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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 1 - Semiconductor Material & Junction Diode"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.1 Page No 51"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The electron drift velocity = 40.00 m/s\n",
+      "The time required for an electron to move across the thickness = 12.50 micro seconds\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "miu = 0.2# m**2/V-s\n",
+    "V = 100# mV\n",
+    "V = V * 10**-3# V\n",
+    "d = 0.5# mm\n",
+    "d = d * 10**-3# m\n",
+    "# mobility, miu = Vd/E and\n",
+    "E = V/d\n",
+    "# Drift velocity,\n",
+    "Vd = miu*E# m/s\n",
+    "print \"The electron drift velocity = %.2f m/s\"%Vd\n",
+    "# Time required,\n",
+    "T = d/Vd# sec\n",
+    "T=T*10**6# µs\n",
+    "print \"The time required for an electron to move across the thickness = %.2f micro seconds\"%T"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.2 Page No 52"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The intrinsic conductivity = 2.24 (ohm-m)**-1\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "q = 1.6*10**-19# C\n",
+    "n_i = 2.5*10**19# /m**3\n",
+    "miu_n = 0.38# m**2/V-s\n",
+    "miu_p = 0.18# m**2/V-s\n",
+    "# The intrinsic conductivity for germanium,\n",
+    "sigma_i = q*n_i*(miu_n+miu_p)# (ohm-m)**-1\n",
+    "print \"The intrinsic conductivity = %.2f (ohm-m)**-1\"%sigma_i"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.3 Page No 52"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The intrinsic carrier concentration = 2.16e+19 per m**3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "rho = 0.50# ohm-m\n",
+    "q = 1.6*10**-19# C\n",
+    "miu_n = 0.39# m**2/V-s\n",
+    "miu_p = 0.19# m**2/V-s\n",
+    "sigma = 1/rho# (ohm-m)**-1\n",
+    "#conductivity of a semiconductor, sigma = q*n_i*(miu_p+miu_n) or\n",
+    "n_i = sigma/(q*(miu_n+miu_p))# /m**3\n",
+    "print \"The intrinsic carrier concentration = %.2e per m**3\"%n_i"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.4 Page No 52"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The conductivity of Si sample = 14.40 (ohm-m)**-1\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "N_D = 10**21# /m**3\n",
+    "N_A = 5*10**20# /m**3\n",
+    "NdasD = N_D-N_A# /m**3\n",
+    "n = NdasD# /m**3\n",
+    "miu_n = 0.18# m**2/V-s\n",
+    "q = 1.6*10**-19# C\n",
+    "# The conductivity of silicon,\n",
+    "sigma = q*n*miu_n# (ohm-m)**-1\n",
+    "print \"The conductivity of Si sample = %.2f (ohm-m)**-1\"%sigma"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.5 Page No 53"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The conductivity of copper = 4.79e+05 mho/cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "At = 63.54## atomic weight of copper\n",
+    "d = 8.9## density = %.2f gm/cm**3\n",
+    "n = 6.023*10**23/At*d# electron/cm**3\n",
+    "q = 1.63*10**-19# C\n",
+    "miu = 34.8# m**2/V-s\n",
+    "# The conductivity of copper,\n",
+    "sigma = n*q*miu# mho/cm\n",
+    "print \"The conductivity of copper = %.2e mho/cm\"%sigma"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.6 Page No 53"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Concentration of holes in a p-type Ge = 3.47e+17 /cm**3\n",
+      "The concentration of electrons in a p-type Ge = 1.80e+09 /cm**3\n",
+      "The concentration of electrons in n-type Si = 4.81e+14 /cm**3\n",
+      "The concentration of holes in n-type Si = 4.68e+05 /cm**3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "sigma = 100# (ohm-m)**-1\n",
+    "n_i = 2.5*10**13# /cm**3\n",
+    "miu_n = 3800# cm**2/V-s\n",
+    "miu_p = 1800# cm**2/V-s\n",
+    "q = 1.6*10**-19# C\n",
+    "# Conductivity of a p-type germanium, sigma = q*p*miu_p or\n",
+    "p = sigma/(q*miu_p)# /cm**3\n",
+    "print \"Concentration of holes in a p-type Ge = %.2e /cm**3\"%p\n",
+    "# The concentration of electrons = %.2f a p-type Ge\n",
+    "n = (n_i**2)/p# /cm**3\n",
+    "print \"The concentration of electrons in a p-type Ge = %.2e /cm**3\"%n\n",
+    "#Given for Si\n",
+    "sigma= 0.1# (ohm m)**-1\n",
+    "miu_n= 1300# cm**2/V-sec\n",
+    "n_i= 1.5*10**10# /cm**3\n",
+    "#sigma = q*n*miu_n\n",
+    "n = sigma/(q*miu_n)# /cm**3\n",
+    "print \"The concentration of electrons in n-type Si = %.2e /cm**3\"%n\n",
+    "# The concentration of holes = %.2f n-type Si\n",
+    "p = (n_i**2)/n# /cm**3\n",
+    "print \"The concentration of holes in n-type Si = %.2e /cm**3\"%p"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.7 Page No 54"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The resistivity of a dopped Ge = 3.72 ohm-cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "miu_n = 3800## cm**2/V-s\n",
+    "miu_p = 1800##  cm**2/V-s\n",
+    "n_i = 2.5*10**13# /cm**3\n",
+    "Nge = 4.41*10**22# /cm**3\n",
+    "q = 1.602*10**-19# C\n",
+    "impurity = 10**8\n",
+    "# The number of donor atoms,\n",
+    "N_D = Nge/impurity##in /cm**3\n",
+    "# The number of holes\n",
+    "p = (n_i**2)/N_D# /cm**3\n",
+    "# Conductivity of an N-type Ge,\n",
