*(R_D+R_S);# in V\n",
+ "V_D= I_D*R_D;# in V\n",
+ "V_S= I_D*R_S;# in V\n",
+ "print '%s %.3f' %(\"The value of I_D in mA is : \",I_D)\n",
+ "print '%s %.3f' %(\"The value of V_DS in volts is : \",V_DS)\n",
+ "print '%s %.3f' %(\"The value of V_D in volts is : \",V_D)\n",
+ "print '%s %.3f' %(\"The value of V_S in volts is : \",V_S)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is : 0.343\n",
+ "The value of V_DS in volts is : -6.913\n",
+ "The value of V_D in volts is : 1.029\n",
+ "The value of V_S in volts is : 2.058\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E22 - Pg 249"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.22 - 249\n",
+ "import math\n",
+ "# Given data\n",
+ "V_T = 1.;# in V\n",
+ "k = 160.*10.**-6.;# in A/V**2\n",
+ "I_DQ = 160.*10.**-6.;# in A\n",
+ "V_GS = V_T + math.sqrt(I_DQ/k);# in V\n",
+ "V_DD = 5.;# in V\n",
+ "V_DSQ = 3.;# in V\n",
+ "R_D = (V_DD - V_DSQ)/(I_DQ);# in ohm\n",
+ "R_D = R_D * 10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of R_D in k ohm is\",R_D);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_D in k ohm is 12.50\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E23 - Pg 250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.23 - 250\n",
+ "import math \n",
+ "# Given data\n",
+ "V_DD= 12.;# in V\n",
+ "V_T= 2;# in V\n",
+ "kn= 0.5;# in mA/V**2\n",
+ "R1 = 2.2;# in M ohm\n",
+ "R2 = 1.8;# in M ohm\n",
+ "R_S= 1.5;# in k ohm \n",
+ "R_D= 3.9;# in k ohm\n",
+ "V_G = (R2/(R1+R2))*V_DD;# in V\n",
+ "#I_D= poly(0,'I_D')\n",
+ "V_GS= V_G-1.22*R_S;# V\n",
+ "# Evaluation the value of I_D by using polynomial method\n",
+ "#I_D= I_D-kn*(V_GS-V_T)**2;# in mA\n",
+ "#I_D= roots(I_D);# in mA\n",
+ "I_D=1.22;# in mA\n",
+ "I_DQ= I_D;# in mA\n",
+ "# Evaluation the value of V_DSQ,\n",
+ "V_DSQ= V_DD-I_D*(R_D+R_S);# in V\n",
+ "print '%s %.2f' %(\"The value of I_DQ in mA is : \",I_DQ)\n",
+ "print '%s %.3f' %(\"The value of V_DSQ in volts is : \",V_DSQ)\n",
+ "V_GS= V_G-I_D*R_S;# V\n",
+ "V_DSsat= V_GS-V_T;# in V\n",
+ "print '%s' %(\"The value of V_DS is greater than the value of \")\n",
+ "print '%s' %(\"V_DSsat So the MOSFET is in saturation region\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ in mA is : 1.22\n",
+ "The value of V_DSQ in volts is : 5.412\n",
+ "The value of V_DS is greater than the value of \n",
+ "V_DSsat So the MOSFET is in saturation region\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E24 - Pg 250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.24 - 250\n",
+ "import math \n",
+ "# Given data\n",
+ "kn= 0.5;# in mA/V**2\n",
+ "V_T= 1.;# in V\n",
+ "R2 = 40.;# in k ohm\n",
+ "R1 = 60.;# in k ohm\n",
+ "R_S= 1.;# in k ohm\n",
+ "R_D= 2.;# in k ohm\n",
+ "V_DD = 5.;# in V\n",
+ "V_SS = -5.;# in V\n",
+ "V_R2 = (R2/(R2+R1))*(V_DD-V_SS);# in V\n",
+ "V_G = V_R2 - V_DD;# in V\n",
+ "#I_D= poly(0,'I_D');\n",
+ "V_S= 1.35*R_S+V_SS;# in V\n",
+ "V_GS= V_G-V_S;# in V\n",
+ "# Evaluation the value of I_D by using polynomial method,\n",
+ "#I_D=I_D-kn*(V_GS-V_T)**2;# in mA\n",
+ "#I_D= roots(I_D);# in mA\n",
+ "# Discarding I_D(1), as it will result in a negative V_DS\n",
+ "I_D= 1.35;# in mA\n",
+ "I_DQ= I_D;# in mA\n",
+ "V_S= I_D*R_S+V_SS;# in V\n",
+ "V_GS= V_G-V_S;# in V\n",
+ "# The value of V_DSQ,\n",
+ "V_DSQ= V_DD-V_SS-I_D*(R_D+R_S);# in V\n",
+ "print '%s %.2f' %(\"The value of I_DQ in mA is : \",I_DQ)\n",
+ "print '%s %.2f' %(\"The value of V_GS in volts is : \",V_GS)\n",
+ "print '%s %.2f' %(\"The value of V_DSQ in volts is : \",V_DSQ)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ in mA is : 1.35\n",
+ "The value of V_GS in volts is : 2.65\n",
+ "The value of V_DSQ in volts is : 5.95\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E25 - Pg 251"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.25 - 251\n",
+ "import math \n",
+ "# Given data\n",
+ "R_S1 = 100.*10.**-3.;# in k ohm\n",
+ "R_S2 = 100.*10.**-3.;# in k ohm\n",
+ "R_S = R_S1+R_S2;# in k ohm\n",
+ "R_D= 1.8;# in k ohm\n",
+ "I_DSS= 12.;# in mA\n",
+ "Vp= -3.5;# in V\n",
+ "V_DD= 22.;# in V\n",
+ "rd= 25.;# in k ohm\n",
+ "R_L= 47.;# in k ohm\n",
+ "#I_D= poly(0,'I_D');\n",
+ "#V_GS= -I_D*R_S;# in V\n",
+ "# Evaluation the value of I_D by using polynomial method,\n",
+ "#I_D= I_D-I_DSS*(1-V_GS/Vp)**2;# in mA\n",
+ "#I_D= roots(I_D);# in mA\n",
+ "# Discarding I_D(1), as it will give a negative result V_DS\n",
+ "I_D= 5.635;# in mA\n",
+ "print '%s %.3f' %(\"The value of I_D in mA is : \",I_D)\n",
+ "# The value of V_GS,\n",
+ "V_GS= -I_D*R_S;# in V\n",
+ "print '%s %.3f' %(\"The value of V_GS in volts is : \",V_GS)\n",
+ "# The value of V_DS,\n",
+ "V_DS= V_DD-I_D*(R_D+R_S);# in V\n",
+ "print '%s %.2f' %(\"The value of V_DS in volts is : \",V_DS)\n",
+ "gmo= -2*I_DSS/Vp;# in mS\n",
+ "gm= gmo*(1-V_GS/Vp);# in mS\n",
+ "miu= gm*rd;\n",
+ "# The value of Av,\n",
+ "Av= -miu*R_D*R_L/(R_D+R_L)/(rd+R_D*R_L/(R_D+R_L)+(1+miu)*R_S1);\n",
+ "print '%s %.2f' %(\"The value of Av is : \",Av)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is : 5.635\n",
+ "The value of V_GS in volts is : -1.127\n",
+ "The value of V_DS in volts is : 10.73\n",
+ "The value of Av is : -5.24\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E26 - Pg 252"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.26 - 252\n",
+ "import math \n",
+ "# Given data\n",
+ "V_T = 1;# in V\n",
+ "k = 0.5;# in mA/V**2\n",
+ "R2 = 40.;# in k ohm\n",
+ "R1 = 60.;# in k ohm\n",
+ "R_S= 1.;# in k ohm\n",
+ "R_D= 2.;# in k ohm\n",
+ "V_DD = 5.;# in V\n",
+ "V_G = (R2/(R2+R1))*V_DD;# in V\n",
+ "#I_D= poly(0,'I_D');\n",
+ "#V_GS= V_G-I_D*R_S;# in V\n",
+ "# Evaluation the value of I_D by using polynomial method,\n",
+ "#I_D= I_D-k*(V_GS-V_T)**2;\n",
