added books

12 files changed, 2784 insertions, 9 deletions

diff --git a/Linear_Integrated_Circuits/Chapter10.ipynb b/Linear_Integrated_Circuits/Chapter10.ipynb
new file mode 100755
index 00000000..db7e9c1b
--- /dev/null
+++ b/Linear_Integrated_Circuits/Chapter10.ipynb
@@ -0,0 +1,305 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : D-A and A-D Converters"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1 Page No.357"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given Data \n",
+      "\n",
+      "n = 9\n",
+      "s = 10.3*10**-3\n",
+      "a = '0b101101111'\n",
+      "\n",
+      "# Solution \n",
+      "# Converting the given value into its equivalent decimal value\n",
+      "x = int(a,2)\n",
+      "#multiplying with the resolution to get the output\n",
+      "\n",
+      "V = round(x*s,2)\n",
+      "\n",
+      "print \"The output value will be =\",V,\"V\" \n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output value will be = 3.78 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2 Page No.357"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from fractions import Fraction \n",
+      "# Given data \n",
+      "\n",
+      "n = 8\n",
+      "v = 10.0\n",
+      "\n",
+      "# Solution \n",
+      "LSB = 1.0/2**n\n",
+      "Vlsb =v/2**n\n",
+      "MSB = v/2\n",
+      "Vo = v - Vlsb\n",
+      "\n",
+      "# Dispalying the output\n",
+      "\n",
+      "print \"The value of LSB                  = \",Fraction(LSB).limit_denominator(256)\n",
+      "print \"The voltage of LSB                = \",int(Vlsb*10**3),\"mV\"\n",
+      "print \"The MSB                           = \",int(MSB)\n",
+      "print \"The value of full scale output is = \",round(Vo,3),\"V\"\n",
+      "     \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of LSB                  =  1/256\n",
+        "The voltage of LSB                =  39 mV\n",
+        "The MSB                           =  5\n",
+        "The value of full scale output is =  9.961 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3 Page No.357"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import numpy as np\n",
+      "V = 10.0                 # Range of the DAC is 0 -10 voltage \n",
+      "\n",
+      "# define a function for performing the DAC action\n",
+      "\n",
+      "def DAC(Vo):\n",
+      "    j = 1\n",
+      "    sum = 0.0\n",
+      "    x = len(Vo)\n",
+      "    for i in range(x):\n",
+      "        sum = sum + ((Vo[i])*(0.5**j))\n",
+      "        j += 1\n",
+      "    \n",
+      "    return (V*sum)\n",
+      "# part 1 \n",
+      "Vo1 = np.array([1, 0])\n",
+      "# part 2\n",
+      "Vo2 = np.array([0, 1, 1, 0])\n",
+      "#part 3\n",
+      "Vo3 = np.array([1, 0, 1, 1, 1, 1, 0, 0])\n",
+      "\n",
+      "# Finding the solution for all 3 parts and printing the outputs \n",
+      "\n",
+      "print \" The output for part 1\",int(DAC(Vo1)),\"V\"\n",
+      "print \" The output for part 2\",(DAC(Vo2)),\"V\"\n",
+      "print \" The output for part 3\",round(DAC(Vo3),2),\"V\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The output for part 1 5 V\n",
+        " The output for part 2 3.75 V\n",
+        " The output for part 3 7.34 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4 Page No.365"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given data \n",
+      "n = 16 \n",
+      "Clockrate = 4*10**6\n",
+      "V = 10.0\n",
+      "c = 0.1*10**-6\n",
+      "va = -8.0\n",
+      "\n",
+      "# Solution \n",
+      "\n",
+      "t21 = (2.0**n)/(Clockrate)\n",
+      "R = ((-V/va)*t21)/c\n",
+      "\n",
+      "print \"The value of (t2-t1)    =\",round(t21*10**3,2),\"ms\"\n",
+      "print \"The value of Resistor R =\",int(round(R*10**-3)),\"kilo Ohms\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of (t2-t1)    = 16.38 ms\n",
+        "The value of Resistor R = 205 kilo Ohms\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5 Page No.365"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "\n",
+      "Va = 4.129\n",
+      "n = 16\n",
+      "Vr = 8\n",
+      "\n",
+      "# Solution \n",
+      "\n",
+      "N = int(round((2**n)*(Va/Vr)))\n",
+      "out = bin(N)    # Converting the voltage value into its binary equivalent\n",
+      "\n",
+      "print \"The binary equivalent is = \",out\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The binary equivalent is =  0b1000010000100001\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.6 Page No.368"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "\n",
+      "d = 3.5 \n",
+      "print \"The last 3 digit can be\",000,\"To\",(10**int(d))-1\n",
+      "print \"Hense the 3 1/2 digit DVM reading varies from\",0000,\"to\",\"1\"+str((10**int(d))-1)\n",
+      "# Reference voltage is 2V\n",
+      "Vref = 2.0\n",
+      "# Resolution R \n",
+      "\n",
+      "R = Vref/2000\n",
+      "print \"The resolution of a 3 1/2 digit DVM is =\",int(R*10**3),\"mV\"\n",
+      "\n",
+      "# Similarly for 4 1/2 digit DVM\n",
+      "\n",
+      "R1 = Vref/20000\n",
+      "\n",
+      "print \"Thus resolution of a 4 1/2 digit DVM is =\",R1*10**3,\"mV\"\n",
+      "\n",
+      "print \"So the resolution of 4 1/2 digit DVM is better than 3 1/2 digit DVM\"\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The last 3 digit can be 0 To 999\n",
+        "Hense the 3 1/2 digit DVM reading varies from 0 to 1999\n",
+        "The resolution of a 3 1/2 digit DVM is = 1 mV\n",
+        "Thus resolution of a 4 1/2 digit DVM is = 0.1 mV\n",
+        "So the resolution of 4 1/2 digit DVM is better than 3 1/2 digit DVM\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Linear_Integrated_Circuits/Chapter2.ipynb b/Linear_Integrated_Circuits/Chapter2.ipynb
new file mode 100755
index 00000000..011589fa
--- /dev/null
+++ b/Linear_Integrated_Circuits/Chapter2.ipynb
@@ -0,0 +1,946 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Operational Amplifier"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 Page No.44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 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": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2 Page No.44"
+     ]
+    },
+    {
+     "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": [
+      "Example 2.3 Page No.49"
+     ]
+    },
+    {
+     "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),\"KiloOhms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of feedback resistor =  40 KiloOhms\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4 Page No.49"
+     ]
+    },
+    {
+     "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": [
+      "Example 2.6 Page No.61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Given Data\n",
+      "from decimal import Decimal, ROUND_HALF_UP\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)),\"kiloOhm\"         #converting the answer into Kilo Ohm\n",
+      "print \" The value of RE =\", x.quantize(Decimal('2.5')),\"MegaOhm\"         #converting the answer into MegaOhm"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of gm = 4 mMho\n",
+        " The value of Rc = 25 kiloOhm\n",
+        " The value of RE = 2.5 MegaOhm\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7 Page No.61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Given Data\n",
+      "RC = 2*10**3\n",
+      "RE = 4.3*10**3\n",
+      "VCC = VEE = 5\n",
+      "beta0 = 200\n",
+      "VBE = 0.7\n",
+      "VT = 25*10**-3                                  # VT = 25 mV \n",
+      "\n",
+      "#calculations \n",
+      "\n",
+      "Ibq  = (((VEE - VBE)/(2*(1+beta0)*RE)))*10**6   # converting the into mA\n",
+      "IBQ = int(Ibq*10)/10.0                          # rounding the answer 2.48 to 2.4\n",
+      "ICQ = beta0 * IBQ*10**-3                        # finding ICQ and convert it into milli Ambere\n",
+      "V01 = V02 = VCC - (RC * ICQ*10**-3)\n",
+      "VCEQ = V01 + VBE\n",
+      "gm = ICQ / VT                                   # Finding the gm value in milli Mho\n",
+      "rpi = float(beta0/gm)                           # finding rpi interms of KiloOhms \n",
+      "ADM = -(gm*RC)/10**3                            # Calculating the gain \n",
+      "ACM = float((-beta0*RC)/(rpi*10**3 + (2*beta0) * RE)) # Calulating the value of ACM and coverting it into a float \n",
+      "CMRR = float(round(ADM,2)/round(ACM,2))               # rounding ACM and ADM to 2 digits after decimals and taking the\n",
+      "                                                      #ratio\n",
+      "CMRRdb = float(20*math.log10(round(CMRR,1)))          # Find the CMRR in dB using equation 20*log(CMRR)\n",
+      "\n",
+      "# Printing all values \n",
+      "\n",
+      "print \"The IBQ        =\",IBQ,\"uA\"\n",
+      "print \"The ICQ        =\",ICQ,\"mA\"\n",
+      "print \"The V01        =\",V01,\"mV\"\n",
+      "print \"The VCEQ       =\",VCEQ,\"V\"\n",
+      "print \"The gm         =\",gm,\"m Mho\"\n",
+      "print \"The rpi        =\",round(rpi,1),\"kilo Ohm\"\n",
+      "print \"The ADM        =\",round(ADM,2)\n",
+      "print \"The ACM        =\",round(ACM,2)\n",
+      "print \"The CMRR       =\",round(CMRR,1)\n",
+      "print \"The CMRR in dB =\",int(CMRRdb*10)/10.0,\"dB\"\n",
+      "                                      "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The IBQ        = 2.4 uA\n",
+        "The ICQ        = 0.48 mA\n",
+        "The V01        = 4.04 mV\n",
+        "The VCEQ       = 4.74 V\n",