+    "sigma = q*N_D*miu_n# (ohm-cm)**-1\n",
+    "# The resistivity of the Ge\n",
+    "rho = 1/sigma# ohm-cm\n",
+    "print \"The resistivity of a dopped Ge = %.2f ohm-cm\"% rho"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.8 Page No 54"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The resistivity of intrinsic silicon = 2.25e+05 ohm-cm\n",
+      "The resistivity of doped silicon = 4.67 ohm-cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "Nsi = 4.96*10**22# /cm**3\n",
+    "n_i = 1.52*10**10# /cm**2\n",
+    "q = 1.6*10**-19# C\n",
+    "miu_n = 0.135#  m**2/V-s\n",
+    "miu_n = miu_n * 10**4#  cm**2/V-s\n",
+    "miu_p = 0.048#  m**2/V-s\n",
+    "miu_p = miu_p * 10**4#  cm**2/V-s\n",
+    "# The conductivity of an intrinsic silicon,\n",
+    "sigma = q*n_i*(miu_n+miu_p)# (ohm-cm)**-1\n",
+    "# The resistivity of intrinsic silicon \n",
+    "rho = 1/sigma# ohm-cm\n",
+    "print \"The resistivity of intrinsic silicon = %.2e ohm-cm\"%rho\n",
+    "\n",
+    "impurity = 50*10**6\n",
+    "# The number of donor atoms,\n",
+    "N_D = Nsi/impurity# /cm**3\n",
+    "# Total free electrons,\n",
+    "n = N_D# /cm**3\n",
+    "# Total holes = %.2f a doped Si,\n",
+    "p = (n_i**2)/n# /cm**3\n",
+    "# Conductivity of a doped Si,\n",
+    "sigma = q*n*miu_n# (ohm-m)**-1\n",
+    "# The resistivity of doped silicon\n",
+    "rho = 1/sigma# ohm-cm\n",
+    "print \"The resistivity of doped silicon = %.2f ohm-cm\"%rho"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.9 Page No 55"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The value of temperature = 0.14 K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "N_D= 5.0*10**28/(2.0*10**8)\n",
+    "# The Fermi level, E_F= E_C if,\n",
+    "N_C= N_D\n",
+    "# Formula N_C= 4.82*10**21*T**(3/2)\n",
+    "T= (N_C/(4.82*10**21.0))**(2.0/3)# K\n",
+    "print \"The value of temperature = %.2f K\"%T"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.10 Page No 55"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The miniority carrier concentration = 0.10 m**2/V-s\n",
+      "The resistivity = 0.60 ohm-m\n",
+      "The position of Fermi level = 0.23 eV\n",
+      "Minority carrier concentration = 9.00e+12 atoms/cm**3\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# Given data\n",
+    "n_i = 1.5*10**16##m**3\n",
+    "impurity = 10**20\n",
+    "minority = (n_i**2)/impurity# atoms/m**3\n",
+    "q = 1.6*10**-19# C\n",
+    "rho = 2*10**3# ohm-m\n",
+    "# The miniority carrier concentration \n",
+    "miu_n = 1/(q*rho*n_i*2)##in m**2/V-s\n",
+    "print \"The miniority carrier concentration = %.2f m**2/V-s\"%miu_n\n",
+    "n = impurity\n",
+    "# The conductivity,\n",
+    "sigma = q*impurity*miu_n# (ohm-m)**-1\n",
+    "# The resistivity \n",
+    "rho = 1/sigma#  ohm-m\n",
+    "print \"The resistivity = %.2f ohm-m\"%rho\n",
+    "kT = 0.026# eV\n",
+    "n_o = n\n",
+    "# The position of Fermi level \n",
+    "E_FdividedEi = kT*math.log(n_o/n_i)# eV\n",
+    "print \"The position of Fermi level = %.2f eV\"%E_FdividedEi\n",
+    "# Minority carrier concentration \n",
+    "M = ((n_i*2)**2)/n_o# atoms/cm**3\n",
+    "print \"Minority carrier concentration = %.2e atoms/cm**3\"%M"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.11 Page No 56"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 41,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The resistivity = 9.62 ohm-cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "d = 5.0*10**22# atoms/cm**3\n",
+    "impurity = 10**8# atoms\n",
+    "N_D = d/impurity\n",
+    "n_i = 1.45*10**10\n",
+    "n = N_D\n",
+    "#Low of mass action,  n*p = (n_i**2)\n",
+    "p = (n_i**2)/n# /cm**3\n",
+    "q = 1.6*10**-19# C\n",
+    "miu_n = 1300# cm/V-s\n",
+    "n_i = n\n",
+    "#The Conductivity\n",
+    "sigma = q*miu_n*n_i# (ohm-cm)**-1\n",
+    "# The resistivity\n",
+    "rho = 1/sigma# ohm-cm\n",
+    "print \"The resistivity = %.2f ohm-cm\"%rho"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.12 Page No 57"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The resistivity = 9.62 ohm-cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "d = 5.0*10**22# atoms/cm**3\n",
+    "impurity = 10**8# atoms\n",
+    "N_D = d/impurity\n",
+    "n_i = 1.45*10**10\n",
+    "n = N_D\n",
+    "#Low of mass action,  n*p = (n_i**2)\n",
+    "p = (n_i**2)/n# /cm**3\n",
+    "q = 1.6*10**-19# C\n",
+    "miu_n = 1300# cm/V-s\n",
+    "n_i = n\n",
+    "#The Conductivity\n",
+    "sigma = q*miu_n*n_i# (ohm-cm)**-1\n",
+    "# The resistivity\n",
+    "rho = 1/sigma# ohm-cm\n",
+    "print \"The resistivity = %.2f ohm-cm\"%rho"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.14 Page No 58"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 40,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The minority carrier concentration = 2.25e+03 holes/cm**3\n",
+      "The location of Fermi level = 0.409 eV\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# Given data\n",
+    "n_i = 1.5*10**10# electrons/cm**3\n",
+    "N_D = 10**17# electrons/cm**3\n",
+    "n = N_D# electrons/cm**3\n",
