Added(A)/Deleted(D) following books

 A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER01.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER02.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER03.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER04.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER07.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER09.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER11.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER15.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER16.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER18.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER19.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER20.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER23.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER24.ipynb
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture02.png
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture04.png
A  Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture20.png
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_4.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_4.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_4.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_4.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_4.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_4.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_4.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_4.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_4.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex1.2_1.png
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex3.7_1.png
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex6.7_1.png
A  Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter02_1.ipynb
A  Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter03_1.ipynb
A  Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter04_1.ipynb
A  Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter05_1.ipynb
A  Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter06_1.ipynb
A  Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter07_1.ipynb
A  Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter08_1.ipynb
A  Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter09_1.ipynb
A  Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter10_1.ipynb
A  Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter11_1.ipynb
A  Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/screenshots/Capture02_1.png
A  Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/screenshots/Capture04_1.png
A  Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/screenshots/Capture10_1.png
A  "sample_notebooks/AjayKumar Verma/Chapter02.ipynb"
A  sample_notebooks/Haseen/Ch2.ipynb
A  sample_notebooks/karansingh/Ch4.ipynb

3 files changed, 1114 insertions, 0 deletions

diff --git a/sample_notebooks/AjayKumar Verma/Chapter02.ipynb b/sample_notebooks/AjayKumar Verma/Chapter02.ipynb
new file mode 100644
index 00000000..d4d9f66e
--- /dev/null
+++ b/sample_notebooks/AjayKumar Verma/Chapter02.ipynb
@@ -0,0 +1,265 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:dbdab30ceee2e893077ec98e42cc102488d52b9f8506602ce6aaa24ef3e22c61"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter02 : Op-amp Fundamentals"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 2.2 : page 79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division \n",
+      "#given data\n",
+      "Iio=20 #in nA\n",
+      "IB=100 #in nA\n",
+      "# Eqn(1) : Iio=IB1-IB2=20\n",
+      "#Eqn(2) : 2*IB=IB1+IB2=200\n",
+      "IB1=(200+20)/2 #in nA\n",
+      "print \"IB1 = %0.f nA\"%IB1\n",
+      "IB2=IB1-Iio #in nA\n",
+      "print \"IB2 = %0.f nA\"%IB2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "IB1 = 110 nA\n",