+        "The gm         = 19.2 m Mho\n",
+        "The rpi        = 10.4 kilo Ohm\n",
+        "The ADM        = -38.4\n",
+        "The ACM        = -0.23\n",
+        "The CMRR       = 167.0\n",
+        "The CMRR in dB = 44.4 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8 Page No.62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# v1 = 15*sin*2*pi(60)t + 5*sin*2*pi(1000)t mV\n",
+      "# v1 = 15*sin*2*pi(60)t - 5*sin*2*pi(1000)t mV\n",
+      "\n",
+      "# 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)\n",
+      "\n",
+      "# derivation part is as follows  \n",
+      "# VDM = (v1 - v2)/2 = 5*sin*2*pi(1000)*t\n",
+      "# VCM = (v1 - v2)/2 = 15*sin*2*pi(60)*t\n",
+      "# VO1 = ADM*VDM + ACM*VCM\n",
+      "# VO2 = ADM*VDM - ACM*VCM\n",
+      "# V01 = -2500*sin*2*pi(1000)t - 0.75*sin*2*pi(60)t mV\n",
+      "# V01 =  2500*sin*2*pi(1000)t - 0.75*sin*2*pi(60)t mV"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ACM is = -0.05\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "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": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10 Page No.67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given Data \n",
+      "VCC = 10.0                  # initializing as floats \n",
+      "beta = 125.0\n",
+      "VBE = 0.7\n",
+      "\n",
+      "# calculations \n",
+      "# Define a function for calculating the value of R1 according to the equation given in 2.67\n",
+      "def R1(I):\n",
+      "    R1 = (beta/(2+beta))*((VCC-VBE)/I)       \n",
+      "    return (R1/1000)\n",
+      "# Cosisdering Ic = 1mA\n",
+      "print \"The value of R1 when I is 1 mA  = \",round(R1(10**-3),2),\"Kilo Ohm\"           # calling the function R1 with arguiment as current value 1 mA\n",
+      "print \"The value of R1 when I is 10 uA = \",int(round(R1(10*10**-6),0)),\"Kilo Ohm\"  # calling the function R1 with arguiment as current value 10 uA"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of R1 when I is 1 mA  =  9.15 Kilo Ohm\n",
+        "The value of R1 when I is 10 uA =  915 Kilo Ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.11 Page No.69"
+     ]
+    },
+    {
+     "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
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.13 Page No.74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "\n",
+      "VCC = 10.0\n",
+      "beta = 100\n",
+      "VBE = 0.7\n",
+      "RE = 10*10**3\n",
+      "\n",
+      "# solution\n",
+      "# fom the KVL loop1\n",
+      "Iref = ((VCC - VBE)/RE)   # Finding the Iref \n",
+      "# Assuming the transistor are identical \n",
+      "# Iref = 2IE\n",
+      "# IE = IC + IB\n",
+      "IC = (beta*Iref/(2*(1+beta)))\n",
+      "I0 = IC\n",
+      "#Displaying the values \n",
+      "\n",
+      "print \"The value of Iref =\",Iref*10**3,\"mA\"             # Converting the value into mA\n",
+      "print \"The value of IC   =\",round(IC*10**3,2),\"mA\"      # Converting the value into mA and rounding the answer\n",
+      "print \"The value of I0   =\",round(I0*10**3,2),\"mA\"      # Converting the value into mA and rounding the answer"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Iref = 0.93 mA\n",
+        "The value of IC   = 0.46 mA\n",
+        "The value of I0   = 0.46 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.14 Page No.75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data  \n",
+      "\n",
+      "VCC = 12.0\n",
+      "Rref = 15*10**3 \n",
+      "R1 = 2.8*10**3\n",
+      "beta = 200\n",
+      "VBE = 0.7\n",
+      "V0 = 6.0\n",
+      "\n",
+      "#Calculations \n",
+      "#Calculations for finding the IC1 and IC2\n",
+      "Iref = (VCC - VBE)/Rref\n",
+      "I1 = VBE / R1\n",
+      "IC1 = (Iref - I1)/(1+(2/beta))\n",
+      "IC2 = IC1 # Due to mirror effect \n",
+      "RC = (VCC - V0)/IC1\n",
+      "\n",
+      "# Displaying the values \n",
+      "\n",
+      "print \"The value of Iref  =\",round(Iref*10**3,2),\"mA\"             # convert into milli amps and rounding the output\n",
+      "print \"The value of I1    =\",round(I1*10**3,2),\"mA\"               # convert into milli amps and rounding the output\n",
+      "print \"The value of IC1   =\",round(IC1*10**3,1),\"mA\"              # convert into milli amps and rounding the output\n",
+      "print \"The value of Rc    =\",int(round(RC*10**-3,0)),\"Kilo Ohm\"   # convert into kilo Ohms and rounding the output"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Iref  = 0.75 mA\n",
+        "The value of I1    = 0.25 mA\n",
+        "The value of IC1   = 0.5 mA\n",
+        "The value of Rc    = 12 Kilo Ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.15 Page No.75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data  \n",
+      "\n",
+      "VCC = 10.0\n",
+      "VBE = 0.75\n",
+      "R1 = 4.7*10**3\n",
+      "\n",
+      "# solution \n",
+      "# I is approximately equals to Ic(2 + (1/beta)), since 1/beta is a small value its considerd as 2Ics \n",
+      "I = (VCC - VBE)/R1\n",
+      "# IE3 is approxiamtly equal to IC3 = I/2\n",
+      "#finding the value of IC \n",
+      "IC = I /2\n",
+      "print \" The value of I  =\",round(I*10**3,2),\"mA\"\n",
+      "print \" The value of IC =\",round(IC*10**3,2),\"mA\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of I  = 1.97 mA\n",
+        " The value of IC = 0.98 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.16 Page No.80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "VCC = 15.0\n",
+      "VBE = 0.7\n",
+      "R = 10*10**3\n",
+      "RC2 = 5*10**3\n",
+      "# consider beta is a large value \n",
+      "I = (VCC - VBE)/R\n",
+      "IC1 = IC2 = I\n",
+      "#V12 = V1 - V2\n",
+      "V12 = VBE + IC2 * RC2 \n",
+      "\n",
+      "# displaying the values\n",
+      "print \"The value of I        =\",round(I*10**3,2),\"mA\"\n",
+      "print \"The value of V1 - V2  =\",V12,\"V\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I        = 1.43 mA\n",
+        "The value of V1 - V2  = 7.85 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.17 Page No.83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Given data\n",
+      "\n",
+      "VEE = 6.0\n",
+      "VCC = 6.0\n",
+      "VD = VD1 = VD2 = VD3 = VD4 = 0.7\n",
+      "R5 = 3.2*10**3\n",
+      "R4 = 1.5*10**3\n",
+      "VBE1 = VBE2 =VBE3 = VBE4 = VBE5 = VBE6 = VBE7 = VBE8 = 0.7\n",
+      "R1 = 2.2*10**3\n",
+      "R2 = 7.75*10**3\n",
+      "R3 = 5.2*10**3\n",
+      "R6 = 1.5*10**3\n",
+      "R7 = 3*10**3\n",
+      "R8 = 3.4*10**3\n",
+      "R9 = 6*10**3\n",
+      "R10 =30*10**3\n",
+      "hfe = 100\n",
+      "VT = 26*10**-3\n",
+      "\n",
+      "# Solution for part A\n",
+      "\n",
+      "VBN1 = round((-VEE+VD+VD)*R5/(R4+R5),2)\n",
+      "I1 = (VEE+VBN1-VBE1)/R1\n",
+      "IQ = I1  # If the base current of Q1 is neglected\n",
+      "IC2 = IC3 = IQ2 = round(IQ/2,6)\n",
+      "VC2 = VC3 = round((VCC - R2*IC2),2)\n",
+      "VE4 = VC2 - VBE4 \n",
+      "I6 = round(VE4/R6,6)\n",
+      "IC4 = IC5 = round((I6/2),6)\n",
+      "VC5 = round(VCC - IC5*R7,2)\n",
+      "VE6 = round(VC5 - VBE6,2)\n",
+      "IC7 = I8 = round((VEE - VD3)/R8,5)\n",
+      "VB8 = VBE8 + VD4 - VEE\n",
+      "I9 = round((VE6 - VB8)/R9,5)\n",
+      "I10 = IC7 - I9\n",
+      "Vo = (I10 * R10) + VB8\n",
+      "V0 = int(Vo)         # Vo value is assumed to be very small\n",
+      "\n",
+      "# Displaying the values \n",
+      "print \"PART A\"\n",
+      "print \"==================\"\n",
+      "print \"the value of VBN1 =\",VBN1,\"V\"\n",
+      "print \"The value of I1   =\",I1*10**3,\"mA\"\n",
+      "print \"The value of IQ   =\",IQ*10**3,\"mA\"\n",
+      "print \"The value of IC2  =\",IC2*10**3,\"mA\"\n",
+      "print \"The value of VC3  =\",VC3,\"V\"\n",
+      "print \"The value of VE4  =\",VE4,\"V\"\n",
+      "print \"The value of IC4  =\",IC4*10**3,\"mA\"\n",
+      "print \"The value of VC5  =\",VC5,\"V\"\n",
+      "print \"The value of VE6  =\",VE6,\"V\"\n",
+      "print \"The value of IC7  =\",IC7*10**3,\"mA\"\n",
+      "print \"The value of VB8  =\",VB8,\"V\"\n",
+      "print \"The value of I9   =\",I9*10**3,\"mA\"\n",
+      "print \"The value of I10  =\",I10*10**3,\"mA\"\n",
+      "print \"The value of V0   =\",V0,\"V\"\n",
+      "\n",
+      "# Solution for part B\n",
+      "VT = 26*10**-3                #the volt equivalent of temperature \n",
+      "IC2 = IC3 = IC4 = IC5 = IC = 0.5*10**-3\n",
+      "hie = (hfe*VT)/IC\n",
+      "RL2 = RL3 = (R2*R3)/(R2+R3)\n",
+      "AV1 = round((hfe*RL2)/hie,0)\n",
+      "AV2 = round(-(hfe*R7)/(2*hie),2)\n",
+      "AV3 = 1                                        # the thrid stage emitter follower differential gain \n",
+      "AV4 = int(-R10/R9)\n",
+      "AV = AV1*round(AV2,1)*AV4                      # again rounding AV2 to get the desired value\n",
+      "\n",
+      "# Displaying the values\n",
+      "print \"PART B\"\n",
+      "print \"==================\"\n",
+      "print \"The value of hie =\",int(hie)\n",
+      "print \"The value of RL2 =\",RL2*10**-3,\"Kilo Ohms\"\n",
+      "print \"The value of AV1 =\",int(AV1)\n",
+      "print \"The value of AV2 =\",round(AV2,1)\n",
+      "print \"The value of AV3 =\",AV3\n",
+      "print \"The value of AV4 =\",AV4\n",
+      "print \"The value of AV  =\",int(AV)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "PART A\n",
+        "==================\n",
