path: root/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch6.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 6 Applications of Operational Amplifier"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.1 Pg 140"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 116,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the input resistance is = 20.00 kohm\n",
      "The resistance R2 is = 100.00 kohm\n"
     ]
    }
   ],
   "source": [
    "# Design an inverting amplifier\n",
    "Av = -5 #\n",
    "#V1 = 0.1 sin wt #\n",
    "V1 = 0.1   # # *sin wt #\n",
    "i = 5*10**-6 #\n",
    "\n",
    "# the input resistance \n",
    "R1 = V1/i / 1000 # kohm\n",
    "print 'the input resistance is = %0.2f'%R1,'kohm'#\n",
    "\n",
    "# The resistance R2\n",
    "#Av = -(R2/R1)#\n",
    "R2 = -(Av*R1)#\n",
    "print 'The resistance R2 is = %0.2f'%R2,'kohm'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.2 Pg 141"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 117,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the input resistance is = 20.00 kohm\n",
      "The resistance R2 is = 80.00 kohm\n"
     ]
    }
   ],
   "source": [
    "# Design an non inverting amplifier\n",
    "Av = 5 #\n",
    "#V1 = 0.1 sin wt #\n",
    "V1 = 0.1 #\n",
    "i = -5*10**-6 #\n",
    "\n",
    "# the input resistance \n",
    "R1 = -V1/i/1000 # kohm\n",
    "print 'the input resistance is = %0.2f'%R1,'kohm'#\n",
    "\n",
    "# The resistance R2\n",
    "#Av = 1+(R2/R1)#\n",
    "R2 = (Av-1)*R1#\n",
    "print 'The resistance R2 is = %0.2f'%R2,'kohm'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.3 Pg 146"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 118,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the cut off frequency of phase shifter is = 723.43 Hz\n",
      "The phase shift is = -15.75\n"
     ]
    }
   ],
   "source": [
    "# To calculate phase shift between two extremes \n",
    "C = 0.22*10**-6 #\n",
    "R = 1*10**3 #\n",
    "f = 1*10**3 #\n",
    "\n",
    "# the cut off frequency of phase shifter \n",
    "fc = 1/(2*pi*R*C) #\n",
    "print 'the cut off frequency of phase shifter is = %0.2f'%fc,'Hz'#\n",
    "f\n",
    "# the phase shift\n",
    "f = 1 # # KHz\n",
    "fc = 7.23 #  # KHz \n",
    "from math import atan ,degrees\n",
    "PS = -2*degrees(atan(f/fc))\n",
    "print 'The phase shift is = %0.2f'%PS"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.4 Pg 146"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 119,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The value of resistance is = 1.92 kohm\n"
     ]
    }
   ],
   "source": [
    "# To design a phase shifter\n",
    "f = 2*10**3 #\n",
    "PS = -135 #\n",
    "# the phase shift\n",
    "# PS = -2*atand(2*pi*R*C)#\n",
    "#RC = 192.1*10**-6 #\n",
    "C = 0.1*10**-6 #\n",
    "R = (192.1*10**-6)/C/1000 # kohm\n",
    "print 'The value of resistance is = %0.2f'%R,'kohm'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.5 Pg 153"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 120,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The value of resistance R1 is = 25.00 kohm\n",
      "The value of resistance R3 is = 25.00 kohm\n",
      "The value of resistance R2 is = 750.00 kohm\n",
      "The value of resistance R4 is = 750.00 kohm\n"
     ]
    }
   ],
   "source": [
    "# Design a difference amplifier\n",
    "Ri = 50 # kohm\n",
    "Ad = 30 \n",
    "\n",
    "R1 = Ri/2 #\n",
    "print 'The value of resistance R1 is = %0.2f'%R1,'kohm'#\n",
    "R3 = R1 #\n",