+        "IB2 = 90 nA\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 2.3 : page 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "G=120 #unitless\n",
+      "To=20 #in degree centigrade\n",
+      "T=50 #in degree centigrade\n",
+      "Dvoff=0.13 #in mV/degree centigrade\n",
+      "#input change\n",
+      "dVin=Dvoff*(T-To) #in mVolt\n",
+      "#output change\n",
+      "Vo=G*dVin #in mVolt\n",
+      "print \"Output voltage = %0.f mV\" %Vo"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output voltage = 468 mV\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 2.4 : page 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "dt=5 #in uSec\n",
+      "Vp=5 #in Volt\n",
+      "dV=(0.9-0.1)*Vp\n",
+      "SR=dV/dt #in V/uSec\n",
+      "print \"Calculated SR = %0.2f V/uSec\"%SR"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Calculated SR = 0.80 V/uSec\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 2.5 : page 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "Vo=10 #in Volt\n",
+      "SR=1 #in V/uSec\n",
+      "dV=(0.9-0.1)*Vo\n",
+      "dt=dV/SR #in uSec\n",
+      "print \"Rise time = %0.f uSec\"%dt"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rise time = 8 uSec\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 2.6 : page 84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "V1=-5 #in Volt\n",
+      "V2=5 #in Volt\n",
+      "SR=0.5 #in V/uSec\n",
+      "dV=V2-V1 # in Volt\n",
+      "dt=dV/SR #in uSec\n",
+      "print \"Rise time = %0.f uSec\"%dt"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rise time = 20 uSec\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 2.7 : page 84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import numpy as np\n",
+      "#given data\n",
+      "fm=50 #in kHz\n",
+      "SR=0.5 #in V/uSec\n",
+      "#formula : SR=2*pie*fm*Vm\n",
+      "Vm=(SR*10**6)/(2*np.pi*fm*10**3) #in Volts\n",
+      "print \"Maximum voltage = %0.2f Volt \"%Vm"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum voltage = 1.59 Volt \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 2.8 : page 84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import numpy as np\n",
+      "#given data\n",
+      "SR=6 #in V/uSec\n",
+      "#formula : SR=2*pie*fm*Vm\n",
+      "# part (i) Vm=1 volt\n",
+      "Vm=1 #in Volts\n",
+      "fm=((SR*10**6)/(2*np.pi*Vm))/1000 #in kHz\n",
+      "print \"when Vm=1 volt the limiting frequency = %0.f kHz\"%fm\n",
+      "# part (ii) Vm=10 volt\n",
+      "Vm=10 #in Volts\n",
+      "fm=((SR*10**6)/(2*np.pi*Vm))/1000 #in kHz\n",
+      "print \"when Vm=10 Volt the limiting frequency = %0.1f kHz\"%fm"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when Vm=1 volt the limiting frequency = 955 kHz\n",
+        "when Vm=10 Volt the limiting frequency = 95.5 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/sample_notebooks/Haseen/Ch2.ipynb b/sample_notebooks/Haseen/Ch2.ipynb
new file mode 100644
index 00000000..7ffe554f
--- /dev/null
+++ b/sample_notebooks/Haseen/Ch2.ipynb
@@ -0,0 +1,452 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:5f773c85e1dfab84f7a94d7a8dae80fafd68f7320fe878d3eb9f67c265c3b4c6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Ch2 : The p-n junction diode"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1: Page 185"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "print \"Part (a)\" \n",
+      "# Applying Thevnin's theorem at XX', in Fig. 2.5(a)\n",
+      "Vth=15*20e3/(10e3+20e3)  # Thevnin equivalent voltage in volts\n",
+      "Zth=10e3*20e3/(10e3+20e3)  # Thevnin equivalent resistance in ohms\n",
+      "# From the figure 2.5(c)\n",
+      "I=Vth/(Zth+20e3)  # Labelled current in amperes\n",
+      "Vo=I*20e3  # Labelled voltage in volts\n",
+      "I=I*1e3  # Labelled current in miliamperes\n",