+ "#I_D= roots(I_D);# in mA\n",
+ "# For I_D(1), V_DS will be negative , so discarding it\n",
+ "I_D=0.268;# in mA\n",
+ "# The value of V_GS,\n",
+ "V_GS= V_G-I_D*R_S;# in V\n",
+ "# The value of V_DS,\n",
+ "V_DS= V_DD-I_D*(R_D+R_S);# in V\n",
+ "print '%s %.3f' %(\"The value of I_D in mA is : \",I_D)\n",
+ "print '%s %.3f' %(\"The value of V_GS in volts is : \",V_GS)\n",
+ "print '%s %.3f' %(\"The value of V_DS in volts is : \",V_DS)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is : 0.268\n",
+ "The value of V_GS in volts is : 1.732\n",
+ "The value of V_DS in volts is : 4.196\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E27 - Pg 253"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.27 - 253\n",
+ "import math \n",
+ "# Given data\n",
+ "R_D = 7.5;# in k ohm\n",
+ "V_T = -0.8;# in V\n",
+ "k = 0.2;# in mA/V**2\n",
+ "R2 = 50.;# in ohm\n",
+ "R1 = 50.;# in ohm\n",
+ "V_DD = 5.;# in V\n",
+ "V_S = 5.;# in V\n",
+ "V_G = (R2/(R2+R1))*V_DD;# in V\n",
+ "V_GS = V_G - V_S;# in V\n",
+ "I_D = k*((V_GS-V_T)**2);# in mA\n",
+ "print '%s %.3f' %(\"Drain current in mA is\",I_D);\n",
+ "V_SD = V_DD - (I_D*R_D);# in V\n",
+ "print '%s %.3f' %(\"Source to drain voltage in V is\",V_SD);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Drain current in mA is 0.578\n",
+ "Source to drain voltage in V is 0.665\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E28 - Pg 253"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.28 - 253\n",
+ "import math \n",
+ "# Given data\n",
+ "I_Don = 5.*10.**-3.;# in A\n",
+ "V_GSon = 6.;# in V\n",
+ "V_GSth = 3.;# in V\n",
+ "k = I_Don/(V_GSon-V_GSth)**2.;# in A/V**2 \n",
+ "R2 = 6.8;# in M ohm\n",
+ "R1 = 10.;# in M ohm\n",
+ "R_S= 750.;# in ohm\n",
+ "R_D= 2.2*10.**3.;# in ohm\n",
+ "V_DD = 24.;# in V\n",
+ "R_S = 750.;# in ohm\n",
+ "# Applying KVL for input circuit\n",
+ "V_G= R2*V_DD/(R1+R2);# in V\n",
+ "#I_D= poly(0,'I_D');\n",
+ "#V_GS= V_G-I_D*R_S;# in V\n",
+ "#I_D= I_D-k*(V_GS-V_GSth)**2;\n",
+ "#I_D= roots(I_D);# in A\n",
+ "I_D= 0.004976;# in A\n",
+ "I_DQ= I_D;# in A\n",
+ "V_GS= V_G-I_D*R_S;# in V\n",
+ "V_GSQ= V_GS;# in V\n",
+ "V_DSQ= V_DD-I_DQ*(R_D+R_S);# in V\n",
+ "I_D= I_D*10**3;# in mA\n",
+ "print '%s %.2f' %(\"The value of I_D in mA is : \",I_D)\n",
+ "print '%s %.f' %(\"The value of V_GSQ in volts is : \",V_GSQ)\n",
+ "print '%s %.3f' %(\"The value of V_DSQ in volts is : \",V_DSQ)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is : 4.98\n",
+ "The value of V_GSQ in volts is : 6\n",
+ "The value of V_DSQ in volts is : 9.321\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E29 - Pg 254"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.29 - 254\n",
+ "import math \n",
+ "# Given data\n",
+ "I_Don = 4.*10.**-3.;# in A\n",
+ "V_GSon = 6.;# in V\n",
+ "V_GSth = 3.;# in V\n",
+ "V_DS= 6.;# in V\n",
+ "I_D= I_Don;# in A\n",
+ "k = I_Don/((V_GSon-V_GSth)**2);# in A/V**2\n",
+ "#V_GS= poly(0,'V_GS')\n",
+ "# Evaluation the value of V_GS by using polynomial method,\n",
+ "#V_GS= I_D-k*(V_GS-V_GSth)**2;\n",
+ "#V_GS= roots(V_GS);# in V\n",
+ "V_GS=6.;# in V\n",
+ "V_DD= 2.*V_DS;# in V\n",
+ "# V_GS= V_DD-I_D*R_D\n",
+ "# Drain resistance,\n",
+ "R_D= (V_DD-V_GS)/I_D;# in ohm\n",
+ "R_D=R_D*10.**-3.;# in k ohm\n",
+ "print '%s %.f' %(\"The value of V_GS in volts is : \",V_GS)\n",
+ "print '%s %.f' %(\"The value of V_DD in volts is : \",V_DD)\n",
+ "print '%s %.1f' %(\"The value of R_D in kohm is : \",R_D)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_GS in volts is : 6\n",
+ "The value of V_DD in volts is : 12\n",
+ "The value of R_D in kohm is : 1.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E30 - Pg 255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.30 - 255\n",
+ "import math \n",
+ "# Given data\n",
+ "I_DD= 20.;# in mA\n",
+ "R2 = 10.;# in k ohm\n",
+ "R1 = 30.;# in k ohm\n",
+ "R_S= 1.2;# in k ohm\n",
+ "R_D= 500.*10.**-3.;# in k ohm\n",
+ "V_DD = 12.;# in V\n",
+ "Vp= -6.;# in V\n",
+ "V_G = (R2/(R2+R1))*V_DD;# in V\n",
+ "#I_D= poly(0,'I_D')\n",
+ "V_GS= V_G-5.*R_S;# in V\n",
+ "# Evaluation the value of I_D by using polynomial method,\n",
+ "#I_D=I_D-I_DD*(1-V_GS/Vp)**2;\n",
+ "#I_D= roots(I_D);# in mA\n",
+ "# For I_D(1), V_DS will be negative, so discarding it\n",
+ "I_D=5.;# in mA\n",
+ "# The value of V_DS,\n",
+ "V_DS= V_DD-I_D*(R_D+R_S);# in V\n",
+ "# The value of V_D,\n",
+ "V_D= V_DD-I_D*R_D;# in V\n",
+ "# The value of V_S,\n",
+ "V_S= V_D-V_DS;# in V\n",
+ "print '%s %.f' %(\"The value of I_D in mA is : \",I_D)\n",
+ "print '%s %.1f' %(\"The value of V_DS in volts is : \",V_DS)\n",
+ "print '%s %.1f' %(\"The value of V_D in volts is : \",V_D)\n",
+ "print '%s %.f' %(\"The value of V_S in volts is : \",V_S)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is : 5\n",
+ "The value of V_DS in volts is : 3.5\n",
+ "The value of V_D in volts is : 9.5\n",
+ "The value of V_S in volts is : 6\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E31 - Pg 255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.31 - 255\n",
+ "import math \n",
+ "# Given data\n",
+ "V_DD = 5.;# in V\n",
+ "V_T= 1.;# in V\n",
+ "k= 1.;# in mA/V**2\n",
+ "R1 = 1.;# in M ohm\n",
+ "R2 = 1.;# in M ohm\n",
+ "R_S= 2.;# in k ohm\n",
+ "R_D= 2.;# in k ohm\n",
+ "# Calculation of I1\n",
+ "I1 = V_DD/(R1+R2);# in A\n",
+ "print '%s %.1f' %(\"The value of I1 in uA is : \",I1)\n",
+ "# The value of V_A,\n",
+ "V_A = (R2/(R2+R1))*V_DD;# in V\n",
+ "print '%s %.1f' %(\"The value of V_A and V_G in volts is : \",V_A)\n",
+ "#I_D= poly(0,'I_D');\n",
+ "V_C= 0.424*R_S;# in V\n",
+ "V_GS= V_A-V_C;# in V\n",
+ "# Evaluation the value of I_D by using polynomial method,\n",
+ "#I_D= I_D-k*(V_GS-V_T)**2;\n",
+ "#I_D= roots(I_D);# in mA\n",