    "print 'The value of resistance R3 is = %0.2f'%R3,'kohm'#\n",
    "\n",
    "# the differential gain\n",
    "#Ad = R2/R1 #\n",
    "R2 = 30*R1 #\n",
    "print 'The value of resistance R2 is = %0.2f'%R2,'kohm'#\n",
    "\n",
    "R4 = R2 #\n",
    "print 'The value of resistance R4 is = %0.2f'%R4,'kohm'# "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.6 Pg 154"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 121,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The common mode rejection ratio is = 26.58  dB\n"
     ]
    }
   ],
   "source": [
    "# Calculate CMRR ratio\n",
    "Ad = 10.24 #\n",
    "Acm = 0.48 #\n",
    "\n",
    "# the common mode rejection ratio CMRR is defined as\n",
    "CMRRdB = 20*log10(Ad/Acm)#\n",
    "print 'The common mode rejection ratio is = %0.2f'%CMRRdB,' dB'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.7 Pg 156"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 122,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " The feedback resistance is = 100.00 kohm\n",
      " The value of resistance R1 is = 100.00 kohm\n"
     ]
    }
   ],
   "source": [
    "# Design current to voltage converter\n",
    "Vo =-10 #\n",
    "Is = 100*10**-6 #\n",
    "\n",
    "# the output voltage of current to voltage converter is defined as\n",
    "#Vo =-1s*R2 \n",
    "R2 = -Vo/Is/1000 #kohm\n",
    "print ' The feedback resistance is = %0.2f'%R2,'kohm'#\n",
    "\n",
    "R1 = R2 #\n",
    "print ' The value of resistance R1 is = %0.2f'%R1,'kohm'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.8 Pg 157"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 123,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The value of resistance R2 is = 44.78 kohm\n"
     ]
    }
   ],
   "source": [
    "# Design high sensitivity current to voltage converter\n",
    "R1 = 5 # kohm\n",
    "Is = 1 #\n",
    "KR = 0.01/10**9 #  #    V / nA\n",
    "\n",
    "# the output voltage of high sensitivity current to voltage converter\n",
    "Vo =-KR*Is #\n",
    "KR = 10*10**6 #\n",
    "R = 1*10**6 #   #we assume   then\n",
    "K = 10 #\n",
    "#1 + (R2/R1)+(R2/R) = 10 #\n",
    "# solving above equation we get\n",
    "\n",
    "R2 = 9*((5*10**6)/(10**3+5))/1000 # kohm\n",
    "print 'The value of resistance R2 is = %0.2f'%R2,'kohm'# "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.9 Pg 160"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 124,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The load current iL is = 5.00 mA\n",
      "The current i3 is = 0.50 mA\n",
      "The current iA is = 5.50 mA\n",
      "The output voltage is = 6.00  V\n",
      "The current i1 is = 49.40  A\n",
      "The current i2 is = 49.40  A\n"
     ]
    }
   ],
   "source": [
    "# Determine a load current in a V to I converter\n",
    "R1 = 10 # kohm\n",
    "R2 = 10  # koohm\n",
    "R3 = 1 # kohm\n",
    "R4 = 1 # kohm\n",
    "VI = -5 #\n",
    "\n",
    "# The Load Current\n",
    "iL = -VI/R3 #\n",
    "print 'The load current iL is = %0.2f'%iL,'mA'#\n",
    "\n",
    "VL = 0.5 #\n",
    "# The  Current i3 and iA\n",
    "i3 = VL/R3 #\n",
    "print 'The current i3 is = %0.2f'%i3,'mA'#\n",
    "\n",
    "iA = i3+iL #\n",
    "print 'The current iA is = %0.2f'%iA,'mA'#\n",
    "\n",
    "# the output voltage \n",
    "Vo = (iA*R3)+VL #\n",
    "print 'The output voltage is = %0.2f'%Vo,' V'#\n",
    "\n",
    "ZL =100 #\n",
    "# The current i1 and i2 \n",
    "#i1 = (VI-iL*ZL)/R1 #\n",
    "i1 = (iL*ZL-Vo)/R2 #\n",
    "print 'The current i1 is = %0.2f'%i1,' A'#\n",
