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   "source": [
    "# Chapter 1: Introduction to Operational Amplifiers"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
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   },
   "source": [
    "## Example 1.1_a"
   ]
  },
  {
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   "execution_count": 2,
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   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Collector current Ic1 is 0.39 mA\n",
      "Voltage Vc1 is 3.38 V\n",
      "Voltage Ve4 is 2.68 V\n",
      "Current Ie4 is 0.297 mA\n",
      "Current Ic5 is 0.297 mA\n",
      "Voltage Vc5 is 4.87 V\n",
      "Voltage Ve6 is 4.17 V\n",
      "Current Ie6 is 0.678 mA\n",
      "Voltage Ve7 is 4.87 V\n",
      "Current I1 is 2.82 mA\n",
      "Current Ie8 is 2.82 mA\n",
      "Voltage Ve8 at the output terminal is -0.35 V\n"
     ]
    }
   ],
   "source": [
    "\n",
    "\n",
    "#Example 1.1_a\n",
    "#The equivalent circuit of the Motorola op-amp MC 1435 is shown in Figure.No-1.2\n",
    "#Determine the collector current in each transistor and the dc voltage at the\n",
    "#output terminal\n",
    "\n",
    "#Variable declaration\n",
    "Vcc=6                 #Voltage in volts\n",
    "Vbe5=0.7              #Voltage in volts\n",
    "Vee=6                 #Voltage in volts\n",
    "Vbe3=6.7              #Voltage in volts\n",
    "Vbe6=0.7              #Voltage in volts\n",
    "Vbe7=0.7              #Voltage in volts\n",
    "Rc1=6.7*10**3         #Resistance in ohms\n",
    "Ic1=0                 #initialization\n",
    "\n",
    "#Calculation\n",
    "Vc1=Vcc-Rc1*Ic1\n",
    "Ve4=Vc1-Vbe5\n",
    "I4=(Ve4+Vee)/(9.1*10**3+5.5*10**3)\n",
    "Vb3=5.5*10**3*I4-Vee\n",
    "Ve3=Vb3-Vbe3\n",
    "Ie3=(Ve3+Vbe3)/3.3*10**3\n",
    "Ic1=1.08*10**-3/2.765 #Since Ie3=2*Ic1\n",
    "Vc1=Vcc-Rc1*Ic1\n",
    "Ve4=Vc1-Vbe5\n",
    "Ie4=(Ve4+Vee)/(29.2*10**3)\n",
    "Ic5=Ie4\n",
    "Vc5=Vcc-3.8*10**3*Ic5\n",
    "Ve6=Vc5-Vbe6\n",
    "Ie6=(Ve6+Vee)/(15*10**3)\n",
    "Ve7=Ve6+Vbe7\n",
    "I1=(Vcc-Ve7)/400\n",
    "Ie8=I1\n",
    "Ve8=-Vee+2*10**3*Ie8\n",
    "\n",
    "#Result\n",
    "print \"Collector current Ic1 is\",round(Ic1*10**3,2),\"mA\"\n",
    "print \"Voltage Vc1 is\",round(Vc1,2),\"V\" \n",
    "print \"Voltage Ve4 is\",round(Ve4,2),\"V\"\n",
    "print \"Current Ie4 is\",round(Ie4*10**3,3),\"mA\"\n",
    "print \"Current Ic5 is\",round(Ic5*10**3,3),\"mA\"\n",
    "print \"Voltage Vc5 is\",round(Vc5,2),\"V\"\n",
    "print \"Voltage Ve6 is\",round(Ve6,2),\"V\"\n",
    "print \"Current Ie6 is\",round(Ie6*10**3,3),\"mA\"\n",
    "print \"Voltage Ve7 is\",round(Ve7,2),\"V\"\n",
    "print \"Current I1 is\",round(I1*10**3,2),\"mA\"\n",
    "print \"Current Ie8 is\",round(Ie8*10**3,2),\"mA\"\n",
    "print \"Voltage Ve8 at the output terminal is\",round(Ve8,2),\"V\"\n"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 1.1_b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Voltage gain of the dual-input,balanced output-differential amplifier is 82.55\n",
      "Voltage gain of the dual-input,unbalanced output-differential amplifier is 22.6\n",
      "Overall gain of the op-amp is 1866.34\n"
     ]
    }
   ],
   "source": [
    "\n",
    "\n",
    "#Example 1.1_b, Figure.No-1.2\n",
    "#Calculate the Voltage gain of the opamp\n",
    "\n",
    "#Variable decclaration\n",
    "Ie1=0.39*10**-3    #Current in amps\n",
    "Ie4=0.298*10**-3   #Current in amps\n",
    "Ie6=0.678*10**-3   #Current in amps\n",
    "Rc1=6.7*10**3      #Resistance in ohms\n",
    "Rc5=3.8*10**3      #Resistance in ohms\n",
    "beta_ac=150\n",
    "\n",
    "#Calculation\n",
    "re1=(25*10**-3)/Ie1\n",
    "re2=re1\n",
    "re4=(25*10**-3)/Ie4\n",
    "re5=re4\n",
    "re6=(25*10**-3)/Ie6\n",
    "k=(Rc1*2*beta_ac*re4)/(Rc1+2*beta_ac*re4)\n",
    "Ad1=k/re1\n",
    "k1=(Rc5*beta_ac*(re6+15*10**3))/(Rc5+beta_ac*(re6+15*10**3))\n",
    "Ad2=k1/(2*re5)\n",
    "Ad=Ad1*Ad2\n",
    "\n",
    "#Result\n",
    "print \"Voltage gain of the dual-input,balanced output-differential amplifier is\",round(Ad1,2)\n",
    "print \"Voltage gain of the dual-input,unbalanced output-differential amplifier is\",round(Ad2,1)\n",
    "print \"Overall gain of the op-amp is\",round(Ad,2)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 1.1_c"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Input resistance Ri is 19.23 kilo ohms\n"
     ]
    }
   ],
   "source": [
    "\n",
    "#Example 1.1_c, Figure.No-1.2\n",
    "#Determine the Input resistance of the opamp\n",
    "\n",
    "#Variable declaration\n",
    "beta_ac=150\n",
    "re1=64.1         #Resistance in ohms\n",
    "\n",
    "#calculation\n",
    "Ri=2*beta_ac*re1\n",
    "\n",
    "#result\n",
    "print \"Input resistance Ri is\",round(Ri/10**3,2),\"kilo ohms\"\n"
   ]
  }
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