{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 4 BJT Fundamentals"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4−1 in page 208"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(a)The value of the Base Current is 3.85e-04 A\n",
      "\n",
      "(b)The value of the Collector Current is  3.615e-03 A \n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Calculate Base and Collector Currents\n",
    "# Given Data\n",
    "alpha=0.90; # Current Gain in CB mode\n",
    "Ico=15*10**-6; # Reverse saturation Current in micro−A\n",
    "Ie=4*10**-3; # Emitter Current in mA\n",
    "# Calculations\n",
    "Ic=Ico+(alpha*Ie);\n",
    "Ib=Ie-Ic;\n",
    "print \"(a)The value of the Base Current is %0.2e A\\n\" %Ib;\n",
    "print \"(b)The value of the Collector Current is %0 .3e A \\n\" %Ic"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4−2 in page 209"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(a)The Current gain alpha for BJT is 0.989 \n",
      "\n",
      "(b)The value of the base Current is 4.44e-05 A\n",
      "\n",
      "(c)The value of the Emitter Current is 4.04e-03 A \n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Calculate alpha using beta\n",
    "# Given Data\n",
    "\n",
    "beta_bjt=90.; # beta gain for the BJT\n",
    "Ic=4*10**-3; # Collector Current in mA\n",
    "# Calculations\n",
    "alpha=beta_bjt/(1.+beta_bjt);\n",
    "Ib=Ic/beta_bjt;\n",
    "Ie=Ic+Ib;\n",
    "print \"(a)The Current gain alpha for BJT is %0.3f \\n\"%alpha\n",
    "print \"(b)The value of the base Current is %0.2e A\\n\"%Ib\n",
    "print \"(c)The value of the Emitter Current is %0.2e A \\n\"%Ie"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4−3 in page 20"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(a)The value of Current gain beta for BJT is 9 \n",
      "\n",
      "(b)The value of the Collector Current is 4.65e-03 A \n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Collector Current in C E mode\n",
    "# Given Data\n",
    "alpha=0.90; # Current Gain of BJT\n",
    "Ico=15*10**-6; # Reverse Saturation Current of BJT in micro−A\n",
    "Ib=0.5*10**-3; # Base Current in C−E mode in mA\n",
    "# Calculations\n",
    "beta_bjt=alpha/(1-alpha);\n",
    "Ic=(beta_bjt*Ib)+(beta_bjt+1)*Ico;\n",
    "print \"(a)The value of Current gain beta for BJT is %0.0f \\n\"%beta_bjt\n",
    "print \"(b)The value of the Collector Current is %0.2e A \\n\"%Ic"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4−4 in page 20"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The Current gain beta for the Device is 250 \n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Calculate beta for the BJT\n",
    "Ib=20*10**-6; # Base current in micro−A\n",
    "Ic=5*10**-3; # Collector Current in mA\n",
    "# Calculations\n",
    "beta_bjt=Ic/Ib;\n",
    "print \"The Current gain beta for the Device is %0.0f \\n\"%beta_bjt;\n",
    "# The Current Gain beta for the Device is 250"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4−5 in page 209"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(a)The value of the Emitter Current is 5.05e-03A \n",
      "\n",
      "(b)The value of beta gain of the BJT is 100 \n",
      "\n",
      "(c)The value of alpha gain of the BJT is 0.990 \n",
      "\n"
     ]
    }
   ],
   "source": [
    "#To Compute Alpha Beta and Emitter Current\n",
    "# Given Data\n",
    "Ib=50*10**-6; # Base Current in mu−A\n",
    "Ic=5*10**-3; # Collector Current in mA\n",
    "# Calculations\n",
    "Ie=Ic+Ib;\n",
    "beta_bjt=Ic/Ib;\n",
    "alpha=Ic/Ie;\n",
    "print \"(a)The value of the Emitter Current is %0.2eA \\n\"%Ie\n",
    "print \"(b)The value of beta gain of the BJT is %0.0f \\n\"%beta_bjt\n",