+        "the value of VBN1 = -3.13 V\n",
+        "The value of I1   = 0.986363636364 mA\n",
+        "The value of IQ   = 0.986363636364 mA\n",
+        "The value of IC2  = 0.493 mA\n",
+        "The value of VC3  = 2.18 V\n",
+        "The value of VE4  = 1.48 V\n",
+        "The value of IC4  = 0.494 mA\n",
+        "The value of VC5  = 4.52 V\n",
+        "The value of VE6  = 3.82 V\n",
+        "The value of IC7  = 1.56 mA\n",
+        "The value of VB8  = -4.6 V\n",
+        "The value of I9   = 1.4 mA\n",
+        "The value of I10  = 0.16 mA\n",
+        "The value of V0   = 0 V\n",
+        "PART B\n",
+        "==================\n",
+        "The value of hie = 5200\n",
+        "The value of RL2 = 3.11196911197 Kilo Ohms\n",
+        "The value of AV1 = 60\n",
+        "The value of AV2 = -28.9\n",
+        "The value of AV3 = 1\n",
+        "The value of AV4 = -5\n",
+        "The value of AV  = 8670\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.18 Page No.87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "VEE = 15.0\n",
+      "VCC = 15.0\n",
+      "VBE1 = VBE2 = VBE3 = VBE4 = VBE5 = VBE= 0.7\n",
+      "R = 28.6*10**3\n",
+      "RC1 = 20*10**3\n",
+      "RC6 = 3*10**3\n",
+      "RC8 = 2.3*10**3\n",
+      "RA = 15.7*10**3\n",
+      "hfe = 100\n",
+      "VT = 25*10**-3\n",
+      "\n",
+      "# Solution for part 1 DC - ANALYSIS\n",
+      "\n",
+      "I = (VEE - VBE3)/R\n",
+      "ICQ4 = I\n",
+      "ICQ1 = ICQ2 = ICQ4/2\n",
+      "# Q1 and Q2 are biased at 0.25mA so their Collector voltages are\n",
+      "IC1 = IC = ICQ1\n",
+      "VCQ1 = VCQ2 = VCC - (IC1*RC1)\n",
+      "VEQ5 = VEQ6 = VCQ1 - VBE\n",
+      "IQ7 = 4*I\n",
+      "# The collector current of Q5 and Q6 are \n",
+      "ICQ5 = ICQ6 = IQ7/2\n",
+      "VCQ6 = VCC - (ICQ6 * RC6)\n",
+      "VEQ8 = VCQ6 + VBE\n",
+      "IEQ8 = (VCC - VEQ8)/RC8\n",
+      "# The voltage VA at the collector of Q8 is \n",
+      "VA = -VCC + (IEQ8 * RA)\n",
+      "IEQ9 = (0-(-VCC))/RC6\n",
+      "\n",
+      "# Displaying all values of Part 1\n",
+      "\n",
+      "print \"The value of I      =\",int(I*10**3),\"mA\"\n",
+      "print \"The value of ICQ1   =\",int(ICQ1*10**3),\"mA\"\n",
+      "print \"The value of VCQ1   =\",int(VCQ1),\"V\"\n",
+      "print \"The value of VEQ5   =\",int(VEQ5),\"V\"\n",
+      "print \"The value of IQ7    =\",int(IQ7*10**3),\"mA\"\n",
+      "print \"The value of ICQ5   =\",int(ICQ5*10**3),\"mA\"\n",
+      "print \"The value of VCQ6   =\",int(VCQ6),\"V\"\n",
+      "print \"The value of VEQ8   =\",int(VEQ8),\"V\"\n",
+      "print \"The value of IEQ8   =\",int(IEQ8*10**3),\"mA\"\n",
+      "print \"The value of VA     =\",VA,\"V\"\n",
+      "print \"The value of IEQ9   =\",int(IEQ9*10**3),\"mA\"\n",
+      "\n",
+      "# Part 2 AC -ANALYSIS\n",
+      "\n",
+      "hieQ12 = (hfe * VT)/IC\n",
+      "# The ac emitter resistance of transister Q5 - Q6\n",
+      "hieQ56 = (hfe * VT)/ICQ5\n",
+      "RL1 = RL2 = (RC1*hieQ56)/(RC1 + hieQ56)          # parallel combination of resistor RC1 and hieQ56\n",
+      "ADM1 = (hfe * RL2)/hieQ12\n",
+      "ADM2 = -(hfe * RC6)/(2*hieQ56)\n",
+      "\n",
+      "\n",
+      "# Displaying the values \n",
+      "\n",
+      "print \"The value of ac emiiter resistace of the transistor Q1-Q2  = \",int(round(hieQ12*10**-3,0)),\"Kilo Ohm\"\n",
+      "print \"The value of ac emiiter resistace of the transistor Q5-Q6  = \",round(hieQ56*10**-3,1),\"Kilo Ohm\"\n",
+      "print \"The value of effective load of Q1-Q2                       = \",round(RL1*10**-3,1),\"Kilo Ohm\"\n",
+      "print \"The value of voltage gain of first differential pair       = \",int(round(ADM1,0)),\"Kilo Ohm\"\n",
+      "print \"The value of voltage gain of second differential pair      = \",int(ADM2),\"Kilo Ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I      = 0 mA\n",
+        "The value of ICQ1   = 0 mA\n",
+        "The value of VCQ1   = 10 V\n",
+        "The value of VEQ5   = 9 V\n",
+        "The value of IQ7    = 2 mA\n",
+        "The value of ICQ5   = 1 mA\n",
+        "The value of VCQ6   = 12 V\n",
+        "The value of VEQ8   = 12 V\n",
+        "The value of IEQ8   = 1 mA\n",
+        "The value of VA     = 0.7 V\n",
+        "The value of IEQ9   = 5 mA\n",
+        "The value of ac emiiter resistace of the transistor Q1-Q2  =  10 Kilo Ohm\n",
+        "The value of ac emiiter resistace of the transistor Q5-Q6  =  2.5 Kilo Ohm\n",
+        "The value of effective load of Q1-Q2                       =  2.2 Kilo Ohm\n",
+        "The value of voltage gain of first differential pair       =  22 Kilo Ohm\n",
+        "The value of voltage gain of second differential pair      =  -60 Kilo Ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Linear_Integrated_Circuits/Chapter3.ipynb b/Linear_Integrated_Circuits/Chapter3.ipynb
new file mode 100755
index 00000000..b6d3c5fe
--- /dev/null
+++ b/Linear_Integrated_Circuits/Chapter3.ipynb
@@ -0,0 +1,306 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : Operational Amplifier Characteristics "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1 Page No.108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "\n",
+      "Ri = 10*10**3\n",
+      "R1 = 10*10**6\n",
+      "Acl = 10\n",
+      "\n",
+      "# Solution \n",
+      "\n",
+      "Rf = Acl * R1\n",
+      "Rt = 47*10**3\n",
+      "Rs = (Rt**2)/(Rf - (2*Rt))\n",
+      "\n",
+      "# Displaying the results\n",
+      "\n",
+      "print \"The value of Rf = \",Rf/10**6,\"Mega Ohms\"\n",
+      "print \"Thevalue of Rs  = \",Rs,\"Ohms\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Rf =  100 Mega Ohms\n",
+        "Thevalue of Rs  =  22 Ohms\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2 Page No.110"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "\n",
+      "Rf = 10*10**3\n",
+      "R1 = 10**3\n",
+      "Vios = 10*10**-3\n",
+      "Ib = 300*10**-9\n",
+      "Ios = 50*10**-9\n",
+      "\n",
+      "# Solution \n",
+      "# Solution for part a\n",
+      "Vot1 = (1 + (Rf/R1))*Vios + Rf*Ib\n",
+      "# Solution for part b \n",
+      "Rcomp = (Rf * R1)/(Rf + R1)\n",
+      "# Solution for part c\n",
+      "Vot2 = (1+(Rf/R1))*Vios + Rf*Ios \n",
+      "\n",
+      "# Displaying the values \n",
+      "\n",
+      "print \"The value of maximum output offset due to Vios and Ib    = \", int( Vot1*10**3),\"mV\"\n",
+      "print \"The value of Rcomp                                       = \",Rcomp,\"Ohms\" # Given answer in the textbook is not correct please make the change \n",
+      "print \"The value of maximum output offset if Rcomp is connected = \",Vot2*10**3,\"mV\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of maximum output offset due to Vios and Ib    =  113 mV\n",
+        "The value of Rcomp                                       =  909 Ohms\n",
+        "The value of maximum output offset if Rcomp is connected =  110.5 mV\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3 Page No.111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "\n",
+      "Acl = 100\n",
+      "Temp0 = 25\n",
+      "Temp1 = 50\n",
+      "Vodrift = 0.15*10**-3\n",
+      "\n",
+      "# Solution \n",
+      "\n",
+      "Vos = Vodrift * (Temp1 - Temp0)\n",
+      "Vo = Vos * Acl\n",
+      "\n",
+      "# Displaying the results \n",
+      "\n",
+      "print \"The value of Input offset voltage = \",Vos*10**3,\"mV\"\n",
+      "print \"The value of Ooutput voltage      = \",int(Vo*10**3),\"mV\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Input offset voltage =  3.75 mV\n",
+        "The value of Ooutput voltage      =  375 mV\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4 Page No.125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "from scipy import signal\n",
+      "import matplotlib.pyplot as plt\n",
+      "import numpy as np\n",
+      "\n",
+      "Vpp = 6.0\n",
+      "Frequency = 2*10**6\n",
+      "\n",
+      "# Solution \n",
+      "\n",
+      "WL = 1.0/Frequency # Calculating the wavelength \n",
+      "SR = Vpp/(WL/2)  # Calculating the slew rate\n",
+      "\n",
+      "# Displaying the results \n",
+      "\n",
+      "print \"The value of slew rate is =\",int(SR*10**-6),\"V/us\"\n",
+      "\n",
+      "# Ploting the result \n",
+      "\n",
+      "Vp = 3\n",
+      "t = np.linspace(0, 1, 500)\n",
+      "plt.plot(t, -Vp*abs(signal.sawtooth(2 * np.pi * 2 * t)))  # Note : Triangular wave is the absolute of sawtooth wave\n",
+      "plt.ylim(-3,0)\n",
+      "plt.title(\"Output wave form of example 3.4\")\n",
+      "plt.ylabel(\"Amplitude (V)\")\n",
+      "plt.xlabel(\"Time(t)\")\n",
+      "plt.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of slew rate is = 24 V/us\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5 Page No.126"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Given data \n",
+      "\n",
+      "Acl = 50\n",
+      "Slew_rate = 0.5\n",
+      "frequency = 20*10**3\n",
+      "# solution \n",
+      "\n",
+      "Vm = round((Slew_rate * 10**6)/(2*math.pi * frequency),2)\n",
+      "Vo = 2 * Vm\n",
+      "Vpeak_to_peak = Vo/Acl\n",
+      "\n",
+      "# Displaying the outputs \n",
+      "\n",
+      "print \"The peak voltage is                   =\",Vm,\"V peak\"\n",
+      "print \"The peak to peak voltage              =\",Vo,\"V peak to peak\"\n",
+      "print \"The value of maximum input voltage is =\",int(Vpeak_to_peak*10**3),\"mVpeak to peak \" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The peak voltage is                   = 3.98 V peak\n",