+    "# The minority carrier concentration\n",
+    "p = (n_i**2)/n# holes/cm**3\n",
+    "print \"The minority carrier concentration = %.2e holes/cm**3\"%p\n",
+    "kT = 0.026\n",
+    "# The location of Fermi level \n",
+    "E_FminusEi = kT*math.log(N_D/n_i)# eV\n",
+    "print \"The location of Fermi level = %.3f eV\"%E_FminusEi"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.15 Page No 59"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The doping level = 1.92e+15 /cm**3\n",
+      "The drift velocity = 650.00 cm/sec\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "V = 1# V\n",
+    "I = 8# mA\n",
+    "I = I * 10**-3# A\n",
+    "R = V/I# ohm\n",
+    "l = 2# mm\n",
+    "l = l * 10**-1# cm\n",
+    "b = 2# mm\n",
+    "b = b * 10**-1# cm\n",
+    "A = l*b# cm**2\n",
+    "L = 2# cm\n",
+    "# R = (rho*L)/A\n",
+    "sigma = L/(R*A)# (ohm-cm)**-1\n",
+    "# n = N_D\n",
+    "miu_n = 1300# cm**2/V-s\n",
+    "q = 1.6 * 10**-19# C\n",
+    "# sigma = n*q*miu_n\n",
+    "N_D = sigma/( miu_n*q )# /cm**3\n",
+    "print \"The doping level = %.2e /cm**3\"%N_D\n",
+    "d = 2.0\n",
+    "E = V/d\n",
+    "# The drift velocity \n",
+    "Vd = miu_n * E# cm/s\n",
+    "print \"The drift velocity = %.2f cm/sec\"%Vd"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.17 Page No 60"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The conductivity = 4.68e+05 mho/m\n",
+      "The mobility = 3.48e-05 m**2/V-s\n",
+      "The drift velocity = 1.79e-04 m/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# Given data\n",
+    "l = 1000# ft\n",
+    "l = l * 12*2.54# cm\n",
+    "R = 6.51# ohm\n",
+    "rho = R/l# ohm/cm\n",
+    "# The conductivity \n",
+    "sigma = 1/rho# mho/cm\n",
+    "sigma = sigma * 10**2# mho/m\n",
+    "D= 1.03*10**-3# m\n",
+    "A= math.pi*D**2/4# m**2\n",
+    "print \"The conductivity = %.2e mho/m\"%sigma\n",
+    "q = 1.6*10**-19# C\n",
+    "n = 8.4*10**28# electrons/m**3\n",
+    "# sigma = n*q*miu\n",
+    "miu = sigma/(n*q)# m**2/V-s\n",
+    "print \"The mobility = %.2e m**2/V-s\"%miu\n",
+    "T = 2\n",
+    "# The drift velocity \n",
+    "V = T/(n*q*A)# m/s\n",
+    "print \"The drift velocity = %.2e m/s\"%V"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.18 Page No 61"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The concentration of holes = 1.50e+16 /cm**3\n",
+      "The concentartion of electrons = 6.67e+07 /cm**3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "N_D = 2*10**16# /cm**3\n",
+    "N_A = 5*10**15# /cm**3\n",
+    "# The concentration of holes \n",
+    "Pp = N_D-N_A# /cm**3\n",
+    "print \"The concentration of holes = %.2e /cm**3\"%Pp\n",
+    "n_i = 10**12\n",
+    "# The concentartion of electrons \n",
+    "n_p = (n_i**2)/Pp# /cm**3\n",
+    "print \"The concentartion of electrons = %.2e /cm**3\"%n_p"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.19 Page No 62"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The hall angle = 1.95 degree\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# Given data\n",
+    "rho = 0.005# ohm-m\n",
+    "Bz = 0.48# Wb/m**2\n",
+    "R_H = 3.55*10**-4# m**3/C\n",
+    "ExByJx= rho\n",
+    "# R_H = Ey/(Bz*Jx)\n",
+    "EyByJx= R_H*Bz\n",
+    "# The hall angle \n",
+    "theta_H = math.degrees(math.atan(EyByJx/ExByJx))# °\n",
+    "print \"The hall angle = %.2f degree\"%theta_H"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.20 Page No 63"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The voltage between contacts = 0.0026 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "R_H = 3.55 * 10**-4# m**3/C\n",
+    "Ix = 15# mA\n",
+    "Ix = Ix * 10**-3# A\n",
+    "A = 15*1# mm\n",
+    "A = A * 10**-6# m**2\n",
+    "Bz = 0.48# Wb/m**2\n",
+    "Jx = Ix/A# A/m**2\n",
+    "# R_H = Ey/(Bz*Jx)\n",
+    "Ey = R_H*Bz*Jx# V/m\n",
+    "# voltage between contacts \n",
+    "Voltage = Ey*Ix# V\n",
+    "print \"The voltage between contacts = %.4f V\"%Voltage"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.21 Page No 63"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The concentration of donor atoms = 4.630e+13 cm**-3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "A = 0.001# cm**2\n",
+    "l = 20# µm\n",
+    "l = l * 10**-4# cm\n",
+    "V = 20# V\n",
+    "I = 100# mA\n",
+    "I = I * 10**-3# A\n",
+    "R = V/I# ohm\n",
+    "# R = l/(sigma*A)\n",
+    "sigma = l/(R*A)# (ohm-cm)**-1\n",
+    "miu_n = 1350# cm**2/V-s\n",
+    "q = 1.6*10**-19# C\n",
+    "# sigma = n*q*miu_n or\n",
+    "# The concentration of donor atoms \n",
+    "n = sigma/(q*miu_n)# cm**-3\n",
+    "print \"The concentration of donor atoms = %.3e cm**-3\"%n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.22 Page No 64"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The doping needed = 8.681e+15 cm**-3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "R = 2# k ohm\n",
+    "R = R * 10**3# ohm\n",
+    "L = 200# µm\n",
+    "L = L * 10**-4# cm\n",
+    "A = 10**-6# cm**2\n",
+    "miu_n = 8000# cm**2/V-s\n",
+    "q = 1.6*10**-19# C\n",
+    "n = '0.9*N_D'\n",
+    "# R = (rho*l)/A= (1/(n*q*miu_n))*(l/A)\n",