+      "print \"Labelled current I = %0.2f mA\"%I \n",
+      "print \"Labelled voltage Vo = %0.2f V\" %Vo\n",
+      "\n",
+      "print \"Part (b)\" \n",
+      "# Applying Thevnin's theorem at XX' and YY', in Fig. 2.5(b)\n",
+      "Vth1=15*10e3/(10e3+10e3)  # Thevnin equivalent voltage at XX' in volts\n",
+      "Zth1=10e3*10e3/(10e3+10e3)  # Thevnin equivalent resistance at YY' in ohms\n",
+      "Vth2=5  # Thevnin equivalent voltage at YY' in volts\n",
+      "Zth2=5e3  # Thevnin equivalent resistance at YY' in ohms\n",
+      "# From the figure 2.5(d)\n",
+      "I=0  # Labelled current in amperes\n",
+      "Vo=5-7.5  # Labelled voltage in volts\n",
+      "print \"Labelled current I = %0.2f mA\"%I \n",
+      "print \"Labelled voltage Vo = %0.2f V\" %Vo "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (a)\n",
+        "Labelled current I = 0.38 mA\n",
+        "Labelled voltage Vo = 7.50 V\n",
+        "Part (b)\n",
+        "Labelled current I = 0.00 mA\n",
+        "Labelled voltage Vo = -2.50 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2: Page 186"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "ID1=1  # Let the initial diode current be 1 A\n",
+      "ID2=15*ID1  # Final diode current\n",
+      "VT=25e-3  # Voltage equivalent to temperatue at room temperature in volts\n",
+      "eta=1  # for Ge\n",
+      "deltaVD=eta*VT*log(ID2/ID1)  # Change in diode voltage in volts\n",
+      "deltaVD=deltaVD*1e3  # Change in diode voltage in milivolts\n",
+      "print \"Change in diode voltage (for Ge) = %0.2f mV\"%deltaVD\n",
+      "eta=2  # for Si\n",
+      "deltaVD=eta*VT*log(ID2/ID1)  # Change in diode voltage in volts\n",
+      "deltaVD=deltaVD*1e3  # Change in diode voltage in milivolts\n",
+      "print \"Change in diode voltage (for Si) = %0.3f mV\" %deltaVD"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Change in diode voltage (for Ge) = 67.70 mV\n",
+        "Change in diode voltage (for Si) = 135.403 mV\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3: Page 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "print \"Part (a)\" \n",
+      "eta=1  # for Ge\n",
+      "T=300  # Room temperature in kelvins\n",
+      "VT=T/11600  # Voltage equivalent to temperatue at room temperature in volts\n",
+      "IS=1  # Let reverse saturation current be 1 A\n",
+      "I=-0.9*IS  # Reverse current\n",
+      "V=eta*VT*log(1+(I/IS))  # Voltagei in volts\n",
+      "V=V*1e3  # Voltage in milivolts\n",
+      "print \"Voltage = %0.2f mV \" %V\n",
+      "\n",
+      "print \"Part (b)\" \n",
+      "V=0.05  # Voltage in volts\n",
+      "If_Ir=(exp(V/(eta*VT))-1)/(exp(-V/(eta*VT))-1)  # Ratio of current in forward bias to that in reverse bias\n",
+      "print \"Ratio of current in forward bias to that in reverse bias = %0.3f\"%If_Ir \n",
+      "\n",
+      "print \"Part (c)\" \n",
+      "IS=10e-6  # Reverse saturation current in amperes\n",
+      "V=0.1  # Voltage in volts\n",
+      "ID=IS*(exp(V/(eta*VT))-1)  # Forward current for 0.1 V in amperes\n",
+      "ID=ID*1e6  # Forward current for 0.1 V in micro-amperes\n",
+      "print \"Forward current for 0.1 V = %0.2f \u03bcA \" %ID\n",
+      "V=0.2  # Voltage in volts\n",
+      "ID=IS*(exp(V/(eta*VT))-1)  # Forward current for 0.1 V in amperes\n",
+      "ID=ID*1e3  # Forward current for 0.1 V in miliamperes\n",
+      "print \"Forward current for 0.1 V = %0.2f mA\"%ID \n",
+      "V=0.3  # Voltage in volts\n",
+      "ID=IS*(exp(V/(eta*VT))-1)  # Forward current for 0.1 V in amperes\n",
+      "print \"Forward current for 0.1 V = %0.2f A\" %ID"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (a)\n",
+        "Voltage = -59.55 mV \n",
+        "Part (b)\n",
+        "Ratio of current in forward bias to that in reverse bias = -6.913\n",
+        "Part (c)\n",