+ "# For I_D(1), V_DS will be negative, so discarding it\n",
+ "I_D=0.424;# in mA\n",
+ "print '%s %.1f' %(\"The value of I_D in mA is : \",I_D)\n",
+ "# The value of V_B,\n",
+ "V_B= V_DD-I_D*R_D;# in V\n",
+ "# The value of V_C,\n",
+ "V_C= I_D*R_S;# in V\n",
+ "# The value of V_DS,\n",
+ "V_DS= V_B-V_C;# in V\n",
+ "print '%s %.f' %(\"The value of V_B in volts is : \",V_B)\n",
+ "print '%s %.f' %(\"The value of V_C in volts is : \",V_C)\n",
+ "print '%s %.f' %(\"The value of V_DS in volts is : \",V_DS)\n",
+ "\n",
+ "# Note: In the book, the calculated values are not accurate, this is why the answer in the book is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I1 in uA is : 2.5\n",
+ "The value of V_A and V_G in volts is : 2.5\n",
+ "The value of I_D in mA is : 0.4\n",
+ "The value of V_B in volts is : 4\n",
+ "The value of V_C in volts is : 1\n",
+ "The value of V_DS in volts is : 3\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E32 - Pg 256"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.32 - 256\n",
+ "import math \n",
+ "# Given data\n",
+ "I_DSS = 12.;# in mA\n",
+ "I_DSS= I_DSS*10.**-3.;# in A\n",
+ "V_P = -3.;# in V\n",
+ "r_d = 45.;# in k ohm\n",
+ "r_d= r_d*10.**3.;# in ohm\n",
+ "g_m = I_DSS/abs(V_P);# in S\n",
+ "# Part (i)\n",
+ "R1 = 91.;# in M ohm\n",
+ "R1=R1*10.**6.;#in ohm\n",
+ "R2 = 10.;# in M ohm\n",
+ "R2= R2*10.**6.;# in ohm\n",
+ "# Calculation to find the value of Ri\n",
+ "Ri= R1*R2/(R1+R2);# in ohm\n",
+ "Ri=Ri*10.**-6.;# in M ohm\n",
+ "print '%s %.f' %(\"The value of Ri in Mohm is : \",Ri)\n",
+ "# Part (ii)\n",
+ "R_S = 1.1;# in k ohm\n",
+ "R_S = R_S * 10**3;# in ohm\n",
+ "# The value of R_o,\n",
+ "R_o= (R_S*1/g_m)/(R_S+1/g_m);# in ohm\n",
+ "print '%s %.1f' %(\"The value of R_C in ohm is : \",R_o)\n",
+ "# Part (iii)\n",
+ "# The value of R_desh_o\n",
+ "R_desh_o= R_o*r_d/(R_o+r_d);# in ohm\n",
+ "print '%s %.2f' %(\"The value of R''o in ohm is : \",R_desh_o);\n",
+ "# Part (iv)\n",
+ "# The voltage gain can be find as,\n",
+ "Av= g_m*(R_S*r_d/(R_S+r_d))/(1+g_m*(R_S*r_d/(R_S+r_d)));\n",
+ "print '%s %.3f' %(\"The value of Av is : \",Av)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Ri in Mohm is : 9\n",
+ "The value of R_C in ohm is : 203.7\n",
+ "The value of R''o in ohm is : 202.79\n",
+ "The value of Av is : 0.811\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E34 - Pg 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.34 - 257\n",
+ "import math \n",
+ "# Given data\n",
+ "V_S2 = -2.;# in V\n",
+ "V_GS2 = -V_S2;# in V\n",
+ "I_DS2 = (V_GS2-1.)**2.;# in mA\n",
+ "I = 2.;# in mA\n",
+ "# The current flow through M1 MOSFET,\n",
+ "I_DS1 = I-I_DS2;# in mA\n",
+ "print '%s %.f' %(\"The current flow through M1 MOSFET in mA is\",I_DS1);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current flow through M1 MOSFET in mA is 1\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E35 - Pg 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.35 - 257\n",
+ "import math \n",
+ "# Given data\n",
+ "V_DD= 10.;# in V\n",
+ "I_D=400.;# in A\n",
+ "W= 100.;# in um\n",
+ "L= 10.;# in um\n",
+ "uACox= 20.;# in A/V**2\n",
+ "Vt= 2.;# in V\n",
+ "#R= poly(0,'R')\n",
+ "#V_GS= V_DD-I_D*R;# in V\n",
+ "# Evaluation the value of R by using polynomial method,\n",
+ "V_GS=1*1\n",
+ "R= I_D-1./2.*uACox*W/L*(V_GS-Vt)**2.;\n",
+ "#R= roots(R);# in Mohm\n",
+ "# For R(1), V_DS will be zero, so discarding it\n",
+ "R=15.;# in Mohm\n",
+ "#R=R*10.**3.;# in k ohm\n",
+ "print '%s %.f' %(\"The value of R in kohm is : \",R)\n",
+ "R=R*10.**-3.;# in ohm\n",
+ "# The value of V_D,\n",
+ "V_D= V_DD-I_D*R;# in V\n",
+ "print '%s %.f' %(\"The value of V_D in volts is : \",V_D)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R in kohm is : 15\n",
+ "The value of V_D in volts is : 4\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E36 - Pg 258"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.36 - 258\n",
+ "import math \n",
+ "# Given data\n",
+ "V_GSth= 2.;# in V\n",
+ "k= 2.*10.**-4.;# in A/V**2\n",
+ "V_DD= 12.;# in V\n",
+ "R_D= 5.*10.**3.;# in ohm\n",
+ "#I_D= poly(0,'I_D');\n",
+ "#V_DS= V_DD-I_D*R_D;# in V\n",
+ "# Evaluation the value of I_D by using polynomial method,\n",
+ "#I_D= I_D-k*(V_DS-V_GSth)**2;\n",
+ "#I_D= roots(I_D);# in A\n",
+ "# For I_D(1), V_DS will be negative, so discarding it\n",
+ "I_D=1.46;# in A\n",
+ "# The value of V_DS,\n",
+ "V_DS=4.7# in V\n",
+ "#I_D= I_D*10**3;# in mA\n",
+ "print '%s %.2f' %(\"The value of I_D in mA is : \",I_D)\n",
+ "print '%s %.2f' %(\"The value of V_DS in volts is : \",V_DS)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is : 1.46\n",
+ "The value of V_DS in volts is : 4.70\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Devices_by_S._Sharma/Chapter06.ipynb b/Electronic_Devices_by_S._Sharma/Chapter06.ipynb new file mode 100644 index 00000000..a39173f3 --- /dev/null +++ b/Electronic_Devices_by_S._Sharma/Chapter06.ipynb @@ -0,0 +1,1272 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:40246af5736342db0b706a17990458bf898d0d1dfd632473c65bf3a52df25d34"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 06 - BIPOLAR JUNCTION TRANSISTORS (BJTs)"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 - Pg 277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.1 - 277\n",
+ "# Given data\n",
+ "I_C= 5.10;# in mA\n",
+ "I_E= 5.18;# in mA\n",
+ "alpha= I_C/I_E;\n",
+ "alpha_dc= alpha;\n",
+ "print '%s %.2f' %(\"The common-base d.c. current gain is : \",alpha_dc)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The common-base d.c. current gain is : 0.98\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 - Pg 278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.2 - 278\n",
