path: root/Applied_Physics_for_Engineers/Chapter_14.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 14: Solid State Electronics"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.1, Page 718"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "e = 1.6e-019;   # Charge on an electron, C\n",
      "mu_h = 0.048;  # Mobility of holes, metre square/volt-s\n",
      "mu_e = 0.135;  # Mobility of electrons, metre square/volt-s \n",
      "\n",
      "#Calculations\n",
      "# For P-type semiconductor\n",
      "rho_p = 1e-01;  # Resistivity of P type silicon, omh-m\n",
      "# As rho_p = 1/(e*N_a*mu_h), solving for N_a\n",
      "N_a = 1/(e*rho_p*mu_h); # Density of acceptor atoms, per metre cube\n",
      "# For N-type semiconductor\n",
      "rho_n = 1e-01;  # Resistivity of N type silicon, omh-m\n",
      "# As rho_n = 1/(e*N_d*mu_h), solving for N_d\n",
      "N_d = 1/(e*rho_n*mu_e); # Density of donor atoms, per metre cube\n",
      "\n",
      "#Results\n",
      "print \"Density of acceptor atoms = %4.2e per metre cube\"%N_a\n",
      "print \"Density of donor atoms = %4.2e per metre cube\"%N_d   #incorrect answer in textbook\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Density of acceptor atoms = 1.30e+21 per metre cube\n",
        "Density of donor atoms = 4.63e+20 per metre cube\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.2, Page 718"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "e = 1.6e-019;   # Charge on an electron, C\n",
      "mu_e = 0.36;    # Mobility of an electron, metre square/V-s\n",
      "mu_h = 0.17;    # Mobility of a hole, metre square/V-s\n",
      "n_i = 2.5e+018; # Intrinsic concentration of Ge sample, per metre cube\n",
      "\n",
      "#Calculations\n",
      "sigma = e*n_i*(mu_h+mu_e);  # Electrical conductivity of Ge sample, mho per metre\n",
      "rho = 1/sigma;  # Electrical resistivity of Ge, ohm-m\n",
      "\n",
      "#Results\n",
      "print \"The electrical conductivity of intrinsic germanium sample = %5.3f mho/m\"%sigma\n",
      "print \"The electrical resistivity of intrinsic germanium sample = %3.1f ohm-m\"%rho\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The electrical conductivity of intrinsic germanium sample = 0.212 mho/m\n",
        "The electrical resistivity of intrinsic germanium sample = 4.7 ohm-m\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.3, Page 719"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "e = 1.6e-019;   # Charge on an electron, C\n",
      "mu_e = 0.13;    # Mobility of an electron, metre square/V-s\n",
      "mu_h = 0.05;    # Mobility of a hole, metre square/V-s\n",
      "n_i = 1.5e+016; # Intrinsic concentration of Si, per metre cube\n",
      "\n",
      "#Calculations\n",
      "# Pure Si\n",
      "sigma = e*n_i*(mu_h+mu_e);  # Electrical conductivity of Si, mho per metre\n",
      "# Pure Si doped with donor impurity\n",
      "n_e = 5e+028/1e+09;     # Concentration of electrons, per metre cube\n",
      "sigma_n = e*n_e*mu_e;   # Electrical conductivity of Si doped with donor impurity, mho per metre\n",
      "# Pure Si doped with acceptor impurity\n",
      "n_h = 5e+028/1e+09;     # Concentration of holes, per metre cube\n",
      "sigma_p = e*n_h*mu_h;   # Electrical conductivity of Si doped with acceptor impurity, mho per metre\n",
      "\n",
      "#Results\n",
      "print \"The electrical conductivity of pure Si = %4.2e mho/m\"%sigma\n",
      "print \"The electrical conductivity of Si doped with donor impurity = %4.2f mho/m\"%sigma_n\n",
      "print \"The electrical conductivity of Si doped with acceptor impurity= %4.2f mho/m\"%sigma_p\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The electrical conductivity of pure Si = 4.32e-04 mho/m\n",
        "The electrical conductivity of Si doped with donor impurity = 1.04 mho/m\n",
        "The electrical conductivity of Si doped with acceptor impurity= 0.40 mho/m\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.4, Page 720"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import *\n",
      "\n",
      "#Variable declaration\n",
      "Nd = 1;     # For simplicity assume donor concentration to be unity, per metre cube\n",
      "Nd_prime = 3*Nd;   # Thrice the donor concentration, per metre cube\n",
      "dE_CF1 = 0.5;  # Energy difference between normal Fermi level and conduction level, eV\n",
      "k_BT = 0.03;    # Thermal energy at room temperature, eV\n",
      "\n",
      "#Calculations\n",
      "# As Nd_prime/Nd = exp((dE_CF1 - dE_CF2))/k_BT, solving for dE_CF2\n",
      "dE_CF2 = dE_CF1-k_BT*log(Nd_prime/Nd);  # Energy difference between new postion of Fermi level and conduction level, eV\n",
      "\n",
      "#Result\n",
      "print \"The new postion of Fermi level when donor concentration is trebled = %5.3f eV\"%dE_CF2\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The new postion of Fermi level when donor concentration is trebled = 0.467 eV\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.5, Page 721"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import *\n",