    "print \"(c)The value of alpha gain of the BJT is %0.3f \\n\"%alpha"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4−6 in page 210"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The value of inverse beta of the BJT is 1 \n",
      "\n",
      "The value of inverse alpha of the BJT is 2 \n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Calculate alpha reverse and beta reverse\n",
    "# Given Data\n",
    "Ie=10.*10**-3; # Emitter Current in mA\n",
    "Ib=5*10**-3; # Base Current in mu−A\n",
    "# Calculations\n",
    "Ic=Ie-Ib;\n",
    "beta_reverse=Ib/Ic;\n",
    "alpha_reverse=Ie/Ic;\n",
    "print \"The value of inverse beta of the BJT is %0.0f \\n\"%beta_reverse\n",
    "print \"The value of inverse alpha of the BJT is %0.0f \\n\"%alpha_reverse"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4−7 in page 210"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Circuit 1:\n",
      "(a)Emitter Current=9.30e-04 A\n",
      "(b)Base Current=9.21e-06 A\n",
      "(c)Collector Voltage=0.792 V\n",
      "\n",
      "\n",
      "Circuit 2:\n",
      "(a)Emitter Current=1.86e-03 A\n",
      "(b) Collector Current=1.842e-03 A\n",
      "(c)Collector Voltage=-5.700 V\n",
      "\n"
     ]
    }
   ],
   "source": [
    "# Calculate Labeled Currents and Voltages\n",
    "# Given Data\n",
    "beta_bjt=100.; # beta gain of BJT\n",
    "Vbe=0.7; # Base−Emitter voltage of BJT in V\n",
    "#Calculation\n",
    "Vcc1=10.;\n",
    "Vee1=-10.;\n",
    "Ve1=-0.7;\n",
    "R1=10*10**3;\n",
    "Ie1=(Vcc1-Vbe)/R1;\n",
    "Ib1=Ie1/(beta_bjt+1);\n",
    "Vc1=Vcc1-R1*(Ie1-Ib1);\n",
    "Vcc2=10.;\n",
    "Vee2=-15.;\n",
    "Ve2=-0.7;\n",
    "R2 =5*10**3;\n",
    "Ie2=(Vcc2-Vbe)/R2;\n",
    "Ic2=(beta_bjt/(beta_bjt+1.))*Ie2;\n",
    "Vc2=Vee2+R2*(Ie2);\n",
    "print \"Circuit 1:\\n(a)Emitter Current=%0.2e A\\n(b)Base Current=%0.2e A\\n(c)Collector Voltage=%0.3f V\\n\\n\"%(Ie1,Ib1,Vc1);\n",
    "print \"Circuit 2:\\n(a)Emitter Current=%0.2e A\\n(b) Collector Current=%0.3e A\\n(c)Collector Voltage=%0.3f V\\n\"%(Ie2,Ic2,Vc2);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4−8 in page 211"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Circuit 1:\n",
      "(a)Base Voltage = 0.0 V\n",
      "(b)Emitter Voltage = -0.7 V\n",
      "\n",
      "Circuit 2:\n",
      "(a)Emitter Voltage = 0.7 V\n",
      "(b) Collector Voltage = -5.7 V\n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Calculate labeled Voltages\n",
    "# Given Data\n",
    "Vbe=0.7; # Base−Emitter voltage of BJT in V\n",
    "Vcc2=10; # DC voltage across Collector in V\n",
    "Vee2=-15; # DC voltage across Emitter in V\n",
    "Rc2=5*10**3; # Collector Resistance in K−ohms\n",
    "# Beta Current Gain of BJT is Infinity\n",
    "# Calculations\n",
    "Vb1=0;\n",
    "Ve1=-0.7;\n",
    "Ve2=0.7;\n",
    "Vc2=Vee2+Rc2*((Vcc2-Vbe)/Rc2);\n",
    "print \"Circuit 1:\\n(a)Base Voltage = %0.1f V\\n(b)Emitter Voltage = %0.1f V\\n\"%(Vb1,Ve1);\n",
    "print \"Circuit 2:\\n(a)Emitter Voltage = %0.1f V\\n(b) Collector Voltage = %0.1f V\\n\"%(Ve2,Vc2);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4−9 in page 211"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Circuit Parameters:\n",
      "(a)Base Voltage = 0.300V\n",
      "(b)Base Current = 1.500e-05 A\n",
      "(c)Emitter Current= 8.000e-04 A\n",
      "(d)Collector Current = 7.850e-04 A\n",
      "(e) Collector Voltage = -1.075 V\n",
      "(f) beta gain = 52.333\n",
      "(g)alpha gain = 0.981\n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Calculating BJT parameters assuming Vbe\n",
    "# Given Data\n",
    "Ve=1.; # Emitter Voltage of BJT in V\n",
    "Vbe=0.7; # Base−Emitter Voltage of BJT in V\n",
    "Rb=20*10**3; # Base Resistance of Circuit in K−ohms\n",
    "Rc=5*10**3; # Collector Resistance of Circuit in K−ohms\n",
    "Re=5*10**3; # Emitter Resistance of Circuit in K−ohms\n",