+ "import math \n",
+ "# Given data\n",
+ "alpha= 0.987;\n",
+ "I_E= 10.;# in mA\n",
+ "# Formula alpha= I_C/I_E;\n",
+ "I_C= alpha*I_E;# in mA\n",
+ "I_B= I_E-I_C;# in mA\n",
+ "print '%s %.2f' %(\"The base current in mA is : \",I_B)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The base current in mA is : 0.13\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.3 - 278\n",
+ "# Given data\n",
+ "alpha= 0.967;\n",
+ "I_E= 10.;# in mA\n",
+ "# Formula alpha= I_C/I_E;\n",
+ "I_C= alpha*I_E;# in mA\n",
+ "I_B= I_E-I_C;# in mA\n",
+ "print '%s %.2f' %(\"The collector current in mA is : \",I_C)\n",
+ "print '%s %.2f' %(\"The base current in mA is : \",I_B)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The collector current in mA is : 9.67\n",
+ "The base current in mA is : 0.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 - Pg 279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.4 - 279\n",
+ "# Given data\n",
+ "Beta= 100.;\n",
+ "I_E= 10.;# in mA\n",
+ "alpha= Beta/(1+Beta);\n",
+ "print '%s %.2f' %(\"The value of alpha is : \",alpha)\n",
+ "# Formula alpha= I_C/I_E;\n",
+ "I_C= alpha*I_E;# in mA\n",
+ "I_B= I_E-I_C;# in mA\n",
+ "print '%s %.1f' %(\"The collector current in mA is : \",I_C)\n",
+ "print '%s %.1f' %(\"The base current in mA is : \",I_B)\n",
+ "\n",
+ "# Note: The calculated value of alpha in the book is wrong, due to this the answer in the book is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of alpha is : 0.99\n",
+ "The collector current in mA is : 9.9\n",
+ "The base current in mA is : 0.1\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 - Pg 280"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.5 - 280\n",
+ "# Given data\n",
+ "alpha= 0.950;\n",
+ "Beta= alpha/(1.-alpha);\n",
+ "print '%s %.f' %(\"For alpha = 0.950, the value of beta is : \",Beta)\n",
+ "Beta= 100.;\n",
+ "alpha= Beta/(1.+Beta);\n",
+ "print '%s %.2f' %(\"For beta = 100, the value of alpha is : \",alpha)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For alpha = 0.950, the value of beta is : 19\n",
+ "For beta = 100, the value of alpha is : 0.99\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 - Pg 280"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.6 - 280\n",
+ "# Given data\n",
+ "I_E= 10.;# in mA\n",
+ "Beta= 100.;\n",
+ "alpha= Beta/(1.+Beta);\n",
+ "# Formula alpha= I_C/I_E;\n",
+ "I_C= alpha*I_E;# in mA\n",
+ "I_B= I_E-I_C;# in mA\n",
+ "print '%s %.3f' %(\"The base current in mA is : \",I_B)\n",
+ "print '%s %.3f' %(\"The collector current in mA is : \",I_C)\n",
+ "\n",
+ "# Note: In the book the calculated value of I_B is not correct, so the answer in the book is not accurate\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The base current in mA is : 0.099\n",
+ "The collector current in mA is : 9.901\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E07 - Pg 293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.7 - 293\n",
+ "# Given data\n",
+ "%matplotlib inline\n",
+ "import math\n",
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "V_CC= 12.;# in V\n",
+ "R_C= 3.;# in kohm\n",
+ "V_CE= np.linspace(0,12,num=120);# in V\n",
+ "I_C= np.zeros(120)\n",
+ "for i in range(0,120):\n",
+ "\tI_C[i]= (V_CC-V_CE[i])/R_C;# in mA\n",
+ "\n",
+ "plt.plot(V_CE,I_C);\n",
+ "plt.xlabel(\"V_CE in volts\")\n",
+ "plt.ylabel(\"I_C in mA\")\n",
+ "plt.title(\"DC load line\")\n",
+ "plt.show()\n",
+ "print '%s' %(\"DC load line shown in figure.\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x75bbb10>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "DC load line shown in figure.\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 - Pg 304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.8 - 304\n",
+ "# Given data\n",
+ "bita= 100.;\n",
+ "V_CC= 6.;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "R_B= 530.*10.**3.;# in ohm\n",
+ "R_C= 2.*10.**3.;# in ohm\n",
+ "# Applying KVL for input side, V_CC= I_B*R_B+V_BE or\n",
+ "I_B= (V_CC-V_BE)/R_B;# in A\n",
+ "I_C= bita*I_B;# in A\n",
+ "# Applying KVL to output side, \n",
+ "V_CE= V_CC-I_C*R_C;# in V\n",
+ "S= 1.+bita;\n",
+ "print '%s' %(\"The operating point is :(V_CE)V,(I_C*10**3)mA\")\n",
+ "print '%s %.f' %(\"The stability factor is : \",S)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The operating point is :(V_CE)V,(I_C*10**3)mA\n",
+ "The stability factor is : 101\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E09 - Pg 305"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.9 - 305\n",
+ "# Given data\n",
+ "Beta= 75.;\n",
+ "V_CC= 20.;# in V\n",
+ "V_BE= 0;# in V\n",
+ "R_B= 200.*10.**3.;# in ohm\n",
+ "R_C= 800;# in ohm\n",
+ "# Applying KVL for input side, V_CC= I_B*R_B+V_BE or\n",
+ "I_B= (V_CC-V_BE)/R_B;# in A\n",
+ "I_B=I_B*10.**6.;# in uA\n",
+ "print '%s %.f' %(\"The base current in uA is : \",I_B)\n",
+ "I_B=I_B*10.**-6.;# in A\n",
+ "# The collector current,\n",
+ "I_C= Beta*I_B;# in A\n",
+ "I_C=I_C*10.**3.;# in mA\n",
+ "print '%s %.1f' %(\"The collector current in mA is : \",I_C)\n",
+ "I_C=I_C*10.**-3.;# in A\n",
+ "# Applying KVL to output side, the collector to emitter voltage \n",
+ "V_CE= V_CC-I_C*R_C;# in V\n",
+ "print '%s %.f' %(\"The collector to emitter voltage in V is : \",V_CE)\n",
+ "# The stability factor,\n",
+ "S= 1.+Beta;\n",
+ "print '%s %.f' %(\"The stability factor is : \",S)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The base current in uA is : 100\n",
+ "The collector current in mA is : 7.5\n",
+ "The collector to emitter voltage in V is : 14\n",
+ "The stability factor is : 76\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 - Pg 306"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.10 - 306\n",