+        "Forward current for 0.1 V = 467.83 \u03bcA \n",
+        "Forward current for 0.1 V = 22.82 mA\n",
+        "Forward current for 0.1 V = 1.09 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4 Page 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "IS=10e-6  # Reverse saturation current in amperes\n",
+      "eta=1  # for Ge\n",
+      "VT=25e-3  # Voltage equivalent to temperatue at room temperature in volts\n",
+      "\n",
+      "print \"Part (a)\" \n",
+      "VD=-24  # Reverse bias in volts\n",
+      "ID=IS*(exp(VD/(eta*VT))-1)  # Current in amperes\n",
+      "ID=ID*1e6  # Current in micro-amperes\n",
+      "print \"Current = %0.2f \u03bcA \"%ID \n",
+      "\n",
+      "print \"Part (b)\" \n",
+      "VD=-0.02  # Reverse bias in volts\n",
+      "ID=IS*(exp(VD/(eta*VT))-1)  # Current in amperes\n",
+      "ID=ID*1e6  # Current in micro-amperes\n",
+      "print \"Current = %0.2f \u03bcA \"%ID \n",
+      "\n",
+      "print \"Part (c)\" \n",
+      "VD=0.3  # Forward bias in volts\n",
+      "ID=IS*(exp(VD/(eta*VT))-1)  # Current in amperes\n",
+      "print \"Current = %0.2f A \"%ID"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (a)\n",
+        "Current = -10.00 \u03bcA \n",
+        "Part (b)\n",
+        "Current = -5.51 \u03bcA \n",
+        "Part (c)\n",
+        "Current = 1.63 A \n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5: Page 188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "T=300  # Operating temperature in kelvins\n",
+      "VT=T/11600  # Voltage equivalent to temperatue at room temperature in volts\n",
+      "ID1=1  # Let the initial diode current be 1 A\n",
+      "ID2=10*ID1  # Final diode current\n",
+      "eta=1  # for Ge\n",
+      "deltaVD=eta*VT*log(ID2/ID1)  # Change in diode voltage in volts\n",
+      "deltaVD=deltaVD*1e3  # Change in diode voltage in milivolts\n",
+      "print \"Change in diode voltage (for Ge) = %0.2f mV \" %deltaVD\n",
+      "eta=2  # for Si\n",
+      "deltaVD=eta*VT*log(ID2/ID1)  # Change in diode voltage in volts\n",
+      "deltaVD=deltaVD*1e3  # Change in diode voltage in milivolts\n",
+      "print \"Change in diode voltage (for Si) = %0.2f mV \" %deltaVD"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Change in diode voltage (for Ge) = 59.55 mV \n",
+        "Change in diode voltage (for Si) = 119.10 mV \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6: Page 188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# In the circuit given in Fig. 2.7\n",
+      "V=50e-3  # Output voltage\n",
+      "VD1=0.7  # Voltage across diode 1 in volts\n",
+      "I1=10e-3  # Current through diode 1 at 0.7 V in amperes\n",
+      "VD2=0.8  # Voltage across diode 2 in volts\n",
+      "I2=100e-3  # Current through diode 2 at 0.8 V in amperes\n",
+      "eta_VT=(VD2-VD1)/log(I2/I1)  # Product of \u03b7 and VT\n",
+      "I=10e-3/(exp(V/eta_VT)+1)  # Current through diode 1 in amperes\n",
+      "R=V/I \n",
+      "print \"R = %0.2f \u03a9 \"%R "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R = 20.81 \u03a9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7: Page 189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "VDD=5  # Applied voltage in volts\n",
+      "VD=0.7  # Diode voltage in volts\n",
+      "I1=1e-3  # Current in amperes at diode voltage = 0.7 V\n",
+      "R=1000  # R in ohms\n",
+      "deltaVD=0.1  # Change in diode voltage in volts for every decade change in current\n",
+      "ratioI=10  # Decade change in current\n",
+      "eta_VT=deltaVD/log(ratioI)  # Product of \u03b7 and VT\n",
+      "ID=(VDD-VD)/R  # Diode current in amperes\n",
+      "VD2=VD+eta_VT*log(ID/I1)  # Diode voltage in volts\n",
+      "ID=ID*1e3  # Diode current in miliamperes\n",
+      "print \"Diode current = %0.2f mA\" %ID\n",