    "\n",
    "i2 = i1 #\n",
    "print 'The current i2 is = %0.2f'%i2,' A'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.10 Pg 163"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 125,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The value of resistance R1f is = 0.0606 K ohm \n",
      "The value of resistance R2 is = 75.5 K ohm \n"
     ]
    }
   ],
   "source": [
    "# Design an instrumentation amplifier\n",
    "#A = 5 to 500 #  adjustable gain\n",
    "VR = 100*10**3 #\n",
    "\n",
    "# the maximum differential gain of instrumentation amplifier is 500 \n",
    "#Amax = (R4/R3)*(1+(2R2/R1))#\n",
    "#by solving above equation we get following equation\n",
    "# 2R2 -249R1f = 0                                                  equation 1\n",
    "\n",
    "# the minimum differential gain of instrumentation amplifier is 5\n",
    "# Amin = (R4/R3)*(1+(2R2/R1)) #\n",
    "#by solving above equation we get following equation\n",
    "# 2R2 -1.5R1f = 150*10**3                                           equation 2\n",
    "\n",
    "#by solving equation 1 and 2 we get\n",
    "print 'The value of resistance R1f is = 0.0606 K ohm '#\n",
    "\n",
    "print 'The value of resistance R2 is = 75.5 K ohm '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.11 Pg 164"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 127,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The value of resistane R1 is = 9.09 kohm\n"
     ]
    }
   ],
   "source": [
    " # To find the value of resistance R1 for instrumentation amplifier\n",
    "A =100 #\n",
    "R2 = 450*10**3 #\n",
    "R3 = 1*10**3 #\n",
    "R4 = 1*10**3 #\n",
    "\n",
    "# The gain of differential amplifier \n",
    "# A = (R4/R3)*(1+(2R2/R1)) #\n",
    "#but  R3 = R4  then\n",
    "# A = 1+(2R2/R1) #\n",
    "R1 = 2*R2/(A-1)/1000 # kohm\n",
    "print 'The value of resistane R1 is = %0.2f'%R1,'kohm'# "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.12 Pg 167"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 130,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " At t = 1 msec the time constant RC is = 0.10  m sec\n",
      " if C = 0.01 uF then R of RC time constant is = 10 K ohm \n",
      " if C = 0.001 uF then R of RC time constant is = 100 K ohm \n"
     ]
    }
   ],
   "source": [
    "# determine the time constant of an integrator\n",
    "Vo = 10 #  # at t= 1 m sec\n",
    "t = 1 # # m sec\n",
    "\n",
    "# the output of integrator \n",
    "#Vo = t/RC # when t is from 0 to 1\n",
    "RC = t/Vo #\n",
    "print ' At t = 1 msec the time constant RC is = %0.2f'%RC,' m sec'#\n",
    "\n",
    "print ' if C = 0.01 uF then R of RC time constant is = 10 K ohm '#\n",
    "\n",
    "print ' if C = 0.001 uF then R of RC time constant is = 100 K ohm '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.13 Pg 168"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 131,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " The feedback resistance Rf is = 100.00 kohm\n",
      " The frequency fa is = 2.00 kHz\n",
      " The value of capacitor C is = 0.8 nF \n"
     ]
    }
   ],
   "source": [
    "# Design an integrator circuit\n",
    "A = 10 #\n",
    "f =20*10**3 #\n",
    "R = 10*10**3 #  # we assume \n",
    "Rf =10*R #\n",
    "\n",
    "print ' The feedback resistance Rf is = %0.2f'%(Rf/1000),'kohm'#\n",
    "\n",
    "# for proper integration f>= 10fa \n",
    "fa = f/10/1000 #\n",
    "print ' The frequency fa is = %0.2f'%fa,'kHz'#\n",
    "\n",
    "# in practical integrator\n",