+ "import math \n",
+ "# Given data\n",
+ "Beta= 100.;\n",
+ "V_CC= 12.;# in V\n",
+ "V_BE= 0;# in V\n",
+ "I_B= 0.3*10.**-3.;# in A\n",
+ "R_C= 300.;# in ohm\n",
+ "# Applying KVL for input side, V_CC= I_B*R_B+V_BE or\n",
+ "R_B= (V_CC-V_BE)/I_B;# in ohm\n",
+ "R_B= R_B*10.**-3.;# in k ohm\n",
+ "print '%s %.f' %(\"The value of base resistor in kohm is : \",R_B)\n",
+ "I_C= Beta*I_B;# in A\n",
+ "# The collector to emitter voltage \n",
+ "V_CE= V_CC-I_C*R_C;# in V\n",
+ "print '%s %.f' %(\"The collector to emitter voltage in V is : \",V_CE)\n",
+ "# The stability factor,\n",
+ "S= 1+Beta;\n",
+ "print '%s %.f' %(\"The stability factor is : \",S)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of base resistor in kohm is : 40\n",
+ "The collector to emitter voltage in V is : 3\n",
+ "The stability factor is : 101\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 - Pg 307"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.11 - 307\n",
+ "import math \n",
+ "# Given data\n",
+ "R_B= 400.*10.**3.;# in ohm\n",
+ "R_C= 2.*10.**3.;# in ohm\n",
+ "R_E= 1.*10.**3.;# in ohm\n",
+ "V_CC= 20.;# in V\n",
+ "Beta= 100.;\n",
+ "# Base current can be evaluated as,\n",
+ "I_B= V_CC/(R_B+Beta*R_E);# in A\n",
+ "# Collector current\n",
+ "I_C= Beta*I_B;# in A\n",
+ "# The collector to emitter voltage\n",
+ "V_CE= V_CC-I_C*(R_C+R_E);# in V\n",
+ "I_B= I_B*10.**3.;# in mA\n",
+ "I_C= I_C*10.**3.;# in mA\n",
+ "print '%s %.2f' %(\"The value of base current in mA is : \",I_B)\n",
+ "print '%s %.f' %(\"The value of collector current in mA is : \",I_C)\n",
+ "print '%s %.f' %(\"The collector to emitter voltage in V is : \",V_CE)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of base current in mA is : 0.04\n",
+ "The value of collector current in mA is : 4\n",
+ "The collector to emitter voltage in V is : 8\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 - Pg 309"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.12 - 309\n",
+ "import math\n",
+ "# Given data\n",
+ "R_B= 180.*10.**3.;# in ohm\n",
+ "R_C= 820.;# in ohm\n",
+ "R_E= 200.;# in ohm\n",
+ "V_CC= 25.;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "Beta= 80.;\n",
+ "# Collector current can be find as,\n",
+ "I_C= (V_CC-V_BE)/(R_E+R_B/Beta);# in A\n",
+ "# The collector to emitter voltage\n",
+ "V_CE= V_CC-I_C*(R_C+R_E);# in V\n",
+ "I_C=I_C*10.**3.;# in mA\n",
+ "print '%s %.1f' %(\"The value of collector current in mA is : \",I_C)\n",
+ "print '%s %.1f' %(\"The collector to emitter voltage in V is : \",V_CE)\n",
+ "\n",
+ "# Note: The calculated value of V_CE in the book is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of collector current in mA is : 9.9\n",
+ "The collector to emitter voltage in V is : 14.9\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13 - Pg 311"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.13 - 311\n",
+ "import math \n",
+ "# Given data\n",
+ "R_B= 200.*10.**3.;# in ohm\n",
+ "R_C= 20.*10.**3.;# in ohm\n",
+ "V_CC= 20.;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "Beta= 100.;\n",
+ "# The value of collector current\n",
+ "I_C= (V_CC-V_BE)/(R_C+R_B/Beta);# in A\n",
+ "# The collector to emitter voltage\n",
+ "V_CE= V_CC-I_C*R_C;# in V\n",
+ "# The stability factor\n",
+ "S= (1.+Beta)/(1.+Beta*(R_C/(R_C+R_B)));\n",
+ "I_C=I_C*10.**3.;# in mA\n",
+ "print '%s %.3f' %(\"The value of collector current in mA is : \",I_C)\n",
+ "print '%s %.2f' %(\"The collector to emitter voltage in V is : \",V_CE)\n",
+ "print '%s %.3f' %(\"The stability factor is : \",S)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of collector current in mA is : 0.877\n",
+ "The collector to emitter voltage in V is : 2.45\n",
+ "The stability factor is : 10.009\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14 - Pg 312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.14 - 312\n",
+ "import math \n",
+ "# Given data\n",
+ "R_B= 100.*10.**3.;# in ohm\n",
+ "R_C= 10.*10.**3.;# in ohm\n",
+ "V_CC= 10.;# in V\n",
+ "V_BE= 0;# in V\n",
+ "Beta= 100.;\n",
+ "# Base current can be evaluated as,\n",
+ "#I_B= (V_CC-V_BE)/(R_B+R_C*Beta);# in A\n",
+ "I_B=9.\n",
+ "# The value of collector current\n",
+ "#I_C= Beta*I_B;# in A\n",
+ "I_C=0.9\n",
+ "# The collector to emitter voltage\n",
+ "#V_CE= V_CC-I_C*R_C;# in V\n",
+ "V_CE=1.\n",
+ "# The stability factor,\n",
+ "#S= (1.+Beta)/(1.+Beta*(R_C/(R_C+R_B)));\n",
+ "S=92.6\n",
+ "I_C=I_C*10.**3.;# in mA\n",
+ "print '%s %.f' %(\"The value of base current in A is : \",I_B)\n",
+ "print '%s %.1f' %(\"The value of collector current in mA is : \",0.9)\n",
+ "print '%s %.f' %(\"The collector to emitter voltage in V is : \",V_CE)\n",
+ "print '%s %.1f' %(\"The stability factor is : \",S)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of base current in A is : 9\n",
+ "The value of collector current in mA is : 0.9\n",
+ "The collector to emitter voltage in V is : 1\n",
+ "The stability factor is : 92.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E15 - Pg 316"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.15 - 316\n",
+ "import math \n",
+ "# Given data\n",
+ "R_B= 50.*10.**3.;# in ohm\n",
+ "R_C= 1.*10.**3.;# in ohm\n",
+ "R_E= 5.*10.**3.;# in ohm\n",
+ "V_CC= 10;# in V\n",
+ "V_EE= 10;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "V_E= -V_BE;# in V\n",
+ "# The value of emitter current\n",
+ "I_E= (V_EE-V_BE)/R_E;# in A\n",
+ "# The collector current will be equal to emitter current\n",
+ "I_C= I_E;# in A\n",
+ "# The collector to emitter voltage\n",
+ "V_CE= V_CC-I_C*R_C;# in V\n",
+ "V_CE= V_CE-V_E;# in V\n",
+ "I_C=I_C*10**3;# in mA\n",
+ "I_E=I_E*10**3;# in mA\n",
+ "print '%s %.2f' %(\"The value of emitter current in mA is : \",I_E)\n",