+    "# rho = L/(R*q*miu_n*A)\n",
+    "n = L/(R*q*miu_n*A)# /cm**-3\n",
+    "# The doping needed \n",
+    "Nd= n/0.9\n",
+    "print \"The doping needed = %.3e cm**-3\"%Nd"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.23 Page No 65"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The position of the Fermi level = 0.29 eV\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# Given data\n",
+    "KT = 26*10**-3\n",
+    "Nd = 10**15\n",
+    "n_i = 1.5*10**10\n",
+    "# The position of the Fermi level \n",
+    "E_FminusE_Fi = KT*math.log(abs( Nd/n_i ))# eV\n",
+    "print \"The position of the Fermi level = %.2f eV\"%E_FminusE_Fi"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.24 Page No 65"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The concentration of donors atoms = 1.2176e+16 cm**-3\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# Given data\n",
+    "Na = 5 * 10**15# cm**-3\n",
+    "Nc = 2.8 * 10**19# cm**-3\n",
+    "E_CminusE_F = 0.215# eV\n",
+    "KT = 26* 10**-3# eV\n",
+    "# The concentration of donors atoms \n",
+    "Nd = Na + Nc * (math.exp( -E_CminusE_F/KT ))# cm**-3\n",
+    "print \"The concentration of donors atoms = %.4e cm**-3\"%Nd"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.25 Page No 65"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The percentage doping efficiency = 78.12 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "Nd = 10**18\n",
+    "R = 10# ohm\n",
+    "A =10**-6# cm**2\n",
+    "L = 10# mm\n",
+    "L = L * 10**-4# cm\n",
+    "miu_n = 800# cm**2/V-s\n",
+    "q = 1.6*10**-19# C\n",
+    "#Formula used, n = L/(q*miu_n*A*R)\n",
+    "n = L/(q*miu_n*A*R)# cm**-3\n",
+    "# The percentage doping efficiency \n",
+    "doping = (n/Nd)*100## % doping efficiency in %\n",
+    "print \"The percentage doping efficiency = %.2f %%\"%doping"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.26 Page No 66"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The current through the diode under forward bias = 10.72 µA\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# Given data\n",
+    "Io = 2*10**-7# A\n",
+    "V = 0.1# V\n",
+    "# Current through the diode under forward bias,\n",
+    "I = Io*( (math.exp(40*V))-1 )# A\n",
+    "I = I * 10**6# µA\n",
+    "print \"The current through the diode under forward bias = %.2f µA\"%I\n",
+    "\n",
+    "# Note: Calculated value of I in the book is wrong."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.28 Page No 67"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The dynamic resistance in forward direction = 3.36 ohm\n",
+      "The dynamic resistance in reverse direction = 0.39 Mohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# Given data\n",
+    "T = 125.0# degree C\n",
+    "T = T + 273.0# K\n",
+    "V_T = T/11600.0\n",
+    "Io = 30# µA\n",
+    "Io = Io * 10**-6# A\n",
+    "V = 0.2# V\n",
+    "# The dynamic resistance = %.2f forward direction,\n",
+    "r_f = V_T/( Io * (math.exp(V/V_T)) )# ohm\n",
+    "print \"The dynamic resistance in forward direction = %.2f ohm\"%r_f\n",
+    "r_f = V_T/( Io * (math.exp(-V/V_T)) )# ohm\n",
+    "# The dynamic resistance = %.2f reverse direction \n",
+    "r_f = r_f * 10**-6# Mohm\n",
+    "print \"The dynamic resistance in reverse direction = %.2f Mohm\"%r_f"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.29 Page No 68"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The voltage = -59.87 mV\n",
+      "The ratio of diode current with a forward bias to current with a reverse bias = -6.842\n",
+      "The value of I1 = 458.13 µA\n",
+      "The value of I2 = 21.90 mA\n",
+      "The value of I3 = 1.03 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# Given data\n",
+    "Eta = 1\n",
+    "V_T = 0.026\n",
+    "# I = Io*( (exp(V/(Eta*V_T))) - 1 ) and  I = -Io\n",
+    "# I = -0.9*Io\n",
+    "# -0.9*Io = Io*( (exp(V/(Eta*V_T))) - 1 )\n",
+    "V = Eta*V_T*math.log(0.1)# V\n",
+    "V = V * 10**3# mV\n",
+    "print \"The voltage = %.2f mV\"%V\n",
+    "V = 0.05# V\n",
+    "# The ratio of diode current with a forward bias to current with a reverse bias \n",
+    "If_by_Ir= ( (math.exp(V/V_T))-1 )/( (math.exp(-V/V_T))-1 )\n",
+    "print \"The ratio of diode current with a forward bias to current with a reverse bias = %.3f\"%If_by_Ir\n",
+    "Io = 10# µA\n",
+    "V = 0.1# V\n",
+    "# The value of I1 \n",
+    "I1 = Io*( (math.exp(V/V_T))-1 )# µA\n",
+    "print \"The value of I1 = %.2f µA\"%I1\n",
+    "V = 0.2# V\n",
+    "# The value of I2\n",
+    "I2 = Io*( (math.exp(V/V_T))-1 )# µA \n",
+    "I2 = I2 * 10**-3# mA\n",
+    "print \"The value of I2 = %.2f mA\"%I2\n",
+    "V = 0.3# V\n",
+    "# The value of I3\n",
+    "I3 = Io*( (math.exp(V/V_T))-1 )# µA\n",
+    "I3 = I3 * 10**-6# A\n",
+    "print \"The value of I3 = %.2f A\"%I3"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.30 Page No 69"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The factor  by which current will get multiplied = 638.025\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# Given data\n",
+    "# Io150 = Io25 * 2**((150-25)/10)\n",