+        "The peak to peak voltage              = 7.96 V peak to peak\n",
+        "The value of maximum input voltage is = 159 mVpeak to peak \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6 Page No.127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "\n",
+      "Vptpi = 500*10**-3  #input peak to peak voltage \n",
+      "Vptpo = 3\n",
+      "Tr = 4*10**-6\n",
+      "\n",
+      "# Solution \n",
+      "\n",
+      "Vdelta  = (0.9 - 0.1) * Vptpo\n",
+      "SR = (Vdelta / Tr ) * 10**-6\n",
+      "\n",
+      "# Display the values \n",
+      "\n",
+      "print \"The value of delta V  = \",Vdelta,\"V\"\n",
+      "print \"The slew rate is      = \",SR,\"V/us\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of delta V  =  2.4 V\n",
+        "The slew rate is      =  0.6 V/us\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Linear_Integrated_Circuits/Chapter4.ipynb b/Linear_Integrated_Circuits/Chapter4.ipynb
new file mode 100755
index 00000000..47b4cf90
--- /dev/null
+++ b/Linear_Integrated_Circuits/Chapter4.ipynb
@@ -0,0 +1,283 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : Operational Amplifier Applications "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 Page No.139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Rf = 50*10**3\n",
+      "R1 = 40*10**3\n",
+      "R2 = 25*10**3\n",
+      "R3 = 10*10**3\n",
+      "R4 = 20*10**3\n",
+      "R5 = 30*10**3\n",
+      "\n",
+      "# Solution \n",
+      "R = (R1 * R2)/(R1 + R2)\n",
+      "Vo11 = (R + Rf)/R\n",
+      "\n",
+      "def Vo(V1,V2,V3,V4):\n",
+      "    \n",
+      "    return -1.25*V1 - 2.0*V2 + 2.32*V3 + 1.16*V4 \n",
+      "\n",
+      "Vo1 = Vo(2,3,4,5)\n",
+      "\n",
+      "#Dispalying the output \n",
+      "\n",
+      "print \"The value of Vo = \",Vo1,\"V\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Vo =  6.58 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3 Page No.167"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%matplotlib inline\n",
+      "import math\n",
+      "from pylab import figure, show\n",
+      "import numpy as np\n",
+      "# Given data\n",
+      "\n",
+      "fmax = 100\n",
+      "Vp = 1\n",
+      "\n",
+      "# solution part a\n",
+      "\n",
+      "C1 = 0.1*10**-6 # Set it as the capacitance value.\n",
+      "Rf = 1/(2*math.pi*fmax*C1)\n",
+      "fb = 10*fmax\n",
+      "R1 = 1/(2*math.pi*fb*C1)\n",
+      "Cf = R1 *C1/Rf \n",
+      "\n",
+      "# Solution part b\n",
+      "fig = figure(1)\n",
+      "t  = np.linspace(0,20*10**-3)\n",
+      "ax1 = fig.add_subplot(211)\n",
+      "ax1.plot(t,np.sin(628*t))\n",
+      "ax1.set_ylabel('Voltage vi (v)')\n",
+      "ax1.set_xlabel('Time(t)')\n",
+      "ax1.set_title('A sine input wave')\n",
+      "\n",
+      "ax2 = fig.add_subplot(212)\n",
+      "ax2.plot(t,-np.cos(628*t))\n",
+      "ax2.set_ylabel('Voltage vi (v)')\n",
+      "ax2.set_xlabel('Time(t)')\n",
+      "ax2.set_title('A cosine output wave')\n",
+      "\n",
+      "show()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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wjRv0TEXZGT0a6NuXBq0x6s7r1zShZnQ0HR/KTt++wMSJwNixfCtRDuQWh/Lq\n1Svk5OTA19cX69atE3eqq6uLxgrmUsMMSs3w9aUR9Mq+b56YCNjbA8nJAPMPkR4/P/rAsXUr30qk\n49Yt4Isv6PhggYw1Q24GJTs7GwCt6S6pjruhoSFnIqSFGZSa8ewZ0LGj8gc6TplC66SvWMG3EtUg\nM5PWCnnwQLndbD08aNmGBQv4VqI8yM2gCIVCiYakVMSTJ0/q3Gl2djbc3Nzw9OlTCIVCHD9+HI0k\nzHBCoRB6enpip4CoqKhK9TCDUjO8vIBOnZT30DIlhSb6e/yYxZ5wyfTpQNOmwMqVfCupG8nJ1Jg8\neQLo6fGtRnngLZcXlyxevBhNmjTB4sWLsW7dOuTk5GDt2rUVrmvdujVu3bpV7WqIGZSac/s2DfhK\nTASUsfrArFlA/fqqdZCsCCQk0PT2T54AHFb5lhtz5tDxvG4d30qUC14MypkzZ3Dp0iUIBAL06dMH\nw4YNk6rTdu3a4eLFizA2NkZmZiYcHR3x4MGDCte1bt0a0dHR1Z7ZMINSO4YPB/r3p3EIykTp1kxc\nHNCsGd9qVA83N3o2NX8+30pqR3Y2zah97x5NN8SoOXI3KL6+vrh58ya8vLxACMHRo0dhZ2eHNWvW\n1LlTAwMD5OTkAKBnNIaGhuL3ZTE3N4e+vj40NTUxbdo0TJkyRfKXYAalVty6RY1KQgJ92lcWFi2i\n0fGqnj2ZL27fBkaMoKvXMjX1FJ7Vq2kVysBAvpUoH3I3KNbW1rh79y40NTUBACKRCLa2trh3716V\nDQ8YMACZmZkVfv79999j3Lhx5QyIoaGh2AmgLBkZGTAxMcGLFy8wYMAAbN26FQ4ODhW/BDMotWbE\nCOpiqSyBgS9f0ozCMTFAixZ8q1FdBg6kh9sTJvCtpGYUFNDVyR9/0AzKjNrB9dxZbS4vgUCA3Nxc\n8bZTbm5upYf1Zfnjjz8q/V3pVlezZs2QkZEBo0pcS0xMTAAATZs2xRdffIGoqCiJBgUA/P39xf/v\n6OgIR0fHajWqM/7+NIfT1KnKsUrZsoUmgWTGRLb4+tLYnnHjlKPu/JYttIgWMyY1IzIyEpGRkTJr\nv9oVSlBQEHx9fcUT9MWLF7F27Vq4S1Eab/HixWjcuDGWLFmCtWvXIjc3t8Kh/Js3byASiaCrq4uC\nggIMHDgQy5cvx8CBAyt+CbZCqRMjR9JKh3Pn8q2kanJz6VNoVBQNwmPIDkKAbt1obMqoUXyrqZrs\nbLpqvXY6svGtAAAgAElEQVQNsLLiW41yIrctr5kzZ8LT0xO9e/dGeno6bt68CYFAgK5du4pXDnUl\nOzsbrq6uSElJKec2nJ6ejilTpiAkJARPnjzBqH9HdHFxMby8vOBXSQUlZlDqxt27tF5KYqJir1JW\nraJuwgcO8K1EPQgLowlF791T7ADBJUvow0ZAAN9KlBe5GZTNmzfj2LFjSE9Ph5ubGzw8PNCpUyfO\nOuYSZlDqzqhRgIMDnUAUkfx8uiq5dEk9aqArAoTQXFijRwMzZvCtRjJpaTRINzaWeXZJg9wP5ZOT\nk3H06FEcO3YMb968gaenJzw8PNCmTRvOREgLMyh1JyaGVu9LTAR0dPhWU5F166j30bFjfCtRL+7c\noavXR48UMy5l+nQawMjikaSD18DGO3fuYMKECbh37x5EIhFnIqSFGRTpGD0a6NVL8eIPMjPpYeuV\nK0DbtnyrUT++/JLWZFe06PmEBFog7OFDli1BWuRuUIqLixEaGoqjR48iIiICTk5O8PDwwIgRIzgT\nIS3MoEhHbCwwaBD9Q23QgG81/zF+PM0txZ5C+SElhabpUbRtJU9PGuD6zTd8K1F+5GZQzp07h6NH\njyIkJATdunWDh4cHhg8frpDlf5lBkR4vL/o0uno130oo16/TldODB4q55aIu+PoCL14AP/3EtxLK\n3bu0Bs7jxyzTNBfIzaD07dsXHh4eGD16tEJlFpYEMyjSk5lJDznPn+ffp7+khKYAmT0b8PbmV4u6\n8+oVdc39809abplvPv+cnvnNmsW3EtVAJZJDcg0zKNwQEAAcPAj89Re/QW179wL799OqjDWIoWXI\nmK1bgdBQ6k7MJ5GRNIL/wQNabZIhPVzPnUoQC8uQF1OmUJfRPXv405CbS/fGt25lxkRRmDaNnq/9\n+Sd/GvLzgcmTgU2bmDFRZNgKhVGOe/dojq979/jJ6DtnDvDuHQtWUzR++QX47jvg5k1+JvSZM2ne\nLhbcyi1sy0sCzKBwi58fLVgUFCTffu/fp8YsLg5o0kS+fTOqhhDqJNGyJbB5s3z7PneOrk5iY5W7\n0qgiwgyKBJhB4ZY3b+jB/I4d9ABUHpSU0Boto0YBPj7y6ZNRO7KzqRvx9u3A0KHy6TM3lzqL/PQT\njd5ncAszKBJgBoV7zp6laTfu35dPBP2KFUB4OE2xolVtDmwGX1y5Qlcqt27JJzZl3DjqHrx9u+z7\nUkeYQZEAMyiywdMTMDamB6Gy5NQpenYSFcUqMSoDq1bRA/qICODfMkky4fRpYOFCGnvCYk5kAzMo\nEmAGRTa8fAn07EnjQWS1DVXqBBAWBtjZyaYPBreIRHR7sm9fYNky2fTx4gXd6jpxAujdWzZ9MHgo\nsMVQX5o0oQeivXvTFCiurty2//IlrRy5ZQszJsqEpibw889Aly60nk4lNe/qTFERMGkSDWplxkS5\nYCsURrXExtIn0qAgoF8/btosKqLlZrt1oxmFGcrH779Td97bt7nzynv/HnBzAwoLgV9/BerV46Zd\nhmRYYCND7pRuPbi708NYLpg3jx72K0ruMEbtGTqUJvDs2ZNm/pWWggJg2DDgo4/o+QkzJsoHMyiM\nGtGnD7B7N/2Df/y47u0UFwPLl9ND3SNHZHuoy5A9K1YAixfTba8//qh7O7m5dMVqZkZXwh99xJ1G\nhvzgxaCcOHECn3zyCTQ1NXH79u1KrwsPD0e7du1gZWWFdWxfpEZERkbKrO0vvgD8/Wmq++vXa//5\n+/dpHYtr16hB0dfnXGI5ZHkvlA1Z3ovJk+kK1tsb2LaNBkHWhhcv6AF/ly40j5usHzLYuJAdvBgU\na2trnDp1Cp999lml14hEIvj4+CA8PBxxcXEICgpCfHy8HFUqJ7L+Y5k6lT6VurhQAxMXV/1niouB\nNWsAJyf6+bNn6ZOorGETx3/I+l706QNcvQrs3EnPVYqKqv8MITSupU8fmpJ+yxb5JCVl40J28GJQ\n2rVrV20J4aioKFhaWkIoFEJbWxvu7u44c+aMnBQyqmLsWFoatlcv6uUzcSKQmlrxOpGIHuj37Elj\nFqKjqUFhSR9VE3NzuvpMSaHOFqVbmwUF5a9LTKQrXUtLmpB03jzg++/ZuFAFFPYMJS0tDS1atBC/\nNzMzQ1paGo+KGGWpX58GnT16BJiYALa21AOsa1fAwgIwMKBJBAcOpAbnjz9oAS+GaqOnBwQH0xVp\nURE1HMbGdKtz7lz6ENKjB5CTAxw/Dvz9NzUqDBWByIj+/fuTTz/9tMIrODhYfI2joyO5deuWxM+f\nPHmSTJ48Wfz+0KFDxMfHR+K1FhYWBAB7sRd7sRd71eJlYWHB6bwvs8DGP6Rx+QBgamqK1DL7KKmp\nqTCrZOM9ISFBqr4YDAaDIT28b3mRSlxC7Ozs8PjxYyQnJ6OwsBDHjh3D8OHD5ayOwWAwGDWFF4Ny\n6tQptGjRAtevX8fnn38OZ2dnAEB6ejo+//xzAICWlha2bduGQYMGoUOHDnBzc0P79u35kMtgMBiM\nGqASqVcYDAaDwT+8b3l9SE2CGWfPng0rKyvY2Njgzp071X42OzsbAwYMQJs2bTBw4EDk5ubK/Htw\ngSzuhb+/P8zMzNCpUyd06tQJ4eHhMv8eXCDNvZg4cSKMjY1hbW1d7np1HBeV3Qt1GxepqalwcnLC\nJ598gk8//RQ//vij+Hp1GxdV3YtajwtOj/ilpLi4mFhYWJCkpCRSWFhIbGxsSFxcXLlrQkJCiLOz\nMyGEkOvXrxN7e/tqP7to0SKybt06Qggha9euJUuWLJHjt6obsroX/v7+ZMOGDfL9MlIizb2YPn06\nmTRpErl9+zb59NNPy31G3cYFIYRcunRJ4r1Qt3GRkZFB7ty5QwghJC8vj7Rp04bEx8cTQtRvXFR1\nL2o7LhRqhVKTYMbg4GCMGzcOAGBvb4/c3FxkZmZW+dmynxk3bhxOnz4t3y9WB2R1L4DKHSFkiaOj\nIwwNDVFYWFjrz0pzL3bu3Im9e/fCwMCgQrt8jQt/f394e3vX6bOS7sWsWbPKtVfZvQAABwcHifcC\n4GdcSENdx0VWVhaaNWsGW1tbAEDDhg3Rvn17cZybOs0X1d0LoHbjQqEMSk2CGSu7Jj09vdLPZmVl\nwdjYGABgbGyMrKwsWX4NTpDVvQCArVu3wsbGBpMmTZLLcj45ORlRUVEwMjJCcHBwrT8vzb2oClUZ\nF69fv672mpoEBct7XEhLXcfFs2fPyl2TnJyMO3fuwN7eHoDqjAsu7gVQu3GhUAZFUMPcCzWxmIQQ\nie0JBIIa98MnXN6LssyYMQNJSUm4e/cuTExMsGDBgrrIqxUHDx5E//794e3tjQMHDlR5bXZ2NiZM\nmABTU1MYGhriiy++EN+LPXv2wMrKCjNmzEBISAgyMjLEn/v7778xdOhQ6Ovro2PHjsjPz4dAIMD4\n8eOx7N+ygvn5+TAzM8PGjRthbGyMV69eITAwEMB/93vhwoVo1aoVmjVrhhkzZuDdu3cSdRJCsGrV\nKgiFQhgbG2PcuHHiiT0yMrLcHy4ACIVCREREIDw8HGvWrMGxY8egq6uLTp06AaArOD8/P9jb20Nf\nXx8jR45ETk5OhfZKdZa2Fxsbi1u3blVor+y4SE9Px/z588XvnZyckJKSIn7fokULfPbZZ0hKSkKf\nPn1w8uRJGBkZwc7ODpcvXxa3oaOjI9YEAHfu3EHTpk0hEokAAPv27UOHDh1gaGiIwYMHl+tDFtT1\nb6Ts5/Lz8zFmzBhs2bIFDSXUGVb1+aK6e1Hb+UKhDEpNghk/vObZs2cwMzOT+HNTU1MA9CmjdMmf\nkZEBIyMjWX4NTuDyXpT9rJGRkfiPZPLkyYiKipLxN6EGxc3NDa6urjh79iyeP39e6bXe3t549+4d\n4uLi8Pz5c8yfPx+mpqa4e/culi5dihMnTmDx4sUwMTGBu7s7AODs2bPIy8vD+vXr8erVK5w4cQIv\nX76EqalphQkhKysLr1+/Rnp6Opo3b46ZM2fi1atXyMjIgEAgQEJCAmJiYpCQkIC0tDSsWLFCos79\n+/fjwIEDiIyMxJMnT5Cfnw+fKuokl+oYPHgwli5dCnd3d+Tl5ZU7MD906BD279+PjIwMaGlpYfbs\n2RXaKf23LW3PwMAAffv2Ldfeh//+7969Q0xMDABqGIqKivD27VsAwJMnT1BQUAAnJycIBALY29vj\n0qVLaNOmDTw9PeHi4oLCwkI0b94cPXr0wC+//CJu98iRI3BxcYGmpibOnDmDNWvW4NSpU3j58iUc\nHBzg4eFR6f3ggrr+jZTOC0VFRRg9ejTGjh2LkSNHiq9Rp/miuntR6/mi7sdA3FNUVETMzc1JUlIS\nef/+fbUHS9euXRMfLFX12UWLFpG1a9cSQghZs2aNUhyyyepepKeniz+/ceNG4uHhIdPv8ddff5F6\n9eqR169fE0IIsbGxIZs2bZJ4bXp6OtHQ0CC5ubnlfl5UVER0dXXJ9OnTxd8nOjqaaGtrk6dPn5Lz\n588TU1NT0qNHDyISicrdi/Hjx5NvvvmGJCUlEaFQSOrXr09EIhEhhI6Lhg0bkhs3bpDVq1cTbW1t\nkpiYKO736tWrpHXr1hK19u3bl+zcuVP8/uHDh0RbW5uIRCJy4cIFYmZmVu56oVBIIiIiCCGELF++\nnIwdO7bc7x0dHYmfn5/4fVxcHPnoo49ISUlJufZK/23NzMxIeHg4sbGxITNnzizXnqRx0aJFC3L7\n9m0SFBREPDw8SP369cmDBw/Ivn37yIgRIyodFwYGBiQ2NpYQQsjevXtJ3759CSGElJSUkBYtWpC/\n/vqLEELI4MGDyU8//SRuQyQSER0dHZKSkiLx/nGBNH8jJSUlxNvbm8ydO7dCu+o2X1R1L2o7XyiU\nQSGEkNDQUNKmTRtiYWFBVq9eTQghZNeuXWTXrl3ia7766itiYWFBOnbsWC4XmKTPEkLIP//8Q/r1\n60esrKzIgAEDSE5Ojvy+kBTI4l54e3sTa2tr0rFjRzJixAiSmZkp0+8wefJkMnz4cPH7VatWEVtb\nW4nX3rhxgzRp0kTi7+zs7IiRkVG576Onp0cWL15MCCHkxx9/JE2bNiUaGhrE0NCQXLp0iRBCDUqH\nDh2IiYkJ0dLSIpqammTfvn2EEDou6tWrR0xNTUmfPn2IQCAgjRo1Er/09fWJrq6uRD3t27cnoaGh\n4vdv374lAoGApKen19mg7NixQ/w+Pz+fCAQC8vz58wrthYaGEi0tLWJiYkJWr15Nli9fTuzt7asc\nF2PHjiUbN24klpaWpFGjRkRDQ4M0atSI9OrVi2zcuFE8LkxMTEjDhg2Jnp6e+Lrz588TQgjJzs4m\n9evXJxkZGSQyMpK0atWq3P1o2LBhufuno6NDrl27JvH+cUVd/0b++usvIhAIiI2NDbG1tSW2trYk\nLCyMEKJ+80VV96K284XCGRSG6vDmzRuip6dHGjZsSJo1a0aaNWtGDAwMiEAgIDExMRWur2yFQggh\nkyZNEhsPQuiEW7pCKcvz58+Jo6MjWbZsGSGEGpTS/69qoi99oi77RFYV/fr1K2cAyq5QoqKiiKGh\nofh3xcXFpEGDBmKD4u/vL9Gg+Pr6it+XXaHUpb0P2bNnDxk2bBixtrYmaWlpJCQkhHh4eJDWrVuL\nJ5ZLly4RIyMjcv/+ffHnDAwMxP0QQsiIESPI5s2bydSpU8vpHTRoEDly5Ej1N46h0ijUGQpDtTh9\n+jS0tLQQHx+PmJgYxMTEID4+Hg4ODjh48GCF601MTODs7IyZM2ciNzcXRUVFuHTpEgDAw8MD+/fv\nR0xMDN6/f4+lS5eie/fuaNmyJaKjo3Hjxg0UFRVBR0cH9erVg+a/Zf8IfWiqVquGhgamTJmCuXPn\n4sWLFwCoV8y5c+ckXu/h4YFNmzYhOTkZ+fn54nMRDQ0NtGnTBu/evUNoaCiKioqwatUqvH//XvzZ\nZs2aITk5uZwuQgh+/vlnxMfH482bN/j222/h4uICgUBQp/Y+pE+fPrhw4QLevXuH5s2bo3fv3ggP\nD0d2drb4ID8vLw9aWlpo0qQJCgsLsWLFigoeZJ6enjhw4AB++eUXeHp6in8+ffp0rF69GnH/Vlwr\nPctiqBfMoDBkxsGDBzFx4kSYmZnByMgIRkZGMDY2ho+PD44cOYKSkpIKnzl06BC0tbXRrl07GBsb\ni6N2+/Xrh5UrV2L06NFo3rw5kpKScPToUQDA69evMXXqVBgaGkIoFKJJkyZYtGgRgIpeOlV5w6xb\ntw6Wlpbo3r079PX1MWDAADx69EjitRMnToS3tzc+++wzmJubQ0dHB1u3bgUA6OvrY8eOHZg8eTLM\nzMzQsGHDcl5fLi4uAIDGjRvDzs5OrMvb2xvjx4+HiYkJCgsLxd+9Lu19iJWVFXR1deHg4AAA0NPT\ng4WFBXr16iW+J4MHD8bgwYPRpk0bCIVC1K9fHy1btizXzvDhw5GQkAATE5Ny0fYjR47EkiVL4O7u\nDn19fVhbW+Ps2bOV3muGasJrLq+JEyciJCQERkZGuHfvnsRrZs+ejbCwMOjo6CAwMFD8NMVgqBJO\nTk7w9vbGxIkT+ZbCYNQZXlcoEyZMqDI3TGhoKBISEvD48WPs3r0bM2bMkKM6BkO+8Phsx2BwAq8G\npao0EEDlqQIYDFVEGQLoGIyqkFnFRi6oLFVAaVoEBkNVuHDhAt8SGAypUWiDAlSdKqAUS0tLJCYm\nyksSg8FgqAQWFhacllBXaC+vqlIFlCUxMVHsHqoOr8JCgkOHCGxsCD75hGD/foKCAoKEBAJPz+XY\nvJlg5kyCbt0IunShP+dbMx+v5cuX865B3uNiwQKCZs0Ixowh+PZbgqAggpgYgq+/Xo6rVwlcXAgM\nDQnmzydITuZfMxsX/L64fhBX6BXK8OHDsW3bNri7u+P69eto1KiR2m93/fwzsHQpYGkJrFkDDB4M\nlC7aLCwAKytgzhz6nhBg2zage3dg+3bA1ZU/3QzZkpoKuLkBBgbA/ftA48blf//rr0CPHvT19Cmw\ndSvQuTPg7Azs3g3o6PCjm6Fa8LpC8fDwQM+ePfHw4UO0aNEC+/btQ0BAAAICAgAAQ4YMgbm5OSwt\nLTFt2jTs2LGDT7m8s3cv8PXXwC+/AOfP08mgqnNcgQCYNQsID6dGaMYM4N+cgAwVIiwM6NoVGDEC\n+O23isbkQ1q1AtavB5KT6RgZPpyNCwZHEBVARb5GlRw8SIiZGSGPH1d93YULFyT+PDeXEFdXQjp2\nJOTBA+71KSKV3QtVoaiIkKVLCTE1JeTixaqvrexeFBcT4ulJyKBBhLx9y71GRUTVx0Vt4Hru5DWw\nkSsEAgFU4GtUyvHjwNy5dFXSrl3d2yEECAgAVqwAoqOB5s2508iQL4QAEyYAKSnA0aOANBnWi4sB\nLy+goICufj/+mDudDMWG67mTGRQF5/RpYPp04I8/gDKZLqRi5Urg3DlqoLS1uWmTIV/27AG2bAFu\n3AAaNJC+vaIiwN2dGpcTJ4CPPpK+TYbiwwyKBFTVoISG0qfQsDB6gMoVJSXAsGFA27bAxo3ctcuQ\nD7dvA4MGAZcv039DrigsBFxcAC0tuuphDxuqD9dzp0K7DaszsbHA+PFAcDC3xgQANDSAQ4fo6ocl\nhFUucnLopL99O7fGBKCrkuPHgTdvqBMHg1Fb2ApFASkupq6+M2cCsswVWPqk+9df0p3NMORDSQkw\nciTQujXd7pIVL17Q7dXgYKBbN9n1w+AftkJRAzZtAho1ottdsqRzZ2DtWmDUKCA/X7Z9MaRn/Xo6\n2f/vf7Ltp2lTYPNm+jBTpuwKg1EtbIWiYDx+DPTsCURF0SdReTBpEt3mOHKk6rgWBn9cvEgDF2/e\nBMqkt5MZhNDVkI0N9QpkqCbsUF4CqmJQSkoAJye6YiiNdpcHb9/SLTZfX8DDQ379MmrG69dAhw7A\nTz/RLUp5kZ4O2NpSD0MbG/n1y5AfbMtLhdm9m3ra+PjIt9/69WmKliVL6EqFoVisWQP07y9fYwLQ\nOKW1a+nWV3GxfPtmKCdshaIgpKbSM42LF+nTKB+4utLD2GXL+OmfUZGkJMDODrh3j59AVEKoIevb\nl65gGaoF2/KSgLIbFEJoXIi9Pb+TeenkFRsLSEjqzOABRTDyycl0XFy+zLwBVQ1mUCSg7AblyBG6\ntRAdzX+E8tKlQFoacOAAvzoYdAL39AQePOA/G/D27TTY8dIl5rihSjCDIgFlNijv3wNt2lCj0qsX\n32qAvDwaMHfmDM1gy+CHkhK6Yp07l+bZ4puSEnowv2YNMHQo32oYXMEO5VWMvXuBTz9VDGMCALq6\nwPff04lMSW20SvDzzzSjgaJ43Wlo0Bxwy5ZR48JgSKJGK5T4+HgkJydDQ0MDrVq1QjsF20hV1hXK\n27e0UFZwMNClC99q/qOkhO6ZL1lCYx8Y8qWggK4Sjx+nMUmKAiF01errC4wZw7caBhfIbcsrKSkJ\nmzZtQmhoKExNTdG8eXMQQpCRkYFnz55h6NChmDdvHoRCIWdi6oqyGpSNG+k++a+/8q2kIpcuAV9+\nCcTHU7dihvzw96fnJkeP8q2kIuHhwPz51OtMU5NvNQxp4XzurKxQiouLCzl37hwpLCys8LvCwkJy\n9uxZ4uLiUpvaKxUICwsjbdu2JZaWlmTt2rUVfn/hwgWip6dHbG1tia2tLVm5cqXEdqr4GgpLXh4h\nRkaExMbyraRyxowhZNUqvlWoF6mphBgaEpKczLcSyZSUENKrFyGHDvGthMEFXM+dvB3Ki0QitG3b\nFn/++SdMTU3RtWtXBAUFoX379uJrIiMjsXHjRgQHB1fZljKuUNasoe65QUF8K6mchARag/zJE3q2\nwpA9M2fSe71uHd9KKicyEpg8ma5eWYp75Ubuh/IdO3bE6tWrkZiYyFmnABAVFQVLS0sIhUJoa2vD\n3d0dZ86cqXCdshmKmvDqFd3u8vfnW0nVWFrSCO1du/hWoh5kZtJtrgUL+FZSNY6OgFDIXMsZFanW\noAQHB0NTUxOurq6ws7PD+vXrkZKSInXHaWlpaFEmy52ZmRnS0tLKXSMQCHD16lXY2NhgyJAhiIuL\nk7pfRWDTJuDzz7mvZyEL/Pyo3nfv+Fai+mzaRF2EpSnnKy9WrqRJI1k2YkZZqjUoQqEQS5Yswa1b\ntxAUFITY2Fi05iANrqAG0VGdO3dGamoqYmJiMGvWLIwcOVLqfvnmn3+ArVuBb7/lW0nN6NiReqAF\nBvKtRLXJyaEu5AsX8q2kZvToQcfGnj18K2EoElo1uSg5ORnHjh3D8ePHoampiR9++EHqjk1NTZGa\nmip+n5qaCjMzs3LX6JbZuHd2dsbMmTORnZ0NQ0PDCu35l9k/cnR0hKOjo9QaZcH69dTl0tycbyU1\nZ+lSGrE9eTItD8vgnu3bafqdVq34VlJzVqygQY4TJ/Ifyc+oGZGRkYiMjJRZ+9Ueytvb26OwsBCu\nrq5wc3ODOUczYXFxMdq2bYuIiAg0b94c3bp1q3Aon5WVBSMjIwgEAkRFRcHV1RXJyckVv4SSHMq/\nfEmj4mNi5FPTgkucnGjdlLFj+VaiehQU0AeMyEigzPBXCkaNAvr0kW+5BQZ3cD13Vvu8eeDAAZkE\nMmppaWHbtm0YNGgQRCIRJk2ahPbt2yMgIAAAMG3aNJw8eRI7d+6ElpYWdHR0cFQRHfNrwc6dwBdf\nKJ8xAehZyrx5dKWiwfIrcMrevUDv3spnTABg8WIazf/VV2z1yqhihRIYGIixY8dCq5JRUlhYiMOH\nD2OCrOvU1gBlWKG8e0c9YyIigE8+4VtN7SmNkv7mG1rJj8ENhYWAhQVw6hTNTqCM9O5NVyguLnwr\nYdQWua1Q8vPz0bVrV7Rr1w52dnYwMTEBIQSZmZmIjo7GgwcPMGXKFM6EqDqHDwOdOimnMQFohtml\nS4HVq4ERI1jGWa74+We6MlFWYwJQN+e1a+nZIBsX6k2VZyiEEFy5cgWXL18Wuwq3atUKvXv3Rs+e\nPWvkqSUPFH2FQghNALllC43rUFZKSqhB3LYN6NePbzXKj0hEi6kFBNDYDmVFJKIu8IGBdLXCUB5Y\n+noJKLpBCQujCfXu3lX+J7iDB2lAW0QE30qUnxMnaIDr1avKPy527KC150+d4lsJozYwgyIBRTco\n/fvTRItffsm3EukpKqIR9L/8otzbNHxDCL1/y5cDw4fzrUZ63ryhZ4RXrgBWVnyrYdQUVg9Fybh7\nl+Y8cnfnWwk3aGsDs2bR7TtG3bl8GcjPV51iVTo6wNSpwObNfCth8AlbociYceNoHW4/P76VcEdO\nDo2biIsDTEz4VqOcjBlDY3u++opvJdyRmUnPhB4/Bho35lsNoybIbcvr0KFD8Pb2xoYNGySKmD9/\nPmcipEVRDUp6Oj2MT0gAJAT3KzUzZtCcU999x7cS5SMlhXr8JSerXhbnSZOA1q2pezlD8ZHblteb\nN28AAHl5ecjPzy/3ysvL40yAKrN1K032p2rGBKDbXgEBLDlgXdi+na5cVc2YALT41vbtLJmoulLt\nltfz589hpODpTxVxhZKfTw8pb9yggWuqyMCBNBWLKjgbyIuCgv/GhTLlc6sNzs40yHHiRL6VMKpD\n7ofyvXv3xsCBA/HTTz8hJyeHs45VncBAmuNIVY0JQKOjt2yhHkuMmnH4MK0Tr6rGBKCBjhs3snGh\njlRrUB49eoSVK1fi/v376NKlC4YOHYpDhw7JQ5vSUlJCt7vmzuVbiWxxdgby8qirKKN6CAF+/FH1\nEyn260e/68WLfCthyJsauQ3b29tj06ZNiIqKgoGBAcaNGydrXUrNH38A9eurftSwhgZzIa4N58/T\nAEYnJ76VyBaBAPDxoQ9VDPWiWoPy6tUrBAYGwtnZGT169ICJiQlu3rwpD21Ky7Zt9A9K2aOfa8K4\ncTRqnoMinirPli3A7NnqMS68vWk6fjYu1ItqD+Vbt26NESNGwM3NDd27d1eY/F1lUaRD+SdPgG7d\n6PjtdAsAABl5SURBVB+SuhQdmjsXqFePJghkSCYxEejeHXj6VL3GhY4OTSjKUEzknnqlpKQEGgpe\nAEORDMrChfQJ9H//41uJ/FDHybK2zJsHfPyxehndR48ABwc6LurV41sNQxIsl5cEFMWgvHkDtGwJ\nREWpthePJIYPp2lEpk7lW4nikZdHXYXv3KHjQ51wdqZph9ixq2LCcnkpMOrgEloZs2fTQ1gFsOsK\nx8GD9CBe3YwJ8N/hPBsX6gGvBiU8PBzt2rWDlZUV1q1bJ/Ga2bNnw8rKCjY2Nrhz546cFdYcQv47\njFdH+vWjmYgvXeJbiWJROi5mzeJbCT8MHkxzv924wbcShjyo1qA8fPgQ/fr1wyf/lhqMjY3FqlWr\npO5YJBLBx8cH4eHhiIuLQ1BQEOLj48tdExoaioSEBDx+/Bi7d+/GjBkzpO5XVly+TNOQKHMBLWko\ndRXdto1vJYrF+fO01vpnn/GthB80NWkCTDYu1INqDcqUKVOwevVqfPTRRwAAa2trBAUFSd1xVFQU\nLC0tIRQKoa2tDXd3d5w5c6bcNcHBweKYF3t7e+Tm5iIrK0vqvmXB1q30D0fB/RdkypdfAn/+CTx7\nxrcSxUGdXMgrY8IEICSEZiNmqDbVTn9v3ryBvb29+L1AIIC2trbUHaelpaFFixbi92ZmZkhLS6v2\nmmcKOFulpdGJVN0PHvX0aDLM3bv5VqIYPH1KtwC9vPhWwi8GBoCrK7BnD99KGLJGq7oLmjZtioSE\nBPH7kydPwoSDIhg1jWf50AOhss/5+/uL/9/R0RGOcizSvWsX4OlJJ1R156uv6AH0119TN1l1Ztcu\numpr2JBvJfzz1VfU48vXlxZpY/BDZGQkIiMjZdZ+tQZl27ZtmDp1Kh48eIDmzZujdevWOHz4sNQd\nm5qaIjU1Vfw+NTUVZmZmVV7z7NkzmJqaSmyvrEGRJ+/f0yevCxd46V7haN+e1oD55RdqZNWVd++A\nn35iec5K6diRlo7+9VfAzY1vNerLhw/b33Fc0KjaLS8LCwtERETg5cuXePjwIa5cuQKhUCh1x3Z2\ndnj8+DGSk5NRWFiIY8eOYfgHxbWHDx+OgwcPAgCuX7+ORo0awdjYWOq+ueTECcDamk6kDAo7nAeO\nHQO6dGH11csyaxbL76XqVBvYuGHDhgrbTPr6+ujSpQtsbW2l6jwsLAxz586FSCTCpEmT4Ofnh4CA\nAADAtGnTAEDsCdagQQPs378fnTt3rvgleAxs7NaNVqf7wBaqNcXFNG3/r7/SSVXdIATo2pVWs/z8\nc77VKA7FxTRG68wZWrGSwT9yj5T39PREdHQ0hg0bBkIIQkJCYG1tjadPn2LMmDFYsmQJZ2LqCl8G\n5cYNwMOD1tDW1JR79wrN2rU09ca+fXwrkT83btDtvkeP2Lj4kDVr6N+LOo4LRUTuBsXBwQFhYWFo\n+O/JYn5+PoYMGYLw8HB06dKlQuwIH/BlUMaOpU9aCxbIvWuF58ULoE0bICEBaNyYbzXyxdsbsLVl\n40ISL1/Ss5THj4GmTflWw5B76pUXL16IY1AAQFtbG1lZWdDR0UE9Nc74lplJfetZmVPJNG0KjBhB\nD6bViaws4PffaewFoyJNmgCjRgF79/KthCELqvXy8vLygr29PUaOHAlCCH777Td4enqioKAAHTp0\nkIdGhSQggHqrGBjwrURx8fEBxoyhT+rqsvWzdy/9zoaGfCtRXGbNog8bixbRLAIM1aFG2YZv3ryJ\nK1euQCAQoFevXrCzs5OHthoj7y2vwkKaPfbcOeoiy6icHj2AxYuBL77gW4nsKSoCWremKxQp/VVU\nnt69aUr/0aP5VqLe8Ja+PisrC+/evRN7fLVUoNSp8jYoQUE09uT8ebl1qbQcP05diNUhaeSRIywm\nqaYcOwbs2MHqzvON3M9QgoODYWVlBXNzczg6OkIoFMLZ2ZkzAcrIjz+qb/bY2jJqFJCcDNy6xbcS\n2UIIsGkTfepmVM+oUbQwW2ws30oYXFKtQfnmm29w7do1tGnTBklJSYiIiCiX20vdiI4GMjKAYcP4\nVqIcaGlR47t5M99KZMvVqzRNO4s7qRna2sD06SzQUdWo1qBoa2ujSZMmKCkpgUgkgpOTE6Kjo+Wh\nTSHZuhWYOZMdJtaGyZOpR1x6Ot9KZMemTcCcOerjfMAFU6cCJ08C//zDtxIGV1RrUAwMDJCXlwcH\nBwd4eXlh9uzZ4pgUdeP5cyA4GJg0iW8lyoWBAQ3027GDbyWyITmZnpswV+HaYWREM0yom2u5KlPt\noXxBQQHq1auHkpISHD58GK9fv4aXlxcaK1C0mrwO5VesAFJSmA99XXj0iHr2PH0K1K/PtxpuWbCA\n1jtZv55vJcrHrVvUAzAxkWUh5gO5H8qvWLECmpqa0NbWxvjx4zF79mz88MMPnAlQFt6+BbZvZ9HP\ndaVNG8DeHvj5Z76VcEteHhAYyJw06kqXLjRy/tgxvpUwuKBag3Lu3LkKPwsNDZWJGEUmMBDo3p1l\nFZaGuXPp4TxPeTxlwv79QL9+QKtWfCtRXhYvBn74QbXGhbpSqUHZuXMnrK2t8fDhQ1hbW4tfQqEQ\nHTt2lKdG3hGJ6HbG4sV8K1Fu+valh9Z//MG3Em4QiYAtW5irsLQMGkT/e/YsvzoY0lPpGcqrV6+Q\nk5MDX19frFu3TrzPpqurq1DnJ4Dsz1BOnKBP1qxYkvTs20c9e1RhkXv6NM2ee/26eteM54JDh+hq\njwULyxe5RcpnZ2cDoCV4JZXdNVSgZEWyNCiltS2WLaP5hxjS8e4dTVtz4YLybx86OtJYCnd3vpUo\nP0VF/9XQUbDMTiqN3AyKUCistH67QCDAkydPOBMhLbI0KBcuADNmAHFxgEa1J06MmuDvD6Sl0TQl\nysrt2/QB48kT5p3EFRs30loy7IBefvCWy4tLsrOz4ebmhqdPn0IoFOL48eNo1KhRheuEQiH09PTE\nXmZRUVES25OlQXF2ptljWewJd/zzD/X6un1beQ+zR40CPvuMOhowuCEvjybXjIqilR0ZsocXg3Lm\nzBlcunQJAoEAffr0wTAp844sXrwYTZo0weLFi7Fu3Trk5ORg7dq1Fa5r3bo1bt26Ve32mqwMSmws\nMHgwkJQEfPwx582rNX5+wKtXyhnsGBtLD5ITEwEdHb7VqBZLlwKvX9OEogzZI3eD4uvri5s3b8LL\nywuEEBw9ehR