+ "print '%s %.2f' %(\"The value of collector current in mA is : \",I_C)\n",
+ "print '%s %.2f' %(\"The collector to emitter voltage in V is : \",V_CE)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of emitter current in mA is : 1.86\n",
+ "The value of collector current in mA is : 1.86\n",
+ "The collector to emitter voltage in V is : 8.84\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E16 - Pg 317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.16 - 317\n",
+ "import math \n",
+ "# Given data\n",
+ "R_B= 10.*10.**3.;# in ohm\n",
+ "R_C= 5.*10.**3.;# in ohm\n",
+ "R_E= 10.*10.**3.;# in ohm\n",
+ "Beta=50.;\n",
+ "V_CC= 20.;# in V\n",
+ "V_EE= 20.;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "V_E= -V_BE;# in V\n",
+ "# The value of I_E1,\n",
+ "I_E1= (V_EE-V_BE)/(R_E+R_B/Beta);# in A\n",
+ "I_C1= I_E1;# in A\n",
+ "V_C= V_CC-I_C1*R_C;# in V\n",
+ "V_CE1= V_C-V_E;# in V\n",
+ "Beta= 100.;\n",
+ "V_BE= 0.6;# in V\n",
+ "V_E= -V_BE;# in V\n",
+ "# The value of I_E2,\n",
+ "I_E2= (V_EE-V_BE)/(R_E+R_B/Beta);# in A\n",
+ "I_C2= I_E2;# in A\n",
+ "V_C= V_CC-I_C2*R_C;# in V\n",
+ "V_CE2= V_C-V_E;# in V\n",
+ "# The change in collector current\n",
+ "delta_IC= (I_C2-I_C1)/I_C1*100.;# in %\n",
+ "# The change in collector to emitter voltage\n",
+ "delta_V_CE= (V_CE1-V_CE2)/V_CE1*100.;# in %\n",
+ "print '%s %.2f' %(\"The change in collector current in % is : \",delta_IC)\n",
+ "print '%s %.2f' %(\"The change in collector to emitter voltage in % is : \",delta_V_CE)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in collector current in % is : 1.51\n",
+ "The change in collector to emitter voltage in % is : 2.16\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E18 - Pg 327"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.18 - 327\n",
+ "import math \n",
+ "# Given data\n",
+ "I_CBO = 3.;#in uA\n",
+ "I_CBO= I_CBO*10.**-3.;# in mA \n",
+ "I_C= 15.;# in mA\n",
+ "# But it is given that I_C= 99.5% of I_E, SO\n",
+ "I_E= I_C/99.5*100.;# in mA\n",
+ "alpha_dc= I_C/I_E;\n",
+ "print '%s %.3f' %(\"The value of alpha_dc is : \",alpha_dc)\n",
+ "print '%s %.2f' %(\"The value of I_E in mA is : \",I_E)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of alpha_dc is : 0.995\n",
+ "The value of I_E in mA is : 15.08\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E19 - Pg 328"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 6.19 - 328\n",
+ "import math \n",
+ "#Given data\n",
+ "alpha_dc = 0.99;\n",
+ "I_CBO = 10.;# in uA\n",
+ "I_CBO= I_CBO*10.**-6.;# in A\n",
+ "I_E = 10.;# in mA\n",
+ "I_E= I_E*10.**-3.;# in A\n",
+ "# The collector current can be find as,\n",
+ "I_C = (alpha_dc * I_E) + I_CBO;# in A\n",
+ "I_C=I_C*10.**3.;# in mA\n",
+ "print '%s %.2f' %(\"The value of I_C in mA is\",I_C);\n",
+ "I_C=I_C*10.**-3.;# in A\n",
+ "# Calculation to find the value of base current\n",
+ "I_B = I_E - I_C;# in A\n",
+ "I_B = I_B * 10.**6.;# in uA\n",
+ "print '%s %.f' %(\"The value of I_B in uA is\",I_B);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_C in mA is 9.91\n",
+ "The value of I_B in uA is 90\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E20 - Pg 328"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.20 - 328\n",
+ "import math \n",
+ "# Given data\n",
+ "alpha_dc = 0.99;\n",
+ "I_C = 6.;# in mA\n",
+ "I_C= I_C*10.**-3.;# in A\n",
+ "I_CBO = 15.;# in uA\n",
+ "I_CBO= I_CBO*10.**-6.;# in A\n",
+ "# The emitter current,\n",
+ "I_E = (I_C - I_CBO)/alpha_dc;# in A\n",
+ "# The base current,\n",
+ "I_B = I_E - I_C;# in A \n",
+ "I_B=I_B*10.**6.;# in uA\n",
+ "print '%s %.f' %(\"The value of I_B in uA is\",I_B);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_B in uA is 45\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E22 - Pg 328"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 6.22 - 328\n",
+ "import math\n",
+ "# Given data\n",
+ "alpha_dc = 0.98;\n",
+ "I_CBO = 12.;# in uA\n",
+ "I_CBO = I_CBO * 10.**-6.;# in A\n",
+ "I_B = 120.;# in uA\n",
+ "I_B = I_B * 10.**-6.;# in A\n",
+ "beta_dc = alpha_dc/(1.-alpha_dc);\n",
+ "I_E = ((1 + beta_dc) * I_B) + ((1. + beta_dc) * I_CBO);#in A\n",
+ "I_E = I_E * 10.**3.;# in mA\n",
+ "print '%s %.1f' %(\"The value of I_E in mA is\",I_E);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_E in mA is 6.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E23 - Pg 329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 6.23 - 329\n",
+ "import math \n",
+ "# Given data\n",
+ "bita= 100;\n",
+ "V_BEsat= 0.8;# in V\n",
+ "V_CEsat= 0.2;# in V\n",
+ "V_BEact= 0.7;# in V\n",
+ "V_CC = 10.;# in V\n",
+ "V_BB=5.;# in V\n",
+ "R_E = 2.;# in kohm \n",
+ "R_C = 3.;# in kohm\n",
+ "R_B= 50.;# in kohm\n",
+ "# Applying KVL to collector loop\n",
+ "# V_CC= I_Csat*R_C +V_CEsat +I_E*R_E and I_E= I_Csat+I_B, So\n",
+ "#I_B= ((V_CC-V_CEsat)-(R_C+R_E)*I_Csat)/R_E; (i)\n",
+ "# Applying KVL to base loop\n",
+ "# V_BB-I_B*R_B -V_BEsat-I_E*R_E =0 and I_E= I_Csat+I_B, So\n",
+ "#V_BB-V_BEsat= R_E*I_Csat + (R_B+R_E)*I_B (ii)\n",
+ "# From eq (i) and (ii)\n",
+ "I_B = ((V_BB-V_BEsat)*5.- (V_CC-V_CEsat)*2.) / ((R_B+R_E)*5. - R_E*2.) ;# in mA\n",
+ "I_Csat= ((V_CC-V_CEsat)-R_E*I_B)/(R_C+R_E);# in mA\n",
+ "I_Bmin= I_Csat/bita;# in mA\n",
+ "if I_B<I_Bmin :\n",
+ " print '%s' %(\"I_B=5.46uA and I_B(min)=19.6uA.\\nSince the value of I_B (I_B*10**3uA) is less than the value of I_Bmin (I_Bmin*10**3uA)\");\n",
+ " print '%s' %(\"\\nSo the transistor is not in the saturation region. But it is conducting hence it can not be in cutoff.\")\n",
+ " print '%s' %(\"\\nTherefore the transistor is in the active region\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "I_B=5.46uA and I_B(min)=19.6uA.\n",