+    "#Io150 = 5800*Io25\n",
+    "T = 150# degree C\n",
+    "T = T  + 273# K\n",
+    "V_T = 8.62*10**-5 * T# V\n",
+    "V = 0.4# V\n",
+    "Eta = 2\n",
+    "Vt = 0.026# V \n",
+    "# The factor  by which current will get multiplied \n",
+    "I150byI25= 5800*math.exp(V/(Eta*V_T))/math.exp(V/(Eta*Vt))\n",
+    "print \"The factor  by which current will get multiplied = %.3f\"%I150byI25"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.31 Page No 69"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 27,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The operating point of the diode is : (0.50V,4.50mA)\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "R = 1# ohm\n",
+    "V = 5# V\n",
+    "V1 = 0.5# V\n",
+    "R1 = 1# k ohm\n",
+    "R1 = R1 * 10**3# ohm\n",
+    "# V-(I_D*R1)-(I_D*R) - V1 = 0\n",
+    "I_D = (V-V1)/(R1+R)# A\n",
+    "I_D = I_D * 10**3# mA\n",
+    "V_D = (I_D*10**-3*R) + V1# V\n",
+    "print \"The operating point of the diode is : (%.2fV,%.2fmA)\"%(V_D,I_D)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.32 Page No 70"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 28,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The voltage drop across the forward biased diode, = 0.0180 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# Given data\n",
+    "Eta = 1\n",
+    "kT = 26# meV\n",
+    "# (%e**((e*V1)/kT)) = 2 or\n",
+    "#The voltage drop across the forward biased diode\n",
+    "V1 = math.log(2)*kT# mV\n",
+    "V1= V1*10**-3# V\n",
+    "print \"The voltage drop across the forward biased diode, = %.4f V\"%V1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.33 Page No 71"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 29,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The space charge capacitance = 70.74 pF\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# Given data\n",
+    "epsilon_Ge = 16/(36*math.pi*10**11)# F/cm\n",
+    "d = 2*10**-4# cm\n",
+    "A = 1# mm**2\n",
+    "A = A * 10**-2# cm**2\n",
+    "epsilon_o = epsilon_Ge# F/cm\n",
+    "# The space charge capacitance \n",
+    "C_T = (epsilon_o*A)/d# F\n",
+    "C_T = C_T * 10**12# pF\n",
+    "print \"The space charge capacitance = %.2f pF\"%C_T"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.34 Page No 71"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 30,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The value of C_T = 61.68 pf/cm**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math \n",
+    "# Given data\n",
+    "D = 0.102# cm \n",
+    "A = (math.pi*(D**2))/4# cm**2\n",
+    "sigma_p = 0.286# (ohm-cm)**-1\n",
+    "q = 1.6*10**-19# C\n",
+    "miu_p = 500\n",
+    "# Formula used, sigma_p = q*miu_p*N_A\n",
+    "N_A = sigma_p/(q*miu_p)# atoms/cm**3\n",
+    "V1 = 5# V\n",
+    "V2 = 0.35# V\n",
+    "Vb = V1+V2# V\n",
+    "# The transition capacitance,\n",
+    "C_T = 2.92*10**-4*((N_A/Vb)**(1./2))*A# pF/cm**2\n",
+    "print \"The value of C_T = %.2f pf/cm**2\"%C_T"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.35 Page No 71"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 31,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The value of C_T for reverse bias = 15.00 pF\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "C_T1 = 15# pF\n",
+    "Vb1 = 8# V\n",
+    "Vb2 = 12# V\n",
+    "# C_T1/C_T2 = (Vb2/Vb1)**(1/2)\n",
+    "C_T2 = C_T1 * ((Vb1/Vb2)**(1/2))# pF\n",
+    "print \"The value of C_T for reverse bias = %.2f pF\"%C_T2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.36 Page No 72"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 32,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The voltage = -59.87 mV\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# Given data\n",
+    "V_T = 0.026# V\n",
+    "Eta = 1\n",
+    "I = '-0.9*Io'\n",
+    "# T = Io*((%e**(V/(Eta*V_T)))-1 )\n",
+    "# I = Io*((%e**(V/(Eta*V_T)))-1 )\n",
+    "V = math.log(0.1)*V_T# V \n",
+    "V = V * 10**3# mV\n",
+    "print \"The voltage = %.2f mV\"%V"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.37 Page No 72"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 33,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Part (a) : The value of I_D for first circuit = 0.97 mA\n",
+      "Part (b) : The value of I_D for second circuit = 0.10 mA\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "Vin = 20# V\n",
+    "Vgamma = 0.7# V\n",
+    "R = 20# k ohm\n",
+    "R = R * 10**3# ohm\n",
+    "# Vin-(I_D*Vin) - Vgamma = 0 or\n",
+    "# The value of I_D,\n",
+    "I_D = (Vin-Vgamma)/R# A\n",
+    "I_D = I_D * 10**3# mA\n",
+    "print \"Part (a) : The value of I_D for first circuit = %.2f mA\"%I_D\n",
+    "\n",
+    "# Part (b)\n",
+    "Vin= 10.# V\n",
+    "Vgamma = 0.7# V\n",
+    "R = 100# k ohm\n",
+    "# Drain current,\n",
+    "I_D= Vin/R# mV\n",
+    "print \"Part (b) : The value of I_D for second circuit = %.2f mA\"%I_D"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.38 Page No 73"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 34,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The value of I_D = 3.10 mA\n",