      "\n",
      "#Variable declaration\n",
      "e = 1.6e-019;   # Charge on an electron, C\n",
      "T = 300;    # Room temperature, K\n",
      "J0 = 300e-03;   # Saturation current density of the pn junction diode, A/metre square\n",
      "J = 1e+05;  # Forward current density of pn junction diode, A/metre square\n",
      "k_B = 1.38e-023;    # Boltzmann constant, J/K\n",
      "eta = 1;    # Ideality factor for Ge diode\n",
      "\n",
      "#Calculations\n",
      "# As J = J0*exp(e*V/(eta*k_B*T)), solving for V\n",
      "V = eta*k_B*T/e*log(J/J0);   # Voltage required to cause a forward current density in pn junction diode, volt\n",
      "\n",
      "#Results\n",
      "print \"The voltage required to cause a forward current density in pn junction diode = %5.3f V\"%V\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The voltage required to cause a forward current density in pn junction diode = 0.329 V\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.6, Page 721"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import *\n",
      "\n",
      "#Variable declaration\n",
      "e = 1.6e-019;   # Charge on an electron, C\n",
      "T = 300;    # Room temperature, K\n",
      "J0 = 200e-03;   # Saturation current density of the pn junction diode, A/metre square\n",
      "J = 5e+04;  # Forward current density of pn junction diode, A/metre square\n",
      "k_B = 1.38e-023;    # Boltzmann constant, J/K\n",
      "eta = 1;    # Ideality factor for Ge diode\n",
      "\n",
      "#Calculations\n",
      "# As J = J0*exp(e*V/(eta*k_B*T)), solving for V\n",
      "V = eta*k_B*T/e*log(J/J0);   # Voltage required to cause a forward current density in pn junction diode, volt\n",
      "\n",
      "#Result\n",
      "print \"The voltage required to cause a forward current density in pn junction diode = %5.3f V\"%V\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The voltage required to cause a forward current density in pn junction diode = 0.322 V\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.7, Page 722"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import *\n",
      "\n",
      "#Variable declaration\n",
      "e = 1.6e-019;   # Charge on an electron, C\n",
      "T = 300;    # Room temperature, K\n",
      "J0 = 300e-03;   # Saturation current density of the pn junction diode, A/metre square\n",
      "J = 1e+05;  # Forward current density of pn junction diode, A/metre square\n",
      "k_B = 1.38e-023;    # Boltzmann constant, J/K\n",
      "eta = 2;    # Ideality factor for Ge diode\n",
      "\n",
      "#Calculations\n",
      "# As J = J0*exp(e*V/(eta*k_B*T)), solving for V\n",
      "V = eta*k_B*T/e*log(J/J0);   # Voltage required to cause a forward current density in pn junction diode, volt\n",
      "\n",
      "#Result\n",
      "print \"The voltage required to cause a forward current density in Si iode = %5.3f V\"%V\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The voltage required to cause a forward current density in Si iode = 0.658 V\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.8, Page 723"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "I = 55e-03;     # Forward current through Si diode, A\n",
      "V = 3;          # Forward bias across Si diode, V\n",
      "eta = 2;        # Ideality factor for Si diode\n",
      "\n",
      "#Calculations\n",
      "R_dc = V/I;     # Static diode resistance, ohm\n",
      "R_ac = 0.026*eta/I;    # Dynamic diode resistance, ohm\n",
      "\n",
      "#Results\n",
      "print \"The static diode resistance = %4.1f ohm\"%R_dc\n",
      "print \"The dynamic diode resistance = %5.3f ohm\"%R_ac\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The static diode resistance = 54.5 ohm\n",
        "The dynamic diode resistance = 0.945 ohm\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.9, Page 723"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import *\n",
      "\n",
      "#Variable declaration\n",
      "R_L = 1000; # Load resistance across HWR, ohm\n",
      "V_rms = 200;    # Rms value of voltage supply, V\n",
      "\n",
      "#Calculations\n",
      "V0 = sqrt(2)*V_rms; # Peak value of voltage, V\n",
      "I0 = V0/(R_L*1e-03);    # Peak value of current, mA\n",
      "I_dc = I0/pi;  # Average value of current, mA\n",
      "I_rms = I0/2;   # Rms value of current, mA\n",
      "V_dc = I_dc*R_L/1e+03;    # Dc output voltage, V\n",
      "PIV = V0;   # Peak inverse voltage, V\n",
      "\n",
      "#Results\n",
      "print \"The average value of current = %2d mA\"%I_dc\n",
      "print \"The rms value of current = %5.1f mA\"%I_rms\n",
      "print \"The dc output voltage = %2d V\"%(V_dc/1)\n",
      "print \"PIV = %5.1f V\"%PIV\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The average value of current = 90 mA\n",
        "The rms value of current = 141.4 mA\n",