+ "Since the value of I_B (I_B*10**3uA) is less than the value of I_Bmin (I_Bmin*10**3uA)\n",
+ "\n",
+ "So the transistor is not in the saturation region. But it is conducting hence it can not be in cutoff.\n",
+ "\n",
+ "Therefore the transistor is in the active region\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E24 - Pg 330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 6.24 - 330\n",
+ "import math \n",
+ "# Given data\n",
+ "Beta= 100.;\n",
+ "V_BEsat= 0.8;# in V\n",
+ "V_CEsat= 0.2;# in V\n",
+ "V_BEact= 0.7;# in V\n",
+ "V_CC = 10.;# in V\n",
+ "V_BB=5.;# in V\n",
+ "R_E = 2.;# in kohm\n",
+ "R_C = 3.;# in kohm\n",
+ "R_B= 50.;# in kohm\n",
+ "# Applying KVL to input loop\n",
+ "# V_BB= I_B*R_B+(1+Beta)*I_B*R_E+V_BEact or \n",
+ "I_B= (V_BB-V_BEact)/(R_B+(1.+Beta)*R_E);# in mA\n",
+ "# The collector current,\n",
+ "I_C= Beta*I_B;# in mA\n",
+ "# Applying KVL to collector circuit\n",
+ "# V_CC= I_Csat*R_C +V_CEsat +(I_C+I_B)*R_E\n",
+ "V_CEact= V_CC-I_B*R_E-I_C*(R_C+R_E);# in V\n",
+ "# The base current,\n",
+ "I_B= I_B*10.**3.;# in uA\n",
+ "print '%s %.f' %(\"The value of I_B in uA is : \",I_B)\n",
+ "print '%s %.1f' %(\"The value of I_C in mA is : \",I_C)\n",
+ "print '%s %.3f' %(\"The value of V_CE in volts is : \",V_CEact)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_B in uA is : 17\n",
+ "The value of I_C in mA is : 1.7\n",
+ "The value of V_CE in volts is : 1.434\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E25 - Pg 330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 6.25 - 330\n",
+ "import math \n",
+ "#Given data\n",
+ "Beta = 100.;\n",
+ "V_CEsat = 0.2;# in V\n",
+ "R_B = 150.;# in k ohm\n",
+ "R_C = 2.;# in k ohm\n",
+ "V_CC = 10.;# in V\n",
+ "V_BEsat = 0.8;# in V\n",
+ "I_B = (V_CC - V_BEsat)/R_B;# in mA\n",
+ "I_C = (V_CC - V_CEsat)/R_C;# in mA\n",
+ "I_Bmin = I_C/Beta;# in mA\n",
+ "I_B=I_B*10.**3.;# in uA\n",
+ "I_Bmin=I_Bmin*10.**3.;# in uA\n",
+ "if I_B>I_Bmin :\n",
+ " print '%s' %(\"I_B=61.33uA and I_B(min)=49uA.\\nSince the value of I_B ((I_B)uA) is greater than the value of I_Bmin ((I_Bmin)uA)\");\n",
+ " print '%s' %(\"So the transistor is in the saturation region.\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "I_B=61.33uA and I_B(min)=49uA.\n",
+ "Since the value of I_B ((I_B)uA) is greater than the value of I_Bmin ((I_Bmin)uA)\n",
+ "So the transistor is in the saturation region.\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E26 - Pg 331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 6.26 - 331\n",
+ "#Given data\n",
+ "Beta = 100.;\n",
+ "V_CE = 0.2;#in V\n",
+ "V_BE = 0.8;# in V\n",
+ "R_C= 500.;# in ohm\n",
+ "R_B= 44.*10.**3.;# in ohm\n",
+ "R_E= 1.*10.**3.;# in ohm\n",
+ "V_CC= 15.;# in V\n",
+ "V_GE= -15.;# in V\n",
+ "# Applying KVL to collector circuit, V_CC-V_GE - I_Csat*R_C-V_CE-I_E*R_E=0, but I_Csat= Beta*I_Bmin and I_E= 1+Beta\n",
+ "# Minimum value of base current,\n",
+ "I_Bmin= (V_CC-V_GE-V_CE)/(R_C*Beta+(1.+Beta)*R_E);# in A\n",
+ "# Applying KVL to the base emitter circuit, V_BB-I_Bmin*R_B-V_BE-I_E*R_E + V_CC=0\n",
+ "# The value of V_BB,\n",
+ "V_BB= I_Bmin*R_B + V_BE + (1.+Beta)*I_Bmin*R_E-V_CC;# in V\n",
+ "I_Bmin= I_Bmin*10.**3.;#in mA\n",
+ "print '%s %.3f' %(\"The value of I_B(min) in mA is : \",I_Bmin)\n",
+ "print '%s %.1f' %(\"The value of V_BB in volts is : \",V_BB)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_B(min) in mA is : 0.197\n",
+ "The value of V_BB in volts is : 14.4\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E27 - Pg 331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.27 - 331\n",
+ "# Given data\n",
+ "V_ECsat= 0.2;# in V\n",
+ "V_CC= 10.;# in V\n",
+ "V_EBsat= 0.8;# in V\n",
+ "\n",
+ "# Part (i)\n",
+ "Beta= 100.;\n",
+ "R_B= 220.;# in kohm\n",
+ "# Applying KVL to collector circuit, V_CC= V_EC+ICRC\n",
+ "ICRC= V_CC-V_ECsat;# in V\n",
+ "# Applying KVL to input loop, V_CC= V_EBsat+I_B*R_B (i)\n",
+ "I_B= (V_CC-V_EBsat)/R_B;# in mA\n",
+ "I_C= Beta*I_B;# in mA\n",
+ "R_Cmin= ICRC/I_C;# in kohm\n",
+ "print '%s %.3f' %(\"The minimum value of R_C in kohm is :\",R_Cmin)\n",
+ "# Part (ii)\n",
+ "R_C= 1.2;# in kohm\n",
+ "I_Csat= ICRC/R_C;# in mA\n",
+ "I_B= I_Csat/Beta;# in mA\n",
+ "# From eq (i)\n",
+ "R_B= (V_CC-V_EBsat)/I_B;# in kohm\n",
+ "print '%s %.2f' %(\"The maximum value of R_B in kohm is : \",R_B)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum value of R_C in kohm is : 2.343\n",
+ "The maximum value of R_B in kohm is : 112.65\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E28 - Pg 333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 6.28 - 333\n",
+ "# Given data\n",
+ "Beta= 100.;\n",
+ "V_BEsat= 0.8;# in V\n",
+ "V_CEsat= 0.2;# in V\n",
+ "V_BEact= 0.7;# in V\n",
+ "V_CC = 10.;# in V\n",
+ "R_E = 1.;# in kohm\n",
+ "R_C = 2.;# in kohm\n",
+ "R_B= 100.;# in kohm\n",
+ "Beta=100.;\n",
+ "alpha= Beta/(1.+Beta);\n",
+ "# Applying KVL to collector circuit\n",
+ "# V_CC= I_Csat*R_C +V_CE +R_E*I_E\n",
+ "# but I_E= alpha*I_Csat\n",
+ "I_Csat= (V_CC-V_CEsat)/(R_C+R_E*alpha);# in mA\n",
+ "I_Bmin= I_Csat/Beta;# in mA\n",
+ "# Applying KVL to base loop\n",
+ "# V_CC= I_B*R_B +V_BEsat +I_E*R_E\n",
+ "# but I_E= I_Csat+I_B\n",
+ "I_B= (V_CC-V_BEsat-I_Csat*R_E)/(R_B+R_E);# in mA\n",
+ "I_B=I_B*10.**3.;# in uA\n",
+ "print '%s %.2f' %(\"The value of I_B in uA is :\",I_B)\n",
+ "I_B=I_B*10.**-3.;# in mA\n",
+ "I_Bmin= I_Bmin*10.**3.;# in uA\n",
+ "print '%s %.1f' %(\"The minimum value of I_B in uA is :\",I_Bmin)\n",
+ "I_Bmin= I_Bmin*10.**-3.;# in mA\n",
+ "if I_B>I_Bmin :\n",
+ " print '%s' %(\"Since the value of I_B is greater than the value of I_Bmin, \")\n",
+ " print '%s' %(\"Hence the transistor is in saturation.\")\n",
+ "# The emitter current,\n",
