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+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3942a48c51f6e66ddeb010a1a5acaeebd4c9f56b964a269d92588d67ccc10453"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Introduction to operational amplifier"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex 2.1 Page no. 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "\n",
+      "R1 = 10*10**3      #R1 input resistance \n",
+      "Rf = 100*10**3     # Rf feedback resistance\n",
+      "vi = float(1)          #input voltage  \n",
+      "RL = 25*10**3\n",
+      "#calculating the values \n",
+      "\n",
+      "i1 = float((vi/R1)*10**3)                 # input resistace id the ratio of input voltage to the input resitance \n",
+      "vo = float(-(Rf/R1)*vi)            # finding the output voltage \n",
+      "iL = float((abs(vo)/RL)*10**3)                  #  calculating the load current \n",
+      "io = float((i1+iL))                 # calculating the output current which is equal to the sum of input current and load current\n",
+      "\n",
+      "#printing the values \n",
+      "\n",
+      "print \"The input current i1  =\",i1,\"mA\"\n",
+      "print \"The output voltage vo =\",vo,\"V\"\n",
+      "print \"The load current iL   =\",iL,\"mA\"\n",
+      "print \"The output current io =\",io,\"mA\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The input current i1  = 0.1 mA\n",
+        "The output voltage vo = -10.0 V\n",
+        "The load current iL   = 0.4 mA\n",
+        "The output current io = 0.5 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex 2.2 Page no.89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "\n",
+      "ACL = 5     # Gain of the amplifier\n",
+      "R1 = 10*10**3 # input resisitance in ohms \n",
+      "\n",
+      "# calculations\n",
+      "\n",
+      "Rf = (5-1) * R1 # calculating the resistance of feedback resistor \n",
+      "\n",
+      "# printing the values \n",
+      "\n",
+      "print \"The value of feedback resistor = \", (Rf/10**3),\"kohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of feedback resistor =  40 kohms\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex 2.4 Page no.92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Given Data\n",
+      "\n",
+      "R1 = 5*10**3\n",
+      "Rf = 20*10**3\n",
+      "vi = 1 \n",
+      "RL = 5*10**3\n",
+      "\n",
+      "# calculating the values \n",
+      "vo = float((1+(Rf/R1))*vi) \n",
+      "ACL = int(vo/vi)\n",
+      "iL = int((vo/RL)*10**3)\n",
+      "i1 = float(((vo - vi)/Rf))*(10**3)\n",
+      "io = iL+i1\n",
+      " \n",
+      " \n",
+      "# printing the values\n",
+      "print \"Output voltage vo = \",vo,\"V\"\n",
+      "print \"Gain ACL          = \",ACL\n",
+      "print \"Load current iL   = \",iL,\"mA\"\n",
+      "print \"The value of i1   = \",i1,\"mA\"\n",
+      "print \"Output current io = \", io,\"mA\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output voltage vo =  5.0 V\n",
+        "Gain ACL          =  5\n",
+        "Load current iL   =  1 mA\n",
+        "The value of i1   =  0.2 mA\n",
+        "Output current io =  1.2 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex 2.5 Page No.94"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given Data\n",
+      "import math\n",
+      "Beta = 200\n",
+      "ICQ = 100*10**-6\n",
+      "ADM = 100\n",
+      "CMRR = 80\n",
+      "\n",
+      "# finding the solution \n",
+      "# for VT =25 milli volt \n",
+      "VT = 25*10**-3\n",
+      "gm = float(ICQ/VT)\n",
+      "Rc = (ADM/gm) \n",
+      "CMRR = 10**(80/20) # log inverse is equal to powers of 10\n",
+      "RE = float((CMRR-1)/gm)\n",
+      "x = Decimal((RE/10**6))\n",
+      "\n",
+      "# printing the values \n",
+      "\n",
+      "print \" The value of gm =\",int(math.ceil((gm*10**3))),\"mMho\"            #converting the answer into milli Mho\n",
+      "print \" The value of Rc =\",int((Rc/10**3)),\"kohm\"         #converting the answer into kohm"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of gm = 4 mMho\n",
+        " The value of Rc = 25 kohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex 2.6 Page no. 95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "\n",
+      "gm = 4*10**-3\n",
+      "RC = 125*10*3\n",
+      "RE = 1.25*10**3\n",
+      "beta0 = 200\n",
+      "\n",
+      "# calculating the values\n",
+      "\n",
+      "rpi = beta0/gm              # value is in ohms \n",
+      "ADM =-500                   # Given Value\n",
+      "ACM = -((200*RC)/(402*RE)+rpi)*10**-6\n",
+      "print \"ACM is =\",round(ACM,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ACM is = -0.05\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9 Page No.63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from fractions import Fraction \n",
+      "# Given data\n",
+      "\n",
+      "beta0 = 100\n",
+      "IQ = 5*10**-4\n",
+      "RC = 10*10**3\n",
+      "RE = 150\n",
+      "VT = 25*10**-3 \n",
+      "\n",
+      "# calculations \n",
+      "\n",
+      "ICQ = float(IQ/2)\n",
+      "gm  = float(ICQ / VT)\n",
+      "rpi = beta0/gm\n",
+      "# calculaing the gain in Differential mode\n",
+      "ADM = ((0.5)*(beta0*RC))/(rpi+((1+beta0)*RE))\n",
+      "# To get the differentila mode gain multiply the value by 2\n",
+      "ADM2 = (ADM*2)\n",
+      "\n",
+      "# print the values \n",
+      "\n",
+      "print \"ICQ value is =\",ICQ*10**3,\"mA\"\n",
+      "print \"gm value is  =\",Fraction(gm).limit_denominator(100),\"Mho\"  \n",
+      "print \"rpi value is =\",int(rpi/10**3),\"kilo Ohm\"\n",
+      "print \"THe gain is  =\",int(math.ceil(ADM2)),\"V/V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ICQ value is = 0.25 mA\n",
+        "gm value is  = 1/100 Mho\n",
+        "rpi value is = 10 kilo Ohm\n",
+        "THe gain is  = 40 V/V\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex 2.8 Page no.97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "import math\n",
+      "I0 = 10*10**-6\n",
+      "VCC =10\n",
+      "VBE = 0.7\n",
+      "beta = 125\n",
+      "VT = 25*10**-3\n",
+      "\n",
+      "# Solution of the problem is \n",
+      "Iref = 10**-3 # Assumption\n",
+      "\n",
+      "R1 = (VCC - VBE)/Iref\n",
+      "# Finding the value RE from the equation 2.74\n",
+      "RE = (VT/(1+(1/beta)*I0))*math.log(Iref/I0)\n",
+      "\n",
+      "# printing the values \n",
+      "\n",
+      "print \"The value of R1 =\",R1/10**3,\"Kilo Ohms\"\n",
+      "print \"The value of RE =\",round(RE*100,1),\"Kilo Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of R1 = 9.3 Kilo Ohms\n",
+        "The value of RE = 11.5 Kilo Ohms\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
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