2dnZYs2ZNnTtt164dLl68CGNjY2RmZsLR0REPHjyocF3r1q0RHR1drROArAzK2LGA\ntTWgAFWOVY4XL4C2benkbGbGt5ra4eJCXchZTBL3ZGQAHTrQQFhW0VH2yN2gWFtb4+7du9D8N0mR\nSCSCra0t7t27V+dODQwMkJOTA4Ae+hsaGorfl8Xc3Bz6+vrQ1NTEtGnTMGXKFMlfQgYG5elToHNn\nukeur89p04x/WbQIeP+eZm9WFu7fB/r3p6uTBg34VqOaTJkCmJrSszaGbOF67qw2xaFAIEBubq54\nlZCbm1vpYX1ZBgwYgMzMzAo///777yu0X1l7V65cgYmJCV68eIEBAwagXbt2cHBwkHitf5nR5+jo\nCEdHx2o1VsWmTfTchBkT2bFwIfX08vMDTEz4VlMzVq0C5s9nxkSWLFwIODjQBw52n7klMjISkZGR\nMmu/2hVKUFAQfH19xRP0xYsXsXbtWrhL4SvZrl07REZGolmzZsjIyICTk5PELa+yfPfdd2jYsCEW\nSNhn4NrKZmYCn3xCt2NMTTlrliGBefNoDMfGjXwrqZ64OMDJia5O1DQ/qtwYM4a6D/v68q1EtZHb\nltfMmTPh6emJ3r17Iz09HTdv3oRAIEDXrl1hIuXj5OLFi9G4cWMsWbIEa9euRW5uboVD+Tdv3kAk\nEkFXVxcFBQUYOHAgli9fjoEDB1b8EhzflBkz6GHrhg2cNcmohPR0Wkb5wQOafVaR8fKiWv38+Fai\n+jx8SJOJPnwIKFDIm8ohN4OyefNmHDt2DOnp6XBzc4OHhwc6derESafZ2dlwdXVFSkpKObfh9PR0\nTJkyBSEhIXjy5AlGjRoFACguLoaXlxf8KvlL5vKmPHoE9OxJB7KCJQRQWWbNogZ83Tq+lVTOw4d0\nGyYxEdDV5VuNejB1KtCoEc3zxZANcj+UT05OxtGjR3Hs2DG8efMGnp6e8PDwQJs2bTgTIS1c3hQX\nF3oYz55C5cezZ4CNDZ20mzT5f3t3HxRV9cYB/Lsg9AZNjTMu4DrhbCDKezExaRkrIm4E0tCM0qCk\nZo1hlmUxjJY6zWw4jjNiZEaT5UsJ+YKiIMPACL5uCC5qwhQVFPI2aqGApCyc3x9H9we6C+zu3b0X\n7vOZ2T+Uc++e++yBs+fec84jdm3MW7SIr51Zu1bsmshHczOfZXnhAjBpkti1GZtEXdhoMBiwePFi\nXLp0CX19fYJVwl5CBeXnn4GkJD5KofUFzrV8Ob+1cd+cDUmor+ej1t9/p0kazpaRwRPbURIux3B6\nh2I0GlFUVITc3FyUlZVBo9EgOTkZ8yS0sZUQQWGMP3BNSeEpa4lz/fUXz43xyy+Al5fYtRls0SK+\nz9S6dWLXRH46OgA/P6Cigq9PIcJyWodSUlKC3NxcFBYW4rnnnkNycjISEhIkmf5XiKAcO8ang166\nRPnixZKezh/S794tdk3+78wZfhu0rg54/HGxayNPmzbxzyE/X+yajD1O61BmzZqF5ORkJCUlSWpn\nYXPsDUpfHxAezlP8JiYKWDFila4u/i10zx6+T5bYjEY+asrIoB2FxdTTw59f/fQT8PzzYtdmbBkT\nm0MKzd6g7N4NfPUVz3dCeS3EdeAAv7VkMIi/i++WLcCRI0BpKbULse3YwbOmVlTQZyEk6lDMsCco\n//0HBATwTsXCInziRIzxDTnnzBF3r6yWFiAkBDh1ircPIi6jkX8emzYBcXFi12bsoA7FDHuC8tln\nwLlzQEGBwJUiNquv57c2LlwQb6eC5GS+hboUZ53JVWEh8N57QE0N7VQgFOpQzLA1KJWVwCuv8Lwc\no23H27Huk0/49G0xki2VlvINCi9fpunjUrNkCeDqOrqTs0mJ0B2KbHMQdnfzKcLZ2dSZSFFGBu/w\nS0ud+763bwNpaUBWFnUmUpSVBZSV0R0FqZLtCOXtt/nzk507HVQpYreCAuDjj/mtL2clONPpAL2e\n/mBJ2enTfPPImhpAqRS7NqMb3fIyw9qgFBTw1K01NbS2QOri4/mGjHbkcxuxX38FZszgz9QmT3b8\n+xHbrVnDv2gcOUKzvuxBt7zs1NbGRye7d1NnMhp8+y1/juLokeTVq3z20MaN1JmMBuvW8eyOOTli\n14QMJKsRCmP8j8azz/LZXWR0qK0FoqKAffuAl14S/vw9PcCsWXzrHZ1O+PMTx6ir4wtgT5/mCx+J\n9eiWlxkjDcqXX/JvuqdPi79ojlintJTnIzl5Utg/Hv39wPz5fLudH34AXGQ3Zh/dsrP53YYTJ5z3\nnG0soVteNvr+ez4q2bOHOpPRaPZsnn43Lg64dk2482Zk8Fsn331HnclolJYG+PoCL78M3Lwpdm3I\nmP8VYowvTtuwgW/bQEPj0WvZMuDVV/nr9m37z5eTAxw8CBw6BDz8sP3nI86nUAA//sh/r6Oi+DNS\nIh5ROpR9+/YhMDAQrq6uOH/+vMVyxcXFCAgIgJ+fHzbakM7PaOR5Nvbv57uVTpliT62JFGRm8lTB\nb7wB3Lpl+3mOHAE+/RQoKpJuUi8yMq6uwLZt/IvGjBk8bw0RhygdSnBwMPLz8zFziC1l+/r6sGLF\nChQXF6O2thZ79+5FXV3diN/j1i2eLOvPP/nIxNtbiJpLX3l5udhVcCgXF37PnDE+nfjoUctlzcWi\ntRVYuBB45x0+OvHzc1xdpWSstwuFgu+ukJ7OH9RXV1suO9ZjISZROpSAgIBhUwhXVlbi6aefhq+v\nL9zc3LBgwQIcPnx42HMzBly8CERH82nBR4/Ka3qwHH5ZHn0UyM0Fvv6a57BJTOQJuu43MBa9vcDm\nzTylrErFZwhNn+68OotNDu0C4Hnot20DtFo+Aaez88EycomFGCT7DKW5uRmTBiSSVqlUaG5utli+\ntBRYuZKvIZg3D0hIAHbtAtzdnVFbIoaYGJ4QLSKCTwXPzAT++ANoaOAdzI0bPC95SQnPWV9Swmf4\nff45bS44liUm8udiubl8c9H4eL79vZCTOYh5DstNGBMTgzYzT8h0Oh3i4+OHPV5h5fLXtWt5J3L0\nKBAYSKtn5eKhh/hn//rrfLv7nBw+Fbi/n6ePzcvjuep1Ov5Fg9qFPEyfzrOw3rjBdynOzwdWrQKe\neQaIjBS7dmMYE1FUVBSrrq42+7OzZ8+y2NhY0791Oh3LzMw0W1atVjMA9KIXvehFLytearVa0L/p\nomdPZxYW1URERKC+vh6NjY3w8fFBXl4e9u7da7bs7zStgxBCRCfKM5T8/HxMmjQJer0ecXFx0Gq1\nAICWlhbE3U3HNm7cOGRnZyM2NhbTpk3D/PnzMXXqVDGqSwghZATGxNYrhBBCxCe5WV4jWcy4cuVK\n+Pn5ITQ0FAaDYdhj//nnH8TExMDf3x9z5sxBR0eHw69DCI6Ixfr166FSqRAeHo7w8HAUFxc7/DqE\nYE8slixZAqVSieDg4EHl5dguLMVCbu2iqakJGo0GgYGBCAoKwtatW03l5dYuhoqF1e1C0CcydjIa\njUytVrOGhgZ2584dFhoaympraweVKSwsZFqtljHGmF6vZ5GRkcMe+9FHH7GNGzcyxhjLzMxk6enp\nTrwq2zgqFuvXr2ebN2927sXYyZ5YMMbYiRMn2Pnz51lQUNCgY+TWLhizHAu5tYvW1lZmMBgYY4x1\ndnYyf39/VldXxxiTX7sYKhbWtgtJjVBGspixoKAAqampAIDIyEh0dHSgra1tyGMHHpOamopDhw45\n98Js4KhYAJYnQkiVPbEAgBdffBFPPvnkA+eVW7sALMcCkE+7aG9vh5eXF8LCwgAAHh4emDp1qmmd\nm5zaxXCxAKxrF5LqUEaymNFSmZaWFovHtre3Q3k3V6hSqUR7e7sjL0MQjooFAHzxxRcIDQ3F0qVL\nR8Vw3p5YDEVu7WI4cmkXV65cGVSmsbERBoMBkXcXqMipXQwXC8C6diGpDmWkixlH0mMyxsyeT6FQ\nWL1oUgxCxmKg5cuXo6GhATU1NfD29saHH35oS/WcytZYWPM5j/V2Mdxxcm0XXV1deO2115CVlQUP\nM9snyKldmIuFte1CUh3KxIkT0dTUZPp3U1MTVCrVkGWuXLkClUpl9v8nTpwIgH/LuDfkb21txYQJ\nExx5GYIQMhYDj50wYYLpl+TNN99EZWWlg6/EfrbG4t7nb4mc2sVwsZBju+jt7UVSUhJSUlKQmJho\nKiPHdmEpFta2C0l1KAMXM965cwd5eXlISEgYVCYhIQG7du0CAOj1ejzxxBNQKpVDHpuQkICdd5OS\n79y5c1DApMpRsWhtbTUdn5+f/8BsHymyJxZDkVu7GIrc2gVjDEuXLsW0adPw/vvvP3CMnNrFULGw\nul3YOqvAUYqKipi/vz9Tq9VMp9Mxxhjbvn072759u6lMWloaU6vVLCQkZNDWLeaOZYyx69evs+jo\naObn58diYmLYv//+67wLsoMjYrFw4UIWHBzMQkJC2Lx581hbW5vzLsgO9sRiwYIFzNvbm7m7uzOV\nSsV27NjBGJNnu7AUC7m1i5MnTzKFQsFCQ0NZWFgYCwsLY8eOHWOMya9dDBULa9sFLWwkhBAiCEnd\n8iKEEDJ6UYdCCCFEENShEEIIEQR1KIQQQgRBHQohhBBBUIdCCCFEENShEGLG9evXTVt2e3t7m7bw\n9vT0xIoVKwR7n9WrV6OiogIAsGXLFvT09Jh+Fh0djc7OTsHeixBHo3UohAxjw4YN8PT0xAcffCDo\neTs7OxEdHW3azmLy5MmoqqrC+PHjAQDffPMNOjs7BX9fQhyFRiiEjMC9713l5eWIj48HwJMPpaam\nYubMmfD19cXBgwexevVqhISEQKvVwmg0AgCqq6sRFRWFiIgIzJ0717RP1OHDhzF79mwAwNatW9HS\n0gKNRoPo6GgAQHx8PHJzc519qYTYjDoUQuzQ0NCA48ePo6CgACkpKYiJicHFixfxyCOPoLCwEL29\nvXj33Xdx4MABVFVVYfHixVizZg0A4NSpU4iIiADAM+n5+PigvLwcZWVlAAAvLy9cu3YN3d3dol0f\nIdYYJ3YFCBmtFAoFtFotXF1dERQUhP7+fsTGxgIAgoOD0djYiN9++w2XL182jUT6+vrg4+MDAPj7\n77/h7e095HsolUo0NTUhICDAsRdDiACoQyHEDu7u7gAAFxcXuLm5mf7fxcUFRqMRjDEEBgbizJkz\nZo/v7+8f8vzMQl4fQqSIbnkRYqORzGeZMmUKrl69Cr1eD4DnnaitrQUAPPXUU6bnKQDg6emJmzdv\nDjq+vb39gbwWhEgVdSiEjMC9UcLADH73Z/O7fyShUCjg5uaG/fv3Iz09HWFhYQgPD8fZs2cBAC+8\n8AKqqqpM5d966y3MnTvX9FC+ra0N48ePx2OPPebQayNEKDRtmBCRdHV1QaPR4Ny5c2Z/npOTg+7u\nbqxatcrJNSPENjRCIUQkHh4e0Gg0OH78uNmf5+XlYdmyZU6uFSG2oxEKIYQQQdAIhRBCiCCoQyGE\nECII6lAIIYQIgjoUQgghgqAOhRBCiCCoQyGEECKI/wE9i3T1dLT66gAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0xb17f280c>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4 Page No.171"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%matplotlib inline\n",
+      "import matplotlib.pylab as plt\n",
+      "import numpy as np\n",
+      "\n",
+      "# Given Data\n",
+      "R1 = 10*10**3\n",
+      "Rf = 100*10**3\n",
+      "Cf = 10*10**-9\n",
+      "\n",
+      "# Solution - Finding the frequency limit of integration fa\n",
+      "\n",
+      "fa = int(1/(2*np.pi*Rf*Cf))\n",
+      "\n",
+      "print \"The value of lower frequency limit of integration is =\",fa,\"Hz\"\n",
+      "\n",
+      "# Studying the response of sine wave input \n",
+      "\n",
+      "V = 1\n",
+      "f = 5*10**3\n",
+      "\n",
+      "vo = round(1/(R1*Cf*2*np.pi*5000),3)\n",
+      "x = np.linspace(-np.pi,np.pi,201)\n",
+      "plt.plot(x,vo*np.cos(2*np.pi*f*x))\n",
+      "plt.xlabel('Time')\n",
+      "plt.ylabel('Voltage')\n",
+      "plt.show()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of lower frequency limit of integration is = 159 Hz\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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Ji9PyQjwH73ctXkao3OpJeJlJfBBWN/ltQxCWwAZhTCYLLBIu4pWD1lFuogpV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\n/H+jHCBTOHebZwAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0xb278fbec>"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 Page No.173 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Given data\n",
+      "\n",
+      "w = 10000\n",
+      "Apeak = 20\n",
+      "A = 3\n",
+      "C = 0.1*10**-6\n",
+      "\n",
+      "# Solution\n",
+      "\n",
+      "Rf = math.sqrt(1/((10**-4)**2))\n",
+      "R1 = Rf/10\n",
+      "\n",
+      "# Displaying the values \n",
+      "\n",
+      "print \"The value of Rf = \",int(Rf/10**3),\"Kilo Ohms\"\n",
+      "print \"The value of R1 = \",int(R1/10**3),\"Kilo Ohms\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Rf =  10 Kilo Ohms\n",
+        "The value of R1 =  1 Kilo Ohms\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9 Page No.178"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import numpy as np\n",
+      "# defining two functions \n",
+      "\n",
+      "def x(t):\n",
+      "    \n",
+      "    return 10*np.sin(3*t)\n",
+      "\n",
+      "def x1(t):\n",
+      "\n",
+      "    return 30*np.cos(3*t)\n",
+      "\n",
+      "print \" The value of x(0) = \",int(x(0))\n",
+      "print \" The value of defferential x(0) =\",int(x1(0))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of x(0) =  0\n",
+        " The value of defferential x(0) = 30\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Linear_Integrated_Circuits/Chapter5.ipynb b/Linear_Integrated_Circuits/Chapter5.ipynb
new file mode 100755
index 00000000..29abc9bb
--- /dev/null
+++ b/Linear_Integrated_Circuits/Chapter5.ipynb
@@ -0,0 +1,215 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Comparators and Waveform generators"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 Page No.212"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%matplotlib inline\n",
+      "import matplotlib.pylab as plt\n",
+      "import numpy as np\n",
+      "# Given Data \n",
+      "\n",
+      "Vz1 = 9.0\n",
+      "Vz2 = 9.0\n",
+      "\n",
+      "x  = np.array([-1, -1, -1, -1, -1, 0, 1, 1, 1, 1, 1]) \n",
+      "plt.plot(9.7*x)\n",
+      "plt.xlabel('Time')\n",
+      "plt.ylabel('Voltage')\n",
+      "plt.show()\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0xb659de2c>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2 Page No.215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "\n",
+      "R2 = 100.0\n",
+      "R1 = 50.0*10**3\n",
+      "Vref = 0\n",
+      "vi = 1\n",
+      "Vsat = 14.0\n",
+      "\n",