+      "The value of Vo = 6.90 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "R1 = 1# k ohm\n",
+    "R1 = R1 * 10**3# ohm\n",
+    "R2 = 2# k ohm\n",
+    "R2 = R2 * 10**3# ohm\n",
+    "V = 10# V\n",
+    "V1 = 0.7# V \n",
+    "# V * (I_D*R1) - (R2*I_D) - V1 = 0\n",
+    "I_D = (V-V1)/(R1+R2)# A\n",
+    "I_D = I_D * 10**3# mA\n",
+    "print \"The value of I_D = %.2f mA\"%I_D\n",
+    "# The output voltage,\n",
+    "Vo = (I_D*10**-3 * R2) +V1# V\n",
+    "print \"The value of Vo = %.2f V\"%Vo"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.39 Page No 73"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 35,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Part (a): The current through resistance = 1.00 A\n",
+      "Part (b) : Current through 10 ohm resistance will be Zero\n",
+      "Part (c): Current will be zero\n",
+      "Part (d): The diode will be ON and current = 1.00 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "V = 10.# V\n",
+    "R = 10# ohm\n",
+    "# Current through resistance,\n",
+    "I = V/R# A\n",
+    "print \"Part (a): The current through resistance = %.2f A\"%I\n",
+    "print \"Part (b) : Current through 10 ohm resistance will be Zero\"\n",
+    "print \"Part (c): Current will be zero\"\n",
+    "print \"Part (d): The diode will be ON and current = %.2f A\"%I"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.40 Page No 74"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 36,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The operating point is  : (0.50V,4.50mA)\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "Vth= 0.5# V\n",
+    "R_F= 1*10**3# ohm\n",
+    "V= 5# V\n",
+    "# Applying KVL for loop, V-Vd-R_F*Ii= 0    (i)\n",
+    "# When Ii=0\n",
+    "Vd= V# V\n",
+    "# When Vd= 0\n",
+    "Ii= V/R_F# A\n",
+    "# From eq(i)\n",
+    "Ii= (V-Vth)/R_F# A\n",
+    "Vd= V-R_F*Ii# V\n",
+    "print \"The operating point is  : (%.2fV,%.2fmA)\"%(Vd,Ii*1000)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.43 Page No 76"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 37,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The voltage at V1 = 6.00 volts\n",
+      "The voltage at V2 = 5.40 volts\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "V_CC = 6# V\n",
+    "Vr = 0.6# V\n",
+    "V1= V_CC##in V\n",
+    "V2 = V1-Vr# V\n",
+    "print \"The voltage at V1 = %.2f volts\"%V1\n",
+    "print \"The voltage at V2 = %.2f volts\"%V2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.44 Page No 76"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 38,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The value of I1 = 1.80 mA\n",
+      "The value of I2 = 1.80 mA\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "V_T = 0.7# V\n",
+    "V = 5# V\n",
+    "R = 2# k ohm\n",
+    "R = R * 10**3# ohm\n",
+    "Vs = 0.7\n",
+    "Vx = Vs+V_T# V\n",
+    "# The value of I1 \n",
+    "I1 = (V-Vx)/R# A\n",
+    "I1 = I1 * 10**3# mA\n",
+    "print \"The value of I1 = %.2f mA\"%I1\n",
+    "# The value of I2 \n",
+    "I2 = I1# mA\n",
+    "print \"The value of I2 = %.2f mA\"%I2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 1.45 Page No 77"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 39,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The value of Vo = 1.00 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given data\n",
+    "Rf = 300.# ohm\n",
+    "V = 0.5# V\n",
+    "R = 600.# ohm\n",
+    "Vi = 2.# V\n",
+    "# The output voltage \n",
+    "Vo = (Vi-V)*( R/(R+Rf) )# V\n",
+    "print \"The value of Vo = %.2f V\"%Vo"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/PraveenKumar/PraveenKumar_version_backup/chapter2.ipynb b/sample_notebooks/PraveenKumar/PraveenKumar_version_backup/chapter2.ipynb
new file mode 100755
index 00000000..3d7aab33
--- /dev/null
+++ b/sample_notebooks/PraveenKumar/PraveenKumar_version_backup/chapter2.ipynb
@@ -0,0 +1,429 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "chapter-2, Economics of generation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1, Page 73"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "#To Determine the Demand and Supply Parameters for 15 bulbs\n",
+      "\n",
+      "W=60  #Wattage of the bulb\n",
+      "N=15  #No. of bulbs\n",
+      "CL=W*N  #Connected Load \n",
+      "Wih=2*(10**3)  #Wattage of immersion heater\n",
+      "Wh=2*(10**3)  #Wattage of heater\n",
+      "\n",
+      "#Usage of Bulbs at different time periods\n",
+      "N1=5 \n",
+      "N2=10 \n",
+      "N3=6\n",
+      "\n",
+      "#Time periods for bulbs\n",
+      "T1=2  #6pm - 8pm\n",
+      "T2=2  #8pm - 10pm\n",
+      "T3=2  #10pm - 12pm\n",
+      "#Time Periods for heaters\n",
+      "T4=4  #1pm - 5pm\n",
+      "T5=3  #8pm - 11pm\n",
+      "\n",
+      "#CASE 1\n",
+      "MD1=W*N2  #Maximum Demand\n",
+      "DF=MD1*100/CL  #Demand Factor\n",
+      "EC1=(N1*W*T1)+(N2*W*T2)+(N3*W*T3)  #Energy Consumed\n",
+      "DLF1=EC1*100/(24*MD1)  #Daily Load Factor\n",
+      "\n",
+      "#CASE 2\n",
+      "MD2=(W*N2)+Wh  #From 8pm - 10pm\n",
+      "EC2=(T4*Wih)+(T5*Wh)+EC1  #Energy Consumed\n",