    "#fa = 1/(2*pi*Rf*C)#\n",
    "\n",
    "C = 1/(2*pi*Rf*fa)*1e6# nF\n",
    "print ' The value of capacitor C is = %0.1f'%C,'nF '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.1 Pg 185"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the value of resistance R1 is = 0.199 Mohm\n",
      "the value of resistance R2 is  = 1 Mohm\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "# design an inverting amplifier with a closed loop voltage gain of Av = -5\n",
    "Av = -5 #\n",
    "Is = 5*10**-6 # # A\n",
    "Rs = 1*10**3 # # ohm\n",
    "# input voltage source Vs = sinwt volts\n",
    "\n",
    "# in an inverting amplifier frequency effect is neglected then i/p volt Vin = 1 V and total resistance equal to Rs+R1\n",
    "\n",
    "# the input current can be written as Iin=Is\n",
    "# Is = (Vin/Rs+R1)#\n",
    "Iin = Is#\n",
    "Vin = 1 # # V\n",
    "R1 = (1-(Iin*Rs))/Iin  #\n",
    "print 'the value of resistance R1 is = %0.3f'%(R1/1e6),'Mohm'#\n",
    "\n",
    "# closed loop voltage gain of an inverting amplifier\n",
    "#Av = -(R2/Rs+R1)\n",
    "R2 = -(Av*(Rs+R1))#\n",
    "print 'the value of resistance R2 is  = %0.f'%(R2/1e6),'Mohm'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.2 Pg 186"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the value of resistance R1 is = 8 kohm\n",
      "the value of resistance R2 is  = 72 kohm\n"
     ]
    }
   ],
   "source": [
    " # design an inverting amplifier with a closed loop voltage gain of Av = 10\n",
    "Av = 10 #\n",
    "Vin = 0.8 #  #V\n",
    "Iin = 100*10**-6 # # A\n",
    "# in an non- inverting amplifier the input voltage Vin=V1=V2 because of vortual short effect then the i/p current In = Vin/R1\n",
    "R1 = Vin/Iin/1e3\n",
    "print 'the value of resistance R1 is = %0.f'%R1,'kohm'#\n",
    "\n",
    "# closed loop voltage gain of an non-inverting amplifier\n",
    "#Av = Vo/Vin =  (1+R2/R1)\n",
    "R2 = (Av-1)*R1 # kohm\n",
    "print 'the value of resistance R2 is  = %0.f'%R2,'kohm'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.3 Pg 187"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the value of resistance R1 is  = 20.00 kohm\n",
      "the value of resistance R2 is  = 80.00 kohm\n",
      "the output current I2 is  = 50 uA\n"
     ]
    }
   ],
   "source": [
    "# design an non-inverting amplifier with colsed loop gain of 5 limited voltage of -5 V <= Vo <= 5 V  and maximum i/p c/n 50 uA\n",
    "R1 = 8*10**3 # # ohm\n",
    "R2 = 72*10**3 # # ohm\n",
    "Iin = 50*10**-6 # # A\n",
    "Vo = 5 # # V \n",
    "\n",
    "# closed loop gain\n",
    "#Av = Vo/Vin = (1+R2/R1)\n",
    "Av = 1+(R2/R1)#\n",
    "# but  \n",
    "Av = 5 #\n",
    "# then\n",
    "# (R2/R1) = 4 #\n",
    "\n",
    "# the output voltage of the amplifier is Vo = 5 V \n",
    "#i.e\n",
    "Vin = 1 # # V\n",
    "# Iin = Vin/R1 #\n",
    "R1 = Vin/Iin/1e3\n",
    "print 'the value of resistance R1 is  = %0.2f'%R1,'kohm'#\n",
    "\n",
    "R2 = 4*R1 #\n",
    "print 'the value of resistance R2 is  = %0.2f'%R2,'kohm'#\n",
    "\n",
    "# the output current I2 is given as\n",
    "I2 = (Vo-Vin)/R2*1e3 # uA\n",
    "print 'the output current I2 is  = %0.f'%I2,'uA'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.4 Pg 188"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the value of resistance Ria is  = 20.00 kohm\n",
      "the value of resistance Rib is  = 15.00 kohm\n",
      "the value of resistance Ric is  = 30.00 kohm\n"
     ]
    }
   ],
   "source": [
    "# Design a op-amp circuit to provide the output voltage Vo = -2(3 V1 +4 V2 +2 V3)\n",