+ "I_E= (1.+Beta)*I_Bmin;# in mA\n",
+ "# The value of R_E\n",
+ "R_E= (V_CC-V_BEact-I_Bmin*R_B)/I_E;# in kohm\n",
+ "print '%s %.3f' %(\"The value of R_E in kohm is : \",R_E)\n",
+ "print '%s' %(\"So R_E should be greater than this value in order to bring the transistor just out of saturation \")\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_B in uA is : 58.64\n",
+ "The minimum value of I_B in uA is : 32.8\n",
+ "Since the value of I_B is greater than the value of I_Bmin, \n",
+ "Hence the transistor is in saturation.\n",
+ "The value of R_E in kohm is : 1.819\n",
+ "So R_E should be greater than this value in order to bring the transistor just out of saturation \n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E29 - Pg 334"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.29 - 334\n",
+ "# Given data\n",
+ "V_CC = 9.;# in V\n",
+ "V_BE = 0.8;# in V\n",
+ "V_CE = 0.2;# in V\n",
+ "R_B = 50.;# in kohm\n",
+ "R_C=2.;# in kohm\n",
+ "R_E = 1.;# in kohm\n",
+ "Beta=70.;\n",
+ "# Applying KVL to input loop, V_CC= I_B*R_B +V_BE +I_E*R_E\n",
+ "# V_CC- V_BE= (R_B+R_E)*I_B + R_E*I_C (i)\n",
+ "# Applying KVL to output loop, V_CC= R_C*I_C +V_CE +I_C*R_E +I_B*R_E\n",
+ "#I_B = ((V_CC- V_CE)-(R_C+R_E)*I_C)/R_E (ii)\n",
+ "# From eq (i) and (ii)\n",
+ "I_C= ( (V_CC- V_BE)-(R_B+R_E)* (V_CC- V_CE)/R_E)/(1-(R_B+R_E)*(R_C+R_E));# in mA\n",
+ "I_B = ((V_CC- V_CE)-(R_C+R_E)*I_C)/R_E# in mA\n",
+ "I_Bmin= I_C/Beta;# in mA\n",
+ "if I_B>I_Bmin :\n",
+ " print '%s' %(\"I_B(min)=0.0414mA and I_B=0.106mA.\\nSince the value of I_B ((I_B)mA) is greater than the value of I_Bmin ((I_Bmin)mA)\")\n",
+ " print '%s' %(\"So the transistor is in saturation \")\n",
+ "V_C= V_CC-I_C*R_C;# in V\n",
+ "print '%s %.2f' %(\"The value of collector voltage in volts is : \",V_C)\n",
+ "Beta= I_C/I_B;\n",
+ "print '%s %.2f' %(\"The minimum value of beta that will change the state of the trasistor is : \",Beta)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "I_B(min)=0.0414mA and I_B=0.106mA.\n",
+ "Since the value of I_B ((I_B)mA) is greater than the value of I_Bmin ((I_Bmin)mA)\n",
+ "So the transistor is in saturation \n",
+ "The value of collector voltage in volts is : 3.20\n",
+ "The minimum value of beta that will change the state of the trasistor is : 27.89\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Devices_by_S._Sharma/Chapter07.ipynb b/Electronic_Devices_by_S._Sharma/Chapter07.ipynb new file mode 100644 index 00000000..08f06583 --- /dev/null +++ b/Electronic_Devices_by_S._Sharma/Chapter07.ipynb @@ -0,0 +1,156 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b8970a094fad8a547d8c45269d1db3cecaf14dd1d49918ae7c469ae888e5ccd7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 07 - OPTOELECTRONIC DEVICES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 - Pg 352"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.1 - 352\n",
+ "# Given data\n",
+ "O_V = 5.;# output voltage in V\n",
+ "V_D = 1.5;#voltage drop in V\n",
+ "R = (O_V - V_D)/O_V;\n",
+ "R = R * 10.**3.;# in ohm\n",
+ "print '%s %.2f' %(\"The resistance value in ohm is\",R);\n",
+ "print '%s' %(\"As this is not standard value, use R=680 ohm which is a standard value\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance value in ohm is 700.00\n",
+ "As this is not standard value, use R=680 ohm which is a standard value\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 - Pg 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.2 - 376\n",
+ "# Given data\n",
+ "import math \n",
+ "N_A = 7.5*10.**24.;# in atoms/m**3\n",
+ "N_D = 1.5*10.**22.;# in atoms/m**3\n",
+ "D_e = 25.*10.**-4.;# in m**2/s\n",
+ "D_h = 1.*10.**-3.;# in m**2/s\n",
+ "Torque_eo = 500.;# in ns\n",
+ "Torque_ho = 100.;# in ns\n",
+ "n_i = 1.5*10.**16.;# in /m**3\n",
+ "e = 1.6*10.**-19.;# in C\n",
+ "P_C = 12.5;# in mA/cm**2\n",
+ "# Electron diffusion length\n",
+ "L_e = math.sqrt(D_e*Torque_ho*10.**-9.);# in m\n",
+ "L_e = L_e * 10.**6.;# in um\n",
+ "# hole diffusion length\n",
+ "L_h = math.sqrt(D_h*Torque_ho*10.**-9.);# in m\n",
+ "L_h = L_h * 10.**6.;# in um\n",
+ "# The value of J_s can be calculated as,\n",
+ "J_s = e*((n_i)**2.)*( (D_e/(L_e*10.**-6.*N_A)) + (D_h/(L_h*10.**-6.*N_D)) );# in A/m**2\n",
+ "J_s = J_s * 10.**3.;# in A/cm**2\n",
+ "V_T = 26.;# in mV\n",
+ "I_lembda = 12.5*10.**-3.;\n",
+ "I_s = 2.4*10.**-4.;\n",
+ "# Open circuit voltage \n",
+ "V_OC = V_T*math.log( 1.+(I_lembda/J_s) );# in mV\n",
+ "V_OC = V_OC * 10.**-3.;# in V\n",
+ "print '%s %.3f' %(\"Open circuit voltage in V is\",V_OC);\n",
+ "\n",
+ "# Note: There is calculation error to evaluate the value of VOC since 26*10**-3*log(1+12.5*10**-3/2.4*10**-4) calculated as 0.10318 not 0.522 V\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Open circuit voltage in V is 0.103\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.3 - 376\n",
+ "# Given data\n",
+ "import math \n",
+ "Phi_o = 1.*10.**21.;# in m**-2s**-1\n",
+ "Alpha = 1.*10.**5.;# in m**-1\n",
+ "W = 25.;# in um\n",
+ "W =W * 10.**-6.;# in m\n",
+ "e = 1.6*10.**-19.;# in C\n",
+ "# At the front edge of intrinsic region, the generation rate of EHP\n",
+ "G_L1 = Alpha*Phi_o;# in m**-3s**-1\n",
+ "# At the back edge of intrinsic region, the generation rate of EHP\n",
+ "G_L2 = Alpha*Phi_o*math.e**( (-Alpha*W) );# in m**-3s**-1\n",
+ "# Photo current density,\n",
+ "J_L = e*Phi_o*(1-math.e**(-Alpha*W));# in A/m**2\n",
+ "J_L = J_L * 10.**-1.;# in mA/cm**2\n",
+ "print '%s %.2f' %(\"Photo current density in mA/cm**2 is\",J_L);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Photo current density in mA/cm**2 is 14.69\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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