+      "DLF2=EC2*100/(24*MD2)  #Daily Load Factor\n",
+      "\n",
+      "print '''i)a) Connected Load is %0.2f W\\nb) The Maximum Demand is %0.2f W\n",
+      "c) The Demand Factor is %0.2f percent\\nd) The Daily Load Factor is %0.2f percent''' %(CL,MD1,DF,DLF1)\n",
+      "print 'ii) The Improved Daily Load Factor is %0.2f percent' %DLF2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)a) Connected Load is 900.00 W\n",
+        "b) The Maximum Demand is 600.00 W\n",
+        "c) The Demand Factor is 66.67 percent\n",
+        "d) The Daily Load Factor is 17.50 percent\n",
+        "ii) The Improved Daily Load Factor is 26.47 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2, Page 74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#To determine the Demand and supply parameter of four consumers\n",
+      "\n",
+      "\n",
+      "#Maximum Demands of various users\n",
+      "MD1=2*(10**3)  #9pm\n",
+      "MD2=2*(10**3)  #12 noon\n",
+      "MD3=8*(10**3)  #5pm\n",
+      "MD4=4*(10**3)  #8pm\n",
+      "MDT=MD1+MD2+MD3+MD4  #Sum of all Maximum Demands\n",
+      "\n",
+      "#Demands of various users\n",
+      "D1=1.6*(10**3)  #8pm\n",
+      "D2=1*(10**3)  #8pm\n",
+      "D3=5*(10**3)  #8pm\n",
+      "\n",
+      "#The Number after the Alphabets represents the Consumer\n",
+      "\n",
+      "#Maximum Demand of the System arises at 8.00 PM\n",
+      "MDS = D1+D2+D3+MD4 \n",
+      "\n",
+      "TDF=MDT/MDS  #Diversity Factor\n",
+      "#Given Values\n",
+      "#Average Loads\n",
+      "AL2=500 \n",
+      "AL4=1000 \n",
+      "#Load Factors\n",
+      "LF1=15/100 \n",
+      "LF3=25/100 \n",
+      "#Calculated Values\n",
+      "#Average Loads\n",
+      "AL1=LF1*MD1 \n",
+      "AL3=LF3*MD3 \n",
+      "#Load Factors\n",
+      "LF2=AL2*100/MD2 \n",
+      "LF4=AL4*100/MD4 \n",
+      "\n",
+      "ALS=AL1+AL2+AL3+AL4   #Combined Average Loads\n",
+      "LFS=ALS*100/MDS  #Combined Load Factor\n",
+      "\n",
+      "#Load Percent\n",
+      "LF1*=100 # %\n",
+      "LF3*=100 # %\n",
+      "\n",
+      "print 'i) The Diversity Factor is %0.2f' %TDF\n",
+      "print 'ii) The Average load and Load factor of:'\n",
+      "print ' Consumer 1 : %0.2f W and %0.2f percent' %(AL1,LF1)\n",
+      "print ' Consumer 2 : %0.2f W and %0.2f percent' %(AL2,LF2)\n",
+      "print ' Consumer 3 : %0.2f W and %0.2f percent' %(AL3,LF3)\n",
+      "print ' Consumer 4 : %0.2f W and %0.2f percent' %(AL4,LF4)\n",
+      "print 'iii) The Combined Load Factor and the Combined Average Load is %0.2f percent and %0.2f W respectively\\n' %(LFS,ALS)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) The Diversity Factor is 1.38\n",
+        "ii) The Average load and Load factor of:\n",
+        " Consumer 1 : 300.00 W and 15.00 percent\n",
+        " Consumer 2 : 500.00 W and 25.00 percent\n",
+        " Consumer 3 : 2000.00 W and 25.00 percent\n",
+        " Consumer 4 : 1000.00 W and 25.00 percent\n",
+        "iii) The Combined Load Factor and the Combined Average Load is 32.76 percent and 3800.00 W respectively\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3, Page 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#To Determine the Yearly Cost of the substation\n",
+      "\n",
+      "Teff=95/100  #Transmission Efficiency\n",
+      "Deff=85/100  #Distribution Efficiency\n",
+      "DFT=1.2  #Diversity Factor For Transmission\n",
+      "DFD=1.3  #Diversity Factor For Distribution\n",
+      "MDGS=100*(10**6)  #Maximum Demand of Generating Station\n",
+      "ALF=40/100  #Annual Load Factor\n",
+      "ACCT=2.5*(10**6)  #Annual Capital Charge for Transmission\n",
+      "ACCD=2*(10**6)  #Annual Capital Charge for Distribution\n",
+      "GCC=100  #Generating Cost per kW demand\n",
+      "GCCU=5/100  # Per Unit Cost\n",
+      "#Fixed Charges from Supply to Substation Annually\n",
+      "GFC=GCC*MDGS/1000  #Generating\n",
+      "TFC=ACCT  #Transmission\n",
+      "TotFCS=GFC+TFC  #Total\n",
+      "#Fixed Charges for supply upto Consumer Annually\n",
+      "DFC=ACCD  #Distribution\n",
+      "TotFCC=TotFCS+DFC  #Total\n",
+      "\n",
+      "AMDS= DFT*MDGS/1000  #Aggregate of Maximum Demand at Supply\n",
+      "AMDC= DFD*AMDS  #Aggregate of Maximum Demand for Consumers\n",
+      "\n",
+      "FCS=TotFCS/AMDS  #Fixed Charges Per KW at substation\n",
+      "CES=GCCU/Teff  #Cost of energy at the substation\n",
+      "\n",
+      "FCC=TotFCC/AMDC  #Fixed Charges per KW at the consumer premises\n",
+      "CEC=CES/Deff  #Cost of Energy at the consumer premises\n",
+      "\n",
+      "CEC*=100  # converting from rupee to paise\n",
+      "\n",
+      "print 'The Yealy Cost per KW demand and the cost per KWhr at:'\n",
+      "print 'a) The substation is %0.2f rupees per KW and %0.2f paise per kWhr'%(FCS,CES)\n",
+      "print 'b) The consumer premises is %g rupees per KW and %g paise per kWhr' %(FCC,CEC)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Yealy Cost per KW demand and the cost per KWhr at:\n",
+        "a) The substation is 104.17 rupees per KW and 0.05 paise per kWhr\n",