+      "print \"Diode voltage = %0.2f V \"%VD2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diode current = 4.30 mA\n",
+        "Diode voltage = 0.76 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8: Page 190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"Part (a)\" \n",
+      "# Since both the diodes are in OFF state\n",
+      "Vo=5  # Output voltage in volts\n",
+      "print \"Output voltage = %0.2f V \"%Vo \n",
+      "\n",
+      "print \"Part (b)\" \n",
+      "#Since diode D1 is in OFF state and diode D2 is in ON state\n",
+      "# From Fig. 2.16(C)\n",
+      "I=(5-0.6)/(4.7e3+300)  # Current flowing through the diode D2 in amperes\n",
+      "Vo=5-I*4.7e3  # Output voltage in volts\n",
+      "print \"Output voltage = %0.2f V \"%Vo\n",
+      "\n",
+      "print \"Part (c)\" \n",
+      "# Since both diodes are in ON state\n",
+      "# Applying KVL in Fig. 2.16(d)\n",
+      "I=(5-0.6)/(2*4.7e3+300)  # Current flowing through diode D1 or diode D2 in amperes\n",
+      "Vo=5-2*I*4.7e3  # Output voltage in volts\n",
+      "print \"Output voltage = %0.2f V \"%Vo"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (a)\n",
+        "Output voltage = 5.00 V \n",
+        "Part (b)\n",
+        "Output voltage = 0.86 V \n",
+        "Part (c)\n",
+        "Output voltage = 0.74 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9 Page 190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "Vy=0.7  # Cut-in voltage in volts\n",
+      "# In the Fig. 2.17\n",
+      "R1=5e3 \n",
+      "R2=10e3 \n",
+      "\n",
+      "print \"Part (a)\" \n",
+      "# Since diode D1 is OFF and diode D2 is ON\n",
+      "ID1=0 # A\n",
+      "ID2=(5-Vy-(-5))/(R1+R2)  # Current through diode D2 in amperes\n",
+      "Vo=5-ID2*R1  # Output voltage\n",
+      "ID2=ID2*1e3  # Current through diode D2 in miliamperes\n",
+      "print \"Output voltage = %0.2f V \" %Vo\n",
+      "print \"Current through diode D1 = %0.2f mA\"%ID1 \n",
+      "print \"Current through diode D2 = %0.2f mA \"%ID2 \n",
+      "\n",
+      "print \"Part (b)\" \n",
+      "# Since both the diodes are ON\n",
+      "VA=4-Vy  # In the fig.\n",
+      "Vo=VA+Vy  # Output voltage\n",
+      "ID2=(5-Vo)/R1  # Current through diode D2 in amperes\n",
+      "IR2=(VA-(-5))/R2  # Current through diode R2 in amperes\n",
+      "ID1=IR2-ID2  # Current through diode D1 in amperes\n",
+      "ID1=ID1*1e3  # Current through diode D1 in miliamperes\n",
+      "ID2=ID2*1e3  # Current through diode D2 in miliamperes\n",
+      "print \"Output voltage = %0.2f V \" %Vo\n",
+      "print \"Current through diode D1 = %0.2f mA\"%ID1 \n",
+      "print \"Current through diode D2 = %0.2f mA \"%ID2 "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (a)\n",
+        "Output voltage = 1.90 V \n",
+        "Current through diode D1 = 0.00 mA\n",
+        "Current through diode D2 = 0.62 mA \n",
+        "Part (b)\n",
+        "Output voltage = 4.00 V \n",
+        "Current through diode D1 = 0.63 mA\n",
+        "Current through diode D2 = 0.20 mA \n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/sample_notebooks/karansingh/Ch4.ipynb b/sample_notebooks/karansingh/Ch4.ipynb
new file mode 100644
index 00000000..5c8ea38c
--- /dev/null
+++ b/sample_notebooks/karansingh/Ch4.ipynb
@@ -0,0 +1,397 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:31068ba7d72c538c2034ff141729c374b5adab63db5f8b55f3d177e0fb459ae4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4:Operational Amplifier"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 : Page 245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# For an op-amp circuit find a) closed loop gain Acl b) input impedance Zin c) output impedance Zo\n",
+      "\n",
+      "from __future__ import division\n",
+      "ro = 85   # ohm\n",
+      "A = 150*10**3   # ohm\n",