    "# Vo = -2(3 V1 + 4 V2+ 2 V3)#      equation 1\n",
    "# the output of the summer circuit is given as\n",
    "# Vo = -R2((Via/Ria)+(Vib/Rib)+(Vic/Ric))    equation 2\n",
    "\n",
    "# compare equation 1 and 2  of Vo we get \n",
    "\n",
    "#  (R2/Ria)= 6 #\n",
    "#  (R2/Rbi=8 #\n",
    "#  (R2/Ric)=4 #\n",
    "\n",
    "R2 = 120*10**3/1e3 #   # we choose then \n",
    "\n",
    "Ria = R2/6 #\n",
    "print 'the value of resistance Ria is  = %0.2f'%Ria,'kohm'#\n",
    "\n",
    "Rib = R2/8 #\n",
    "print 'the value of resistance Rib is  = %0.2f'%Rib,'kohm'#\n",
    "\n",
    "Ric = R2/4 #\n",
    "print 'the value of resistance Ric is  = %0.2f'%Ric,'kohm'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.5 Pg 188"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the value of resistance Ria is  = 90.00  kohm\n",
      "the value of resistance Rib is  = 42.00  kohm\n",
      "the value of resistance Ric is  = 63.00  kohm\n",
      "the value of resistance Rid is  = 210.00 kohm\n"
     ]
    }
   ],
   "source": [
    " # Design a summing amplifier circuit to provide the output voltage Vo = -(7 V11 + 15 V12 + 10 V13 + 3 V14)\n",
    "R2 = 630# kohm # Assume feedback resistance\n",
    "# Vo = -(7 V11 + 15 V12 + 10 V13 + 3 V14)#                    equation 1\n",
    "# the output of the summer circuit is given as\n",
    "# Vo = -R2((Via/Ria)+(Vib/Rib)+(Vic/Ric)+(Vid/Rid))           equation 2\n",
    "\n",
    "# compare equation 1 and 2  of Vo we get \n",
    "\n",
    "#  (R2/Ria)= 7 #\n",
    "#  (R2/Rbi= 15 #\n",
    "#  (R2/Ric)= 10 #\n",
    "#  (R2/Rid)= 3 #\n",
    "\n",
    "Ria = R2/7 #\n",
    "print 'the value of resistance Ria is  = %0.2f'%Ria,' kohm'#\n",
    "\n",
    "Rib = R2/15 #\n",
    "print 'the value of resistance Rib is  = %0.2f'%Rib,' kohm'#\n",
    "\n",
    "Ric = R2/10 #\n",
    "print 'the value of resistance Ric is  = %0.2f'%Ric,' kohm'#\n",
    "\n",
    "Rid = R2/3 #\n",
    "print 'the value of resistance Rid is  = %0.2f'%Rid,'kohm'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.6 Pg 190"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the value of resistance R2 is  = 300.00 kohm\n",
      "the value of resistance R4 is  = 33333.33 kohm\n"
     ]
    }
   ],
   "source": [
    "# Design a op-amp circuit to provide the output voltage Vo = V2 - 3 V1 with Ri1 =Ri2 = 100*10**3\n",
    "Ri1 = 100 # # kohm\n",
    "Ri2 = 100  # # kohm\n",
    "# the i/p resistance \n",
    "R1 = Ri1 #\n",
    "R3 = Ri2 #\n",
    "\n",
    "# Vo = V2 - 3 V1#                                    equation 1\n",
    "# the output of the summer circuit is given as\n",
    "# Vo = [(R4/(R3+R4)*(1+(R2/R1))*Vi2-(R2/R1)*Vi1]        equation 2\n",
    "\n",
    "# compare equation 1 and 2  of Vo we get \n",
    "# (R4/(R3+R4)*(1+(R2/R1)) = 1 #                           equation  3\n",
    "# R2/R1 = 3 #                                           equation 4\n",
    "\n",
    "# by subsituting the value of R1 and R3 in equation 3 and 4\n",
    "\n",
    "# from equation  4\n",
    "R2 = 3*R1 #\n",
    "print 'the value of resistance R2 is  = %0.2f'%R2,'kohm'#\n",
    "\n",
    "# from equation  3\n",
    "R4 = (100*10**3)/3 #\n",
    "print 'the value of resistance R4 is  = %0.2f'%R4,'kohm'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.7 Pg 191"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The load current iL is  = 5 mA\n",
      "The voltage across load VL is  = 1.00  V\n",
      "The non-inverting current across i3 is = 1.00 mA\n",