+        "b) The consumer premises is 92.9487 rupees per KW and 6.19195 paise per kWhr\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex4, Page 78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#To determine the Load factor and suitable units for 24 hr operation of the plant\n",
+      "\n",
+      "\n",
+      "#Demands at Various Time Periods starting from 12PM to 12PM\n",
+      "D1=500*(10**3)  \n",
+      "D2=800*(10**3) \n",
+      "D3=2000*(10**3) \n",
+      "D4=1000*(10**3) \n",
+      "D5=2500*(10**3) \n",
+      "D6=2000*(10**3) \n",
+      "D7=1500*(10**3) \n",
+      "D8=1000*(10**3) \n",
+      "\n",
+      "MD=D5  #Maximum Demand\n",
+      "#Time Periods of demands from 12PM\n",
+      "T1=5 \n",
+      "T2=5 \n",
+      "T3=2 \n",
+      "T4=2 \n",
+      "T5=3 \n",
+      "T6=3 \n",
+      "T7=2 \n",
+      "T8=2 \n",
+      "\n",
+      "#Total Energy Demand in 24hrs\n",
+      "TED=(T1*D1)+(T2*D2)+(T3*D3)+(D4*T4)+(T5*D5)+(D6*T6)+(D7*T7)+(T8*D8) \n",
+      "\n",
+      "LF=TED*100/(24*MD) \n",
+      "\n",
+      "C1000=3*1000*(10**3)  #1000 unit \n",
+      "C500=1*500*(10**3)  #500 Unit\n",
+      "\n",
+      "TCP=C1000+C500  #Total capacity of the plant\n",
+      "PCF=TED*100/(24*TCP)  #Plant Capacity Factor\n",
+      "\n",
+      "#Operating Schedule, Units operated can be seen in the textbook\n",
+      "G1=500*(10**3)   \n",
+      "G2=1000*(10**3) \n",
+      "G3=2000*(10**3) \n",
+      "G4=1000*(10**3) \n",
+      "G5=2500*(10**3) \n",
+      "G6=2000*(10**3) \n",
+      "G7=1500*(10**3) \n",
+      "G8=1000*(10**3) \n",
+      "\n",
+      "TEG=(T1*G1)+(T2*G2)+(T3*G3)+(G4*T4)+(T5*G5)+(G6*T6)+(G7*T7)+(T8*G8) #Total Energy Generated\n",
+      "PUF=TED*100/(TEG)  #Plant Use Factor\n",
+      "\n",
+      "print 'a) The Reserve Capacity is a 1000kW Unit and Load Factor is %0.2f percent' %LF\n",
+      "print 'b) The Plant Capacity Factor is %0.2f percent' %PCF\n",
+      "print 'c) The Plant Use Factor is %0.2f percent' %PUF"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Reserve Capacity is a 1000kW Unit and Load Factor is 51.67 percent\n",
+        "b) The Plant Capacity Factor is 36.90 percent\n",
+        "c) The Plant Use Factor is 96.88 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5, Page 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#To determine the Plant use factore of each unit\n",
+      "\n",
+      "\n",
+      "MDS=25*(10**6)  #Maximum Demand on the System\n",
+      "U1=15*(10**6)  #Load Supplied By Unit 1\n",
+      "U2=12.5*(10**6)  #Load Supplied By Unit 2\n",
+      "#Running Time Factor of the Unit\n",
+      "T1=1 \n",
+      "T2=40/100 \n",
+      "\n",
+      "#Energy generated by each unit\n",
+      "E1=1*(10**8) \n",
+      "E2=1*(10**7) \n",
+      "Et=E1+E2  #Total Energy\n",
+      "\n",
+      "#Maximum Demands on Each Units\n",
+      "MD1=U1 \n",
+      "MD2=MDS-U1 \n",
+      "\n",
+      "#Annual Load Factor for the Units\n",
+      "ALF1=E1*1000*100/(MD1*8760) \n",
+      "ALF2=E2*1000*100/(MD2*8760) \n",
+      "\n",
+      "LF2=E2*1000*100/(MD2*0.4*8760)  #Load Factor for the it is loaded\n",
+      "\n",
+      "\n",
+      "PUF1=ALF1  #Plant Use Factor\n",
+      "PCF1=ALF1  # Plant Capacity Factor\n",
+      "\n",
+      "PCF2=E2*1000*100/(U2*8760)  #Plant Capacity Factor for Unit 2\n",
+      "PUF2=E2*1000*100/(U2*0.4*8760) #Plant Use Factor for Unit 2\n",
+      "\n",
+      "LFP=Et*100*1000/(MDS*8760)  #Annual Load Factor of the Complete Plant\n",
+      "\n",
+      "print 'The Load Factor, Plant Capacity Factor, Plant Use Factor of:'\n",
+      "print 'Unit 1 : %0.2f percent, %0.2f percent, %0.2f percent' %(ALF1,PCF1,PUF1)\n",
+      "print 'Unit 2 : %0.2f percent, %0.2f percent, %0.2f percent' %(ALF2,PCF2,PUF2)\n",
+      "print 'The Annual Load Factor of the Entire Plant is %0.2f percent' %LFP"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Load Factor, Plant Capacity Factor, Plant Use Factor of:\n",
+        "Unit 1 : 76.10 percent, 76.10 percent, 76.10 percent\n",
+        "Unit 2 : 11.42 percent, 9.13 percent, 22.83 percent\n",
+        "The Annual Load Factor of the Entire Plant is 50.23 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex6, Page 91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#To determine the most economic power factor\n",
+      "\n",
+      "from numpy import sqrt\n",
+      "\n",
+      "P=200*(10**3)  #Maximum Demand\n",
+      "pf=0.707  #Power Factor Lagging\n",
+      "\n",
+      "a=100  #Tariff per kVA per year\n",
+      "\n",
+      "b=200  #Power factor improvement cost Per kVA.\n",
+      "r=20  #Interest Depriciation, maintenance and cost of losses amount to  20% of capital cost per year\n",
+      "\n",
+      "# Economic PF = sqrt(1-((b1/a)**2))\n",
+      "\n",
+      "b1=r*b/100 # b' term accrding to the equation above\n",
+      "\n",
+      "pfeco=sqrt(1-((b1/a)**2))  #Economic Power Factor\n",
+      "\n",
+      "print 'The Economic Power Factor is %0.2f ' %pfeco\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Economic Power Factor is 0.92 \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}