+      "# Solution \n",
+      "\n",
+      "VUT = (R2*Vsat)/(R1+R2)\n",
+      "VLT = (R2*-Vsat)/(R1+R2)\n",
+      "\n",
+      "# Displaying the values\n",
+      "\n",
+      "print \"The value of Upper threshold voltage =  \",int(round(VUT*10**3,0)),\"mV\"\n",
+      "print \"The value of Lower threshold voltage = \",int(round(VLT*10**3,0)),\"mV\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Upper threshold voltage =   28 mV\n",
+        "The value of Lower threshold voltage =  -28 mV\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 Page No.216"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import numpy as np \n",
+      "# Given data\n",
+      "\n",
+      "VUT = 0\n",
+      "VH = 0.2\n",
+      "f = 10.0**3\n",
+      "# Solution \n",
+      "\n",
+      "VLT = VUT - VH\n",
+      "theta = round(np.arcsin(VH/2),1)\n",
+      "T = 1/f\n",
+      "Ttheta = theta/(2*np.pi*f) \n",
+      "T1 = (T/2) + Ttheta\n",
+      "T2 = (T/2) - Ttheta \n",
+      "\n",
+      "# Displayig the values \n",
+      "\n",
+      "print \"The value of Lower threshold voltage = \",VLT,\"V\"\n",
+      "print \"The value of angle theta             = \",theta,\"Radian\"\n",
+      "print \"The value of Timeperiode             = \",round(Ttheta*10**3,3),\"ms\"\n",
+      "print \"The value of T1                      = \",round(T1*10**3,3),\"ms\"\n",
+      "print \"The value of T2                      = \",round(T2*10**3,3),\"ms\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Lower threshold voltage =  -0.2 V\n",
+        "The value of angle theta             =  0.1 Radian\n",
+        "The value of Timeperiode             =  0.016 ms\n",
+        "The value of T1                      =  0.516 ms\n",
+        "The value of T2                      =  0.484 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4 Page No.225 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Given data \n",
+      "\n",
+      "f = 100\n",
+      "C = 0.1*10**-6\n",
+      "\n",
+      "# Solution \n",
+      "\n",
+      "R = 1/(math.sqrt(6)*2*math.pi*(10**-7)*100)\n",
+      "R1 = 10*R\n",
+      "Rf = 29*R1\n",
+      "\n",
+      "# Displaying the solutions \n",
+      "\n",
+      "print \"The value of R  =\",round(R*10**-3,3),\"Kilo Ohms\"\n",
+      "print \"The value of R1 =\",int(round(R1*10**-3,0)),\"Kilo Ohms\"\n",
+      "print \"The value of Rf =\",int(round(Rf*10**-3,0)),\"Kilo Ohms\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of R  = 6.497 Kilo Ohms\n",
+        "The value of R1 = 65 Kilo Ohms\n",
+        "The value of Rf = 1884 Kilo Ohms\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Linear_Integrated_Circuits/Chapter6.ipynb b/Linear_Integrated_Circuits/Chapter6.ipynb
new file mode 100755
index 00000000..4ddd3b93
--- /dev/null
+++ b/Linear_Integrated_Circuits/Chapter6.ipynb
@@ -0,0 +1,147 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : Voltage Regulators"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1 Page No.246"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "\n",
+      "Veb = 1.0\n",
+      "beta = 15.0\n",
+      "Vc = 5.0\n",
+      "Io = 1.0   # for 7805 maximum value of I b is 1 Ampere \n",
+      "R1 = 7.0\n",
+      "# Solution \n",
+      "\n",
+      "# for Load = 100ohms \n",
+      "\n",
+      "Rl = 100.0\n",
+      "\n",
+      "Il=Ic=Ii = Vc/Rl\n",
+      "# voltage across R1\n",
+      "V1 = R1*(50*10**-3)\n",
+      "# for load = 5 ohms\n",
+      "\n",
+      "Rl1 = 5.0\n",
+      "Il1 = Vc/Rl1\n",
+      "V12 = Il1 * Rl1 # Finding the Voltage drop across R1 when Rl = 5 Ohms \n",
+      "# Finding the value of Io\n",
+      "Io = (Il1 + (beta*(Veb/R1)))/(beta + 1)\n",
+      "Ic = beta*(Io - (Veb/R1))\n",
+      "\n",
+      "# for load = 1 Ohm\n",
+      "\n",
+      "Rl2 = 1.0\n",
+      "Il2 = Vc/Rl2\n",
+      "# Finding the value of Io\n",
+      "Io1 = (Il2 + (beta*(Veb/R1)))/(beta + 1)\n",
+      "Ic1 = beta*(Io1 - (Veb/R1))\n",
+      "\n",
+      "print \" The value of load current when Rl = 100 ohms is =\",int(Il*10**3),\"mA\"\n",
+      "print \" The voltage across R1 when load is 100 ohms     =\",int(V1*10**3),\"mV\"\n",
+      "print \" The value of load current when Rl = 5 ohms      =\",int(Il1),\"A\"\n",
+      "print \" The voltage across R1 when load is 5 Ohms       =\",int(V12),\"V\"\n",
+      "print \" The value of output current when load is 5 ohms =\",int(Io*10**3),\"mA\"\n",
+      "print \" The value of collector current is               =\",int(round(Ic*10**3)),\"mA\"\n",
+      "print \" The value of output current when load is 1 Ohm  =\",int(Io1*10**3),\"mA\"\n",
+      "print \" The value of collector current is               =\",round(Ic1,2),\"A\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of load current when Rl = 100 ohms is = 50 mA\n",
+        " The voltage across R1 when load is 100 ohms     = 350 mV\n",
+        " The value of load current when Rl = 5 ohms      = 1 A\n",
+        " The voltage across R1 when load is 5 Ohms       = 5 V\n",
+        " The value of output current when load is 5 ohms = 196 mA\n",
+        " The value of collector current is               = 804 mA\n",
+        " The value of output current when load is 1 Ohm  = 446 mA\n",
+        " The value of collector current is               = 4.55 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2 Page No.247"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "\n",
+      "Vo = 7.5\n",
+      "Iq = 4.2*10**-3\n",
+      "Ir1 = 25*10**-3\n",
+      "Vr = 5\n",
+      "\n",
+      "# Solution \n",
+      "\n",
+      "R1 = Vr/Ir1\n",
+      "R2 = 2.5/(Ir1 + Iq)\n",
+      "\n",
+      "# Displaying the outputs \n",
+      "\n",
+      "print \"The value of R1 =\",int(R1),\"Ohms\" \n",
+      "print \"The value of R2 =\",int(R2),\"Ohms\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of R1 = 200 Ohms\n",
+        "The value of R2 = 85 Ohms\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Linear_Integrated_Circuits/Chapter7.ipynb b/Linear_Integrated_Circuits/Chapter7.ipynb
new file mode 100755
index 00000000..9a6ddc8f
--- /dev/null
+++ b/Linear_Integrated_Circuits/Chapter7.ipynb
@@ -0,0 +1,448 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Active Filters"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1 Page No.269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import numpy as np\n",
+      "import math\n",
+      "# Given Data\n",
+      "fh = 10**3\n",
+      "C = 0.1*10**-6\n",
+      "Rf = 5.86*10**3\n",
+      "Ri = 10**4\n",
+      "\n",
+      "# Solution \n",
+      "\n",
+      "R = 1/(2*np.pi*C*fh)\n",
+      "Ao = (1 + Rf/Ri)\n",
+      "\n",
+      "a = np.array([100, 200, 500, 1000, 5000, 10000])\n",
+      "print \" The value of R is =\",round(R/1000,1),\"Kilo ohms\"\n",
+      "print \" The value of Ao   =\",Ao\n",
+      "print \" Frequency in Hz            Gain magnitude in dB 2o log (Vo/Vi)\"\n",
+      "print \" ========================================================\" \n",
+      "for i in a:\n",
+      "\n",
+      "    val = round(20 * math.log10(Ao/(math.sqrt(1 + (i/fh)**4))),3)\n",
+      "    print i,\"            ||               \",val\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of R is = 1.6 Kilo ohms\n",
+        " The value of Ao   = 1.586\n",
+        " Frequency in Hz            Gain magnitude in dB 2o log (Vo/Vi)\n",
+        " ========================================================\n",
+        "100             ||                4.006\n",
+        "200             ||                4.006\n",
+        "500             ||                4.006\n",
+        "1000             ||                0.996\n",
+        "5000             ||                -23.96\n",
+        "10000             ||                -35.994\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2 Page No.270"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Design of a 4th order Butter worth Low pss filter \n",
+      "\n",
+      "fh = 10**3\n",
+      "C = 0.1*10**-6\n",
+      "a1 = 0.765\n",
+      "a2 = 1.848\n",
+      "\n",
+      "Ao1 = 3 - a1\n",
+      "Ao2 = 3 - a2\n",
+      "\n",
+      "# let Rf1 = 12.35 kilo Ohm and Rf2 = 15.2 kilo ohm finding the Ri1 and Ri2 \n",
+      "\n",
+      "Rf1 = 12.35*10**3\n",
+      "Rf2 = 15.2*10**3\n",
+      "\n",
+      "Ri1 = Rf1/(Ao1 - 1)\n",
+      "Ri2 = Rf2/(Ao2 - 1)\n",
+      "\n",
+      "print \" The value of Ao1 = \",Ao1\n",
+      "print \" The value of Ao2 = \",Ao2\n",
+      "print \" The value of Rf1 = \",Rf1/1000,\"Kilo Ohm\"\n",
+      "print \" The value of Rf2 = \",Rf2/1000,\"Kilo Ohm\"\n",
+      "print \" The value of Ri1 = \",int(Ri1/1000),\"Kilo Ohm\"\n",
+      "print \" The value of Ri2 = \",int(Ri2/1000),\"Kilo Ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of Ao1 =  2.235\n",
+        " The value of Ao2 =  1.152\n",
+        " The value of Rf1 =  12.35 Kilo Ohm\n",
+        " The value of Rf2 =  15.2 Kilo Ohm\n",
+        " The value of Ri1 =  10 Kilo Ohm\n",
+        " The value of Ri2 =  100 Kilo Ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3 Page No.271 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Solution to fing out the value of n or order\n",
+      "\n",
+      "# We know the equation is 20 log|(H(jw)| = 20 log (Ao/(math.sqrt(1+(w/wh)**n))\n",
+      "# putig the values into the equation we will get 0.01**2 = 1/(1 + 2**2*n)\n",
+      "# solving the equation to get the value of n\n",
+      "\n",
+      "# 2**(2*n) = 1/(1 + (0.01)**2)\n",