      "The non-inverting current across i4 is  = 6.00 mA\n",
      "The output voltage of given voltage to current converter is  = 6.00  V\n"
     ]
    }
   ],
   "source": [
    " # determine the load current and output voltage\n",
    "Vin = -5 #  # V\n",
    "ZL = 200 # # ohm\n",
    "R1 = 10*10**3 # # ohm\n",
    "R2 = 10*10**3 # # ohm\n",
    "R3 = 1*10**3 # # ohm\n",
    "R4 = 1*10**3 # # ohm\n",
    "\n",
    "# the load c/n of the given voltage to c/n converter circuit is given by\n",
    "iL =-Vin/(R1*R4)*R2*1e3 # m\n",
    "print 'The load current iL is  = %0.f'%iL,'mA'#\n",
    "\n",
    "# the voltage across the load \n",
    "VL = iL/1e3*ZL#\n",
    "print 'The voltage across load VL is  = %0.2f'%VL,' V'#\n",
    "\n",
    "# the non-inverting current across i3 and i4 are\n",
    "i3 = VL/R3*1000 #mA\n",
    "print 'The non-inverting current across i3 is = %0.2f'%i3,'mA'#\n",
    "\n",
    "i4 = iL+i3 # mA\n",
    "print 'The non-inverting current across i4 is  = %0.2f'%i4,'mA'#\n",
    "\n",
    "# the output voltage of given voltage to current converter is given by\n",
    "Vo = (iL/1e3*R3)+VL #\n",
    "print 'The output voltage of given voltage to current converter is  = %0.2f'%Vo,' V'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.8 Pg 192"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 33,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The common mode rejection ratio CMRR is  = 120.50  \n",
      "The common mode rejection ratio CMRR in decibel is  = 41.62  dB \n"
     ]
    }
   ],
   "source": [
    "from math import log10\n",
    "# determine the common mode rejection ratio CMRR\n",
    "# R2/R1 = 10 #\n",
    "# R4/R3 = 11 #\n",
    "\n",
    "# the output of the difference amplifier is given by\n",
    "# Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n",
    "\n",
    "# putting R1 R2 R3 R4 value in above equation we get Vo as\n",
    "\n",
    "# Vo =(121/12)*VI2-10VI1 #                                 equation 1\n",
    "\n",
    "# the differential mode input of difference amplifier is given by\n",
    "# Vd = VI2-VI1 #                                               eqution 2\n",
    "\n",
    "# the common mode input of difference amplifier is given by\n",
    "# VCM = (VI2+VI1)/2 #                                         equation 3\n",
    "\n",
    "# from equation 2 and 3 \n",
    "\n",
    "# VI1 = VCM-Vd/2 #                                          equation 4\n",
    "\n",
    "# VI2 = VCM+Vd/2 #                                          equation 5\n",
    "\n",
    "# substitute equation 4 and 5 in 1  we get \n",
    "# Vo = (VCM/12)+(241Vd/24)#                               equation6\n",
    "\n",
    "# Vd = Ad*Vd+ACM*VCM #                                  equation 7\n",
    "\n",
    "#equation from equation 6 and 7 we get\n",
    "\n",
    "Ad = 241/24 #\n",
    "ACM = 1/12 #\n",
    "\n",
    "# the common mode rejection ratio CMRR is \n",
    "CMRR = abs(Ad/ACM)#\n",
    "print 'The common mode rejection ratio CMRR is  = %0.2f'%CMRR,' '#\n",
    "\n",
    "# in decibal it can be expressed as\n",
    "\n",
    "CMRR = 20*log10(CMRR)#\n",
    "print 'The common mode rejection ratio CMRR in decibel is  = %0.2f'%CMRR,' dB '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.9 Pg 194"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 34,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The output of the difference amplifier is  = -8.12  V \n",
      "The output of the difference amplifier is  = 0.12  V \n",
