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{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 10 - Other Power Amplifiers"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 10.1 Page No 425"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The value of P_DQ = 11.25 mW\n",
      "The value of P_Dmax = 112.50 mW\n",
      "The value of P_Lmax = 562.50 mW\n"
     ]
    }
   ],
   "source": [
    "# given data\n",
    "V_CEQ= 7.5##  V\n",
    "R_L= 50##  Ω\n",
    "I_Csat= V_CEQ/R_L##  A\n",
    "I_CQ= 0.01*I_Csat##  A\n",
    "P_DQ= V_CEQ*I_CQ##  W\n",
    "PP= 2*V_CEQ##  V\n",
    "P_Dmax= PP**2/(40*R_L)##  W\n",
    "P_Lmax= PP**2/(8*R_L)##  W\n",
    "# The value of P_DQ \n",
    "P_DQ= P_DQ*10**3##  mW\n",
    "# The value of P_Dmax \n",
    "P_Dmax= P_Dmax*10**3##  mW\n",
    "# The value of P_Lmax \n",
    "P_Lmax= P_Lmax*10**3##  mW\n",
    "print \"The value of P_DQ = %.2f mW\"%P_DQ\n",
    "print \"The value of P_Dmax = %.2f mW\"%P_Dmax\n",
    "print \"The value of P_Lmax = %.2f mW\"%P_Lmax"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 10.2 Page No 425"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The efficiency of amplifier = 74.03 %\n"
     ]
    }
   ],
   "source": [
    "# given data\n",
    "V_CC= 15##  V\n",
    "I_Csat= 150##  mA\n",
    "P_Lmax= 563##  mW\n",
    "I= 0.02*I_Csat##  mA\n",
    "Idc= 0.318*I_Csat##  mA\n",
    "I_CC= I+Idc##  mA\n",
    "P_CC= V_CC*I_CC##  mW\n",
    "# The efficiency of amplifier \n",
    "Eta= P_Lmax/P_CC*100##  %\n",
    "print \"The efficiency of amplifier = %.2f %%\"%Eta\n",
    "\n",
    "# Note: The answer in the book is not accurate"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 10.3 Page No 426"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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CWA/c3nX7jua+bgW8NMl3klyV5OgWxyNpzFgL42XvFp+7l92Cvg0cWlW7kpwEXAEctdiC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\nJwRJEuCEIElqOCFIkgD4/2yFPh7tHE6jAAAAAElFTkSuQmCC\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x7f863864ffd0>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "AC load line shown in figure\n"
     ]
    }
   ],
   "source": [
    "from numpy import arange\n",
    "%matplotlib inline\n",
    "from matplotlib import pyplot as plt\n",
    "# given data\n",
    "V_CC= 40.0##  V\n",
    "V_CEQ= 20.0##  V\n",
    "R_L= 10.0##  Ω\n",
    "I_Csat= V_CEQ/R_L##  A\n",
    "V_CEcutoff= V_CEQ##  V\n",
    "V_CE= arange(0,0.1+V_CEcutoff,0.1) #  V\n",
    "I_C= (V_CEQ-V_CE)/R_L##  A\n",
    "# The plot of ac load line,\n",
    "plt.plot(V_CE,I_C)\n",
    "plt.xlabel(\"VCE in volts\")\n",
    "plt.ylabel(\"IC in A\")\n",
    "plt.title(\"AC load line\")\n",
    "plt.show()\n",
    "print \"AC load line shown in figure\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 10.4 Page No 427"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The value of P_DQ = 0.39 W\n",
      "The value of P_Lmax = 20.00 W\n",
      "The value of P_Dmax = 4.00 W\n"
     ]
    }
   ],
   "source": [
    "# given data\n",
    "V_CC= 40##  V\n",
    "V_BE= 0.7##  V\n",
    "R= 1*10**3##  Ω\n",
    "R_L= 10##  Ω\n",
    "V_CEQ= 20##  V\n",
    "I_CQ= (V_CC-2*V_BE)/(2*R)##  A\n",
    "# The value of P_DQ\n",
    "P_DQ= V_CEQ*I_CQ##  W\n",
    "print \"The value of P_DQ = %.2f W\"%P_DQ\n",