+      "# taking log to the base 2 on both sides\n",
+      "n =  math.log(((10**4)-1),2)/2\n",
+      "\n",
+      "print \" The order of the filter will be =\",int(math.ceil(n))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The order of the filter will be = 7\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4 Page No.272"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import numpy as np\n",
+      "import math\n",
+      "# Given Data\n",
+      "fl = 10**3\n",
+      "C = 0.1*10**-6\n",
+      "Rf = 5.86*10**3\n",
+      "Ri = 10**4\n",
+      "\n",
+      "# Solution \n",
+      "\n",
+      "R = 1/(2*np.pi*C*fl)\n",
+      "Ao = (1 + Rf/Ri)\n",
+      "\n",
+      "a = np.array([100, 200, 500, 1000, 5000, 10000])\n",
+      "print \" The value of R is =\",round(R/1000,1),\"Kilo ohms\"\n",
+      "print \" The value of Ao   =\",Ao\n",
+      "print \" Frequency in Hz            Gain magnitude in dB 2o log (Vo/Vi)\"\n",
+      "print \" ========================================================\" \n",
+      "for i in a:\n",
+      "\n",
+      "    val = round(Ao/math.sqrt(1 + (fl/i)**4),3)\n",
+      "    print i,\"            ||               \",val\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of R is = 1.6 Kilo ohms\n",
+        " The value of Ao   = 1.586\n",
+        " Frequency in Hz            Gain magnitude in dB 2o log (Vo/Vi)\n",
+        " ========================================================\n",
+        "100             ||                0.016\n",
+        "200             ||                0.063\n",
+        "500             ||                0.385\n",
+        "1000             ||                1.121\n",
+        "5000             ||                1.586\n",
+        "10000             ||                1.586\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5 Page No.276"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Given Data \n",
+      "\n",
+      "fl = 400\n",
+      "fh = 2*10**3\n",
+      "Ao = 4\n",
+      "\n",
+      "# For design top get  Ao = 2 the values of Rf = Ri\n",
+      "\n",
+      "Rf = Ri = 10*10**3\n",
+      "\n",
+      "# for LPF \n",
+      "\n",
+      "C1 = 0.01*10**-6\n",
+      "R1 = 1/(2*math.pi*fh*C1)\n",
+      "\n",
+      "# for HPF \n",
+      "\n",
+      "C2 = 0.01*10**-6\n",
+      "R2 = 1/(2*math.pi*fl*C2)\n",
+      "\n",
+      "fo = math.sqrt(fh*fl)\n",
+      "Q = fo/(fh - fl)\n",
+      "\n",
+      "print \"The value of c1 =\",C1*10**6,\"uF\"\n",
+      "print \"The value of R1 =\",round(R1/1000,1),\"kilo Ohms\"\n",
+      "print \"The value of c2 =\",C2*10**6,\"uF\"\n",
+      "print \"The value of R2 =\",round(R2/1000,1),\"Kilo Ohms\"\n",
+      "print \"The value of Q  =\",round(Q,2)\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of c1 = 0.01 uF\n",
+        "The value of R1 = 8.0 kilo Ohms\n",
+        "The value of c2 = 0.01 uF\n",
+        "The value of R2 = 39.8 Kilo Ohms\n",
+        "The value of Q  = 0.56\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6 Page No.279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Given Data \n",
+      "\n",
+      "fo = 50\n",
+      "C = 0.1*10**-6\n",
+      "\n",
+      "# Solution \n",
+      "\n",
+      "R = 1/(2*math.pi*fo*C)\n",
+      "\n",
+      "print \"The value of Capasitor = \",C*10**6,\"uF\"\n",
+      "print \"The value of Resistor  = \",round(R/1000,1),\"Kilo Ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Capasitor =  0.1 uF\n",
+        "The value of Resistor  =  31.8 Kilo Ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7 Page No.280"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Given Data \n",
+      "\n",
+      "fl = 400\n",
+      "fh = 2*10**3\n",
+      "Ao = 4\n",
+      "\n",
+      "# For design top get  Ao = 2 the values of Rf = Ri\n",
+      "\n",
+      "Rf = Ri = 10*10**3\n",
+      "\n",
+      "# for LPF \n",
+      "\n",
+      "C1 = 0.1*10**-6\n",
+      "R1 = 1/(2*math.pi*fh*C1)\n",
+      "\n",
+      "# for HPF \n",
+      "\n",
+      "C2 = 0.1*10**-6\n",
+      "R2 = 1/(2*math.pi*fl*C2)\n",
+      "\n",
+      "fo = math.sqrt(fh*fl)\n",
+      "Q = fo/(fh - fl)\n",
+      "\n",
+      "print \"The value of c1 =\",C1*10**6,\"uF\"\n",
+      "print \"The value of R1 =\",round(R1/1000,1),\"kilo Ohms\"\n",
+      "print \"The value of c2 =\",C2*10**6,\"uF\"\n",
+      "print \"The value of R2 =\",round(R2/1000),\"Kilo Ohms\"\n",
+      "print \"The value of Q  =\",round(Q,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of c1 = 0.1 uF\n",
+        "The value of R1 = 0.8 kilo Ohms\n",
+        "The value of c2 = 0.1 uF\n",
+        "The value of R2 = 4.0 Kilo Ohms\n",
+        "The value of Q  = 0.56\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.10 Page No.298 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data \n",
+      "\n",
+      "fc = 400\n",
+      "Ao = -2\n",
+      "Vcc = 5\n",
+      "R1 = 10*10**3\n",
+      "HoLP = 2\n",
+      "\n",
+      "R2 = HoLP * R1 \n",
+      "\n",
+      "# for second order butterworth filter Q = 0.707\n",
+      "Q = 0.707\n",
+      "R3 = Q * R2\n",
+      "\n",
+      "fclock = 50 * fc\n",
+      "\n",
+      "# Dispalying the outputs \n",
+      "\n",
+      "print \"The value of R1 =\",R1/1000,\"Ohms\"\n",
+      "print \"The value of R2 =\",R2/1000,\"Kilo Ohms\"\n",
+      "print \"The value of R3 =\",R3/1000,\"Kilo Ohms\"\n",
+      "print \"The value of clock frequency =\",fclock/1000,\"Kilo Hertz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of R1 = 10 Ohms\n",
+        "The value of R2 = 20 Kilo Ohms\n",
+        "The value of R3 = 14.14 Kilo Ohms\n",
+        "The value of clock frequency = 20 Kilo Hertz\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Linear_Integrated_Circuits/Chapter8.ipynb b/Linear_Integrated_Circuits/Chapter8.ipynb
new file mode 100755
index 00000000..fe8ef90b
--- /dev/null
+++ b/Linear_Integrated_Circuits/Chapter8.ipynb
@@ -0,0 +1,125 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 - 555 Timer "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1 Page No.315"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "\n",
+      "R = 100*10**3\n",
+      "T = 100*10**-3\n",
+      "\n",
+      "# Solution \n",
+      "\n",
+      "C = T/(1.1*R)\n",
+      "\n",
+      "print \"The value of C =\",round(C*10**6,1),\"uF\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of C = 0.9 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2 Page No.322"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given Data\n",
+      "\n",
+      "Ra = 6.8*10**3\n",
+      "Rb = 3.3*10**3\n",
+      "C = 0.1*10**-6\n",
+      "\n",
+      "# Solution \n",
+      "# Solution of part a\n",
+      "\n",
+      "tHIGH = round(0.69*(Ra + Rb)*C,5)*10**3\n",
+      "\n",
+      "# Solution pf part b\n",
+      "\n",
+      "tLOW = round(0.69*(Rb)*C,5)*10**3\n",
+      "\n",
+      "# Solution of part c\n",
+      "\n",
+      "f = int(1.45/((Ra + 2*Rb)*C))*10**-3\n",
+      "\n",
+      "# Solution for part d\n",
+      "\n",
+      "D = round(Rb/(Ra + 2*Rb),2)\n",
+      "\n",
+      "\n",
+      "# Displaying the answers \n",
+      "\n",
+      "print \"The value of tHIGH        =\",tHIGH,\"ms\"\n",
+      "print \"The value of tLOW         =\",tLOW,\"ms\"\n",
+      "print \"The value of frequency f  =\",f,\"kHz\"\n",
+      "print \"The value of Duty cycle D =\",D,\"or\",D*100,\"%\"  \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of tHIGH        = 0.7 ms\n",
+        "The value of tLOW         = 0.23 ms\n",
+        "The value of frequency f  = 1.082 kHz\n",
+        "The value of Duty cycle D = 0.25 or 25.0 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Linear_Integrated_Circuits/README.txt b/Linear_Integrated_Circuits/README.txt
index f34d66f4..59462772 100755
--- a/Linear_Integrated_Circuits/README.txt
+++ b/Linear_Integrated_Circuits/README.txt
@@ -1,10 +1,10 @@
-Contributed By: Sagar Shinde
-Course: btech
-College/Institute/Organization: Vishwakarma institute of Technology, Pune.
-Department/Designation: ENTC
+Contributed By: Aswin  Venu
+Course: be
+College/Institute/Organization: Pavai College of technology 
+Department/Designation: ECE
 Book Title: Linear Integrated Circuits
-Author: S Shalivahanan,   V S  Kanchana Bhaaskaran 
-Publisher: Tata McGraw-Hill  Publishing Company Limited, New Dehli
-Year of publication: 2008
-Isbn: 0-07-064818-2
-Edition: 1st
\ No newline at end of file
+Author: D.Roy Choudhury, Shail B. Jain
+Publisher: New age international publishers,New Delhi
+Year of publication: 2012
+Isbn: 9788122430981
+Edition: 4th
\ No newline at end of file
diff --git a/Linear_Integrated_Circuits/screenshots/design_of_voltage_regulator.png b/Linear_Integrated_Circuits/screenshots/design_of_voltage_regulator.png
new file mode 100755
index 00000000..bc470f78
--- /dev/null
+++ b/Linear_Integrated_Circuits/screenshots/design_of_voltage_regulator.png
diff --git a/Linear_Integrated_Circuits/screenshots/filter_design.png b/Linear_Integrated_Circuits/screenshots/filter_design.png
new file mode 100755
index 00000000..8823d86e
--- /dev/null
+++ b/Linear_Integrated_Circuits/screenshots/filter_design.png
diff --git a/Linear_Integrated_Circuits/screenshots/op-amp_differentator_design.png b/Linear_Integrated_Circuits/screenshots/op-amp_differentator_design.png
new file mode 100755
index 00000000..4edec474
--- /dev/null
+++ b/Linear_Integrated_Circuits/screenshots/op-amp_differentator_design.png