+      "R2 = 350*10**3   # ohm  # Feedback resistance\n",
+      "R1 = 10*10**3   # ohm  # Input resistance\n",
+      "\n",
+      "# a) closed loop gain\n",
+      "# ACL = abs(Vo/Vin) = abs(R2/R1)\n",
+      "ACL = abs(R2/R1)  \n",
+      "print '  closed loop gain of an op-amp is = ',ACL,'  '  # 1/beta = ACL\n",
+      "beta = (1/ACL)  \n",
+      "\n",
+      "# b) the input impedance Zin\n",
+      "Zin = R1  \n",
+      "print '  the input impedance Zin = ',Zin,' ohm ' \n",
+      "\n",
+      "# c0 the output impedance Z0\n",
+      "Z0 = (ro)/(1+(beta*A)) \n",
+      "print '  the output impedance Z0 = %0.3f'%Z0,' ohm ' "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  closed loop gain of an op-amp is =  35.0   \n",
+        "  the input impedance Zin =  10000  ohm \n",
+        "  the output impedance Z0 = 0.020  ohm \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 : Page 245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Determine the differece voltage and open loop gain of an op-amp\n",
+      "  \n",
+      "V1 = -5   # volt  # input voltage\n",
+      "V2 = 5   # volt\n",
+      "Vo = 20   #volt  # output voltage\n",
+      "\n",
+      "# the difference voltage is given by \n",
+      "Vd = V2-V1  \n",
+      "print '  The difference voltage is = ',Vd,' V ' \n",
+      "\n",
+      "# open loop gain \n",
+      "A = (Vo/Vd) \n",
+      "print '  The open loop gain is = ',A,'  ' "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  The difference voltage is =  10  V \n",
+        "  The open loop gain is =  2.0   \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3 : Page 246"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Determine the differece voltage and open loop gain of an op-amp\n",
+      "  \n",
+      "V1 = -5   # volt # input voltage\n",
+      "V2 = 0   # volt  # GND\n",
+      "Vo = 20   #volt  # output voltage\n",
+      "\n",
+      "# the difference voltage is given by \n",
+      "Vd = V2-V1  \n",
+      "print '  The difference voltage is = ',Vd,' V ' \n",
+      "\n",
+      "# open loop gain \n",
+      "A = (Vo/Vd) \n",
+      "print '  The open loop gain is = ',A,'  ' "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  The difference voltage is =  5  V \n",
+        "  The open loop gain is =  4.0   \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4 : Page 247"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Determine the differece voltage and open loop gain of an op-amp\n",
+      "  \n",
+      "V1 = 0   # volt # input voltage  # GND\n",
+      "V2 = 5   # volt  \n",
+      "Vo = 20   #volt  # output voltage\n",
+      "\n",
+      "# the difference voltage is given by \n",
+      "Vd = V2-V1  \n",
+      "print '  The difference voltage is = ',Vd,' V ' \n",
+      "\n",
+      "# open loop gain \n",
+      "A = (Vo/Vd) \n",
+      "print '  The open loop gain is = ',A,'  ' "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  The difference voltage is =  5  V \n",
+        "  The open loop gain is =  4.0   \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 : Page 247"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Determine the differece voltage and open loop gain of an op-amp\n",
+      "  \n",
+      "V1 = 5   # volt # input voltage  # GND\n",
+      "V2 = -5   # volt  \n",
+      "Vo = -20   #volt  # output voltage\n",
+      "\n",
+      "# the difference voltage is given by \n",
+      "Vd = V2-V1  \n",
+      "print '  The difference voltage is = ',Vd,' V ' \n",
+      "\n",
+      "# open loop gain \n",
+      "A = (Vo/Vd) \n",
+      "print '  The open loop gain is = ',A,'  ' "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  The difference voltage is =  -10  V \n",