    "PP= 2*V_CEQ##  V\n",
    "# The value of P_Lmax\n",
    "P_Lmax= PP**2/(8*R_L)##  W\n",
    "# The value of P_Dmax\n",
    "P_Dmax= PP**2/(40*R_L)##  W\n",
    "print \"The value of P_Lmax = %.2f W\"%P_Lmax\n",
    "print \"The value of P_Dmax = %.2f W\"%P_Dmax"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 10.5 Page No 428"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The voltage gain of the driver stage = 9.36\n",
      "On ignoring the value of Zin and r'e, the voltage gain = 10.00\n"
     ]
    }
   ],
   "source": [
    "# given data\n",
    "V_E= 1.43##  V\n",
    "R_E= 100##  Ω\n",
    "R_L= 100##  Ω\n",
    "R_C= 1*10**3##  Ω\n",
    "bita= 200#\n",
    "Vt= 25*10**-3##  V\n",
    "I_E= V_E/R_E##  A\n",
    "I_CQ= I_E##  A\n",
    "Zin= bita*R_L##  Ω\n",
    "r_desh_e= Vt/I_CQ##  Ω\n",
    "# The voltage gain of the driver stage \n",
    "A= (R_C*Zin/(R_C+Zin))/(R_E+r_desh_e)#\n",
    "print \"The voltage gain of the driver stage = %.2f\"%A\n",
    "# On ignoring Zin and r_desh_e,\n",
    "A= R_C/R_E#\n",
    "print \"On ignoring the value of Zin and r'e, the voltage gain = %.2f\"%A"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 10.6 Page No 429"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The value of PP = 30.00 volts\n",
      "The value of P_Lmax = 1.12 W\n"
     ]
    }
   ],
   "source": [
    "# given data\n",
    "V_CC= 30.0##  V\n",
    "PP= V_CC##  V\n",
    "R_L= 100.0##  Ω\n",
    "# The value of P_Lmax \n",
    "P_Lmax= PP**2/(8*R_L)##  W\n",
    "print \"The value of PP = %.2f volts\"%PP\n",
    "print \"The value of P_Lmax = %.2f W\"%P_Lmax"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 10.7 Page No 430"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The overall voltage gain = 2000.00\n"
     ]
    }
   ],
   "source": [
    "# given data\n",
    "R_C= 1*10**3##  Ω\n",
    "r_desh_e= 2.5##in Ω\n",
    "Zin= 1.0*10**3##  Ω\n",
    "A2= 10## unit less\n",
    "A3= 1## unit less\n",
    "A1= (R_C*Zin/(R_C+Zin))/r_desh_e## unit less\n",
    "# The overall voltage gain \n",
    "A= A1*A2*A3#\n",
    "print \"The overall voltage gain = %.2f\"%A"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 10.8 Page No 431"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The minimum base current that produces saturation = 108.89 mA\n"
     ]
    }
   ],
   "source": [
    "# given data\n",
    "V_CC= 50.0##  V\n",
    "V_CEsat= 1.0##  V\n",
    "R_L= 5##  Ω\n",
    "bita_dc= 90## unit less\n",
    "I_Csat= (V_CC-V_CEsat)/R_L##  A\n",
    "# The minimum base current that produces saturation \n",
    "I_Bsat= I_Csat/bita_dc##  A\n",
    "I_Bsat= I_Bsat*10**3##  mA\n",
    "print \"The minimum base current that produces saturation = %.2f mA\"%I_Bsat"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 10.9 Page No 432"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The input voltage = 2.85 volts\n"
     ]
    }
   ],
   "source": [
    "# given data\n",
    "I_Csat= 109*10**-3##  A\n",
    "bita_dc= 200#\n",
    "R_B= 1*10**3##  Ω\n",
    "V_BE1= 0.7##  V\n",
    "V_BE2= 1.6##  V\n",
    "# The base current,\n",
    "I_Bsat= I_Csat/bita_dc##  A\n",
    "# The input voltage\n",
    "Vin= I_Bsat*R_B+V_BE1+V_BE2##  V\n",
    "print \"The input voltage = %.2f volts\"%Vin"
   ]
  }
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