    "Vcc=5.; # DC voltage across Collector in V\n",
    "Vee=-5; # DC voltage across Emitter in V\n",
    "# Calculations\n",
    "Vb=Ve-Vbe;\n",
    "Ib=Vb/Rb;\n",
    "Ie=(Vcc -1)/Re;\n",
    "Ic=Ie-Ib;\n",
    "Vc=(Rc*Ic)-Vcc;\n",
    "beta_bjt=Ic/Ib;\n",
    "alpha=Ic/Ie;\n",
    "print \"Circuit Parameters:\\n(a)Base Voltage = %0.3fV\\n(b)Base Current = %0.3e A\\n(c)Emitter Current= %0.3e A\\n(d)Collector Current = %0.3e A\\n(e) Collector Voltage = %0.3f V\\n(f) beta gain = %0.3f\\n(g)alpha gain = %0.3f\\n\"%(Vb,Ib,Ie,Ic,Vc, beta_bjt ,alpha);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4−10 in page 211"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(a)Change in Emitter voltage is +0.40 V\n",
      "\n",
      "(b)Change in Collector Voltage is 0.00 V\n",
      "\n"
     ]
    }
   ],
   "source": [
    "# Measurement of Circuit Voltage changes\n",
    "# Given Data\n",
    "Vb=-5; # Base Voltage of BJT in V\n",
    "Rc=1*10**3; # Collector Resistance in K−ohms\n",
    "Ie=2*10**-3; # Emitter Current of BJT in mA\n",
    "delB=+0.4; # Change in Base Voltage\n",
    "# Calculations\n",
    "delE =+0.4;\n",
    "delC=0;\n",
    "print \"(a)Change in Emitter voltage is +%0.2f V\\n\"%delE\n",
    "print \"(b)Change in Collector Voltage is %0.2f V\\n\"%delC"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4−11 in page 212"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Assume active mode for circuit 1\n",
      "(a)Ve = 1.30 V\n",
      "(b)Ic = 0.00e+00 A\n",
      "(c)Ve = 3.03 V\n",
      "\n",
      "Thus the circuit operates in an active mode\n",
      "\n",
      "\n",
      "For circuit 2,assume active mode\n",
      "\n",
      "(a)Ve = 1.7 V\n",
      "(b)Ie = 4.30e-04 A\n",
      "(c)Vc = 4.30 V\n",
      "\n",
      "This circuit operates in a saturated mode\n",
      "\n",
      "\n",
      "For circuit 3,assume active mode\n",
      "\n",
      "(a)Ve = -4.3 V\n",
      "(b)Ie = 6.9000e-05 A\n",
      "(c)Ic = 0.000e+00 A\n",
      "(d)Vc = -40.2 V\n",
      "\n",
      "The circuit operates in an active mode\n",
      "\n",
      "\n",
      "For circuit 4,assume active mode\n",
      "\n",
      "(a)Ie = 1.86e-03 A\n",
      "(b)Vc = -10.00 V\n",
      "\n",
      "The circuit operates in an active mode\n"
     ]
    }
   ],
   "source": [
    "# Determine mode of operation of BJT\n",
    "# Given Data\n",
    "Vbe=0.7; # Base−Emitter Voltage in V\n",
    "beta_bjt=100; # beta gain of BJ\n",
    "# Calculation\n",
    "print \"Assume active mode for circuit 1\"\n",
    "Vb1=2;\n",
    "Ve_1=Vb1-Vbe;\n",
    "Ie1 =1*10** -3;\n",
    "Ic1=Ie1*(beta_bjt/(1+beta_bjt));\n",
    "Ve1=6-(3*0.99);\n",
    "print \"(a)Ve = %0.2f V\\n(b)Ic = %0.2e A\\n(c)Ve = %0.2f V\\n\"%(Ve_1,Ic1,Ve1);\n",
    "print \"Thus the circuit operates in an active mode\\n\\n\"\n",
    "print \"For circuit 2,assume active mode\\n\"\n",
    "Vcc=1;\n",
    "Ve2=Vcc+Vbe;\n",
    "Ie2=(6-Ve2)/(10*10**3);\n",
    "Vc=0+(10*0.43);\n",
    "print \"(a)Ve = %0.1f V\\n(b)Ie = %0.2e A\\n(c)Vc = %0.2f V\\n\"%(Ve2,Ie2,Vc);\n",
    "print \"This circuit operates in a saturated mode\\n\\n\"\n",
    "print \"For circuit 3,assume active mode\\n\"\n",
    "Ve3=-5+Vbe;\n",
    "Ie3=(9.5-Ve3)/(200*10**3);\n",
    "Ic=Ie3*(beta_bjt/(1+beta_bjt));\n",
    "Vc3=-50+(0.492*20);\n",
    "print \"(a)Ve = %0.1f V\\n(b)Ie = %0.4e A\\n(c)Ic = %0.3e A\\n(d)Vc = %0.1f V\\n\"%(Ve3,Ie3,Ic,Vc3);\n",
    "print \"The circuit operates in an active mode\\n\\n\"\n",
    "print \"For circuit 4,assume active mode\\n\"\n",
    "Ve4 = -20.7;\n",
    "Ie4=(30+Ve4)/(5*10**3);\n",
    "Vc4=(-Ie4*(beta_bjt/(1+beta_bjt))*(2*10**3))-10;\n",
    "print \"(a)Ie = %0.2e A\\n(b)Vc = %0.2f V\\n\"%(Ie4,Vc4)\n",
    "print \"The circuit operates in an active mode\""
   ]
  }
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