+        "  The open loop gain is =  2.0   \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6 : Page 248"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# To find closed loop gain and output voltage Vo of an inverting op-amp\n",
+      "R1 = 10   #kilo ohm  # input resistance\n",
+      "R2 = 25   # kilo ohm  # feedback resistance\n",
+      "Vin = 10   #volt  # input voltage\n",
+      "\n",
+      "# Closed loop gain of an inverting op-amp\n",
+      "Ac = -(R2/R1)  \n",
+      "print 'The Closed loop gain of an inverting op-amp is = ',Ac,'  ' \n",
+      "Ac = abs(Ac) \n",
+      "print 'The |Ac| Closed loop gain of an inverting op-amp is = ',Ac,'  ' \n",
+      "\n",
+      "# the output voltage of an inverting op-amp\n",
+      "Vo = -(R2/R1)*Vin  \n",
+      "print 'The output voltage of an inverting op-amp is = ',Vo,' V ' "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Closed loop gain of an inverting op-amp is =  -2.5   \n",
+        "The |Ac| Closed loop gain of an inverting op-amp is =  2.5   \n",
+        "The output voltage of an inverting op-amp is =  -25.0  V \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7 : Page 248"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# To find closed loop gain and output voltage Vo of an non-inverting op-amp\n",
+      "R1 = 10   #kilo ohm  # input resistance\n",
+      "R2 = 25   # kilo ohm  # feedback resistance\n",
+      "Vin = 10   #volt  # input voltage\n",
+      "\n",
+      "# Closed loop gain of an non-inverting op-amp\n",
+      "Ac = 1+(R2/R1)  \n",
+      "Ac = abs(Ac) \n",
+      "print 'The Closed loop gain of an non-inverting op-amp is = ',Ac,'  ' \n",
+      "\n",
+      "# the output voltage of an inverting op-amp\n",
+      "Vo = (1+R2/R1)*Vin  \n",
+      "print 'The output voltage of an non-inverting op-amp is = ',Vo,' V ' "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Closed loop gain of an non-inverting op-amp is =  3.5   \n",
+        "The output voltage of an non-inverting op-amp is =  35.0  V \n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8 : Page 249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# to find out closed loop gain and output voltage Vo\n",
+      "R1 = 10   #kilo ohm  # input resistance\n",
+      "R3 = 10   #kilo ohm  # input resistance\n",
+      "R2 = 25   # kilo ohm  # feedback resistance\n",
+      "R4 = 25   # kilo ohm  # feedback resistance\n",
+      "Vin2 = 10   #volt  # input voltage\n",
+      "Vin1 = -10   #volt  # input voltage\n",
+      "\n",
+      "# closed loop gain of differntial op-amp is given by\n",
+      "Ac = (R2/R1)  \n",
+      "Ac = abs(Ac)  \n",
+      "print 'The closed loop gain of differntial op-amp is = ',Ac,'  ' \n",
+      "\n",
+      "# the output voltage of an non-inverting op-amp is given by\n",
+      "Vo = (R2/R1)*(Vin2-Vin1)  \n",
+      "print 'The output voltage of an non-inverting op-amp is= ',Vo,' V ' "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The closed loop gain of differntial op-amp is =  2.5   \n",
+        "The output voltage of an non-inverting op-amp is=  50.0  V \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9 : Page 249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Determine the non-inverting input voltage\n",
+      "R1 = 10   #kilo ohm  # input resistance\n",
+      "R2 = 25   #kilo ohm # feedback resistance\n",
+      "Voh = 10   # volt #output voltage\n",
+      "Vol = -10   # volt # output voltage\n",
+      "\n",
+      "# upper voltage\n",
+      "V = (R1/(R1+R2)*Voh)  \n",
+      "print '  The upper voltage is = %0.3f'%V,' V ' \n",
+      "\n",
+      "# Lower voltage\n",
+      "V = (R1/(R1+R2)*Vol)  \n",
+      "print '  The lower voltage is = %0.3f'%V,' V ' "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  The upper voltage is = 2.857  V \n",
+        "  The lower voltage is = -2.857  V \n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file