        "The dc output voltage = 90 V\n",
        "PIV = 282.8 V\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.10, Page 724"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import *\n",
      "\n",
      "#Variable declaration\n",
      "R_L = 980; # Load resistance across FWR, ohm\n",
      "R_F = 20.;   # Internal resistance of two crystal diodes in FWR, ohm\n",
      "V_rms = 50;    # Rms value of voltage supply, V\n",
      "\n",
      "#Calculations\n",
      "V0 = sqrt(2)*V_rms; # Peak value of voltage, V\n",
      "I0 = V0/((R_L+R_F)*1e-03);    # Peak value of current, mA\n",
      "I_dc = 2*I0/pi;  # Average value of current, mA\n",
      "I_rms = I0/sqrt(2);   # Rms value of current, mA\n",
      "V_dc = I_dc*R_L/1e+03;    # Dc output voltage, V\n",
      "eta = 81.2/(1+R_F/R_L);     # Rectification efficiency\n",
      "PIV = 2*V0;   # Peak inverse voltage, V\n",
      "\n",
      "#Results\n",
      "print \"The average value of current = %2d mA\"%I_dc\n",
      "print \"The rms value of current = %2d mA\"%I_rms\n",
      "print \"The dc output voltage = %4.1f V\"%(V_dc/1)\n",
      "print \"The rectification efficiency = %4.1f percent\"%eta\n",
      "print \"PIV = %5.1f V\"%PIV\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The average value of current = 45 mA\n",
        "The rms value of current = 50 mA\n",
        "The dc output voltage = 44.1 V\n",
        "The rectification efficiency = 79.6 percent\n",
        "PIV = 141.4 V\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.11, Page 725"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "delta_IC = 1e-03;   # Change in collector current, A\n",
      "delta_IB = 50e-06;   # Change in base current, A\n",
      "\n",
      "#Calculations\n",
      "bta = delta_IC/delta_IB;  # Base current amplification factor\n",
      "alpha = bta/(1+bta);    # Emitter current amplification factor\n",
      "\n",
      "#Results\n",
      "print \"Alpha of BJT = %4.2f\"%alpha\n",
      "print \"Beta of BJT = %2d\"%bta\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Alpha of BJT = 0.95\n",
        "Beta of BJT = 20\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.12, Page 725"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "I_E = 2;    # Emitter current, mA\n",
      "alpha = 0.88;   # Emitter current amplification factor\n",
      "\n",
      "#Calculations\n",
      "I_C = alpha*I_E;    # Collector current, mA\n",
      "I_B = I_E - I_C;        # Base current of BJT in CB mode, mA\n",
      "\n",
      "#Result\n",
      "print \"The base current of BJT in CB mode = %4.2f mA\"%I_B\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The base current of BJT in CB mode = 0.24 mA\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.13, Page 725"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "I_CBO = 12.5e-03;   # Reverse saturation current, mA\n",
      "I_E = 2;    # Emitter current, mA\n",
      "I_C = 1.97; # Collector current, mA\n",
      "\n",
      "#Calculations\n",
      "# As I_C = alpha*I_E+I_CBO, solving for alpha\n",
      "alpha = (I_C - I_CBO)/I_E;  # Emitter current gain\n",
      "I_B = I_E - I_C;    # Base current, mA\n",
      "\n",
      "#Results\n",
      "print \"The emitter current gain = %5.3f\"%alpha\n",
      "print \"The base current = %4.2f mA\"%I_B\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The emitter current gain = 0.979\n",
        "The base current = 0.03 mA\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.14, Page 726"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "alpha = 0.98;   # Emitter current amplification factor\n",
      "I_CBO = 5e-06;  # Reverse saturation current, A\n",
      "\n",
      "#Calculations\n",
      "bta = alpha/(1-alpha);  # Emitter current amplification factor\n",
      "I_CEO = 1/(1-alpha)*I_CBO;  # Leakage current of BJT in CE mode, mA\n",
      "\n",
      "#Results\n",
      "print \"The base current gain = %2g\"%bta\n",
      "print \"The leakage current of BJT in CE mode = %4.2f mA\"%(I_CEO/1e-03)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The base current gain = 49\n",
        "The leakage current of BJT in CE mode = 0.25 mA\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.15, Page 726"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "R_i = 50;       # Dynamic input resistance of PNP transistor, ohm\n",
      "R_L = 5e+03;    # Load resistance in collector circuit, ohm\n",
      "alpha = 0.96;   # Emitter current amplification factor\n",
      "\n",
      "#Calculations\n",
      "A_v = alpha*R_L/R_i;    # Voltage gain\n",
      "A_p = alpha*A_v;    # Power gain\n",
      "\n",
      "#Results\n",
      "print \"The voltage gain = %2g\"%A_v\n",
      "print \"The power gain = %2d\"%A_p\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The voltage gain = 96\n",
        "The power gain = 92\n"
       ]
      }
     ],
     "prompt_number": 16
    }
   ],
   "metadata": {}
  }
 ]
}