      "the common mode input of difference amplifier is = 2.00  \n",
      "the common mode gain ACM of difference amplifier is = 0.06  \n",
      "the differential gain of the difference amplifier is = 2.00  \n",
      "The common mode rejection ratio CMRR is  = 32.00  \n",
      "The common mode rejection ratio CMRR in decibel is  = 30.10  dB \n"
     ]
    }
   ],
   "source": [
    "# determine Vo when 1) VI1 = 2 V VI2 = -2 V and 2) VI1 = 2 V VI2 = 2 V\n",
    "#                                      and common mode rejection ratio CMRR\n",
    "R1 = 10*10**3 # # ohm\n",
    "R2 = 20*10**3 # # ohm\n",
    "R3 = 10*10**3 # # ohm\n",
    "R4 = 22*10**3 # # ohm\n",
    "\n",
    "\n",
    "# the output of the difference amplifier is given by\n",
    "# Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n",
    "\n",
    "# Case 1 when VI1 = 2 V VI2 = -2 V\n",
    "VI1 = 2 #\n",
    "VI2 = -2 #\n",
    "\n",
    "Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n",
    "print 'The output of the difference amplifier is  = %0.2f'%Vo,' V '#\n",
    "\n",
    "# case 2 when VI1 = 2 V VI2 = 2 V\n",
    "VI1 = 2 #\n",
    "VI2 = 2 #\n",
    "\n",
    "Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n",
    "print 'The output of the difference amplifier is  = %0.2f'%Vo,' V '#\n",
    "\n",
    "# the common mode input of difference amplifier is given by\n",
    "VCM = (VI2+VI1)/2 #\n",
    "print 'the common mode input of difference amplifier is = %0.2f'%VCM,' '#\n",
    "\n",
    "# the common mode gain ACM of difference amplifier is given by\n",
    "ACM = Vo/VCM\n",
    "print 'the common mode gain ACM of difference amplifier is = %0.2f'%ACM,' '#\n",
    "\n",
    "# the differential gain of the difference amplifier is given \n",
    "Ad = R2/R1 # \n",
    "print 'the differential gain of the difference amplifier is = %0.2f'%Ad,' '#\n",
    "\n",
    "# the common mode rejection ratio CMRR is \n",
    "CMRR = abs(Ad/ACM)#\n",
    "print 'The common mode rejection ratio CMRR is  = %0.2f'%CMRR,' '#\n",
    "\n",
    "# in decibal it can be expressed as\n",
    "CMRR = 20*log10(CMRR)#\n",
    "print 'The common mode rejection ratio CMRR in decibel is  = %0.2f'%CMRR,' dB '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.10 Pg 195"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 35,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the maximum differential voltage gain of the instrumentation amplifier is = 101.00  \n",
      "the minimum differential voltage gain of the instrumentation amplifier is = 5.00  \n",
      " the range of the differential voltage gain of the instrumentation amplifier is \n",
      "                    5 <= Av <= 101 \n"
     ]
    }
   ],
   "source": [
    "# To determine the range of the differential voltage gain\n",
    "#R1 = 1 K ohm to 25 K ohm #\n",
    "R2 = 50 # # K ohm\n",
    "R3 = 10 #  # K ohm\n",
    "R4 = 10 # # K ohm\n",
    "\n",
    "# the output of instrumentation amplifier is given by\n",
    "#Vo = (R4/R3)*(1+(2*R2/R1))*(VI@-VI1)#\n",
    "\n",
    "# the differential voltage gain of the instrumentation amplifier can be written as\n",
    "#Av = (Vo/(VI2-VI1)) = (R4/R3)*(1+(2R2/R1))#\n",
    "\n",
    "# For R1 = 1 K ohm   the maximum differential voltage gain of the instrumentation amplifier is\n",
    "R1 = 1 # # K ohm\n",
    "Av = (R4/R3)*(1+(2*R2/R1))#\n",
    "print 'the maximum differential voltage gain of the instrumentation amplifier is = %0.2f'%Av,' '#\n",
    "\n",
    "# For R1 = 25 K ohm   the mminimum differential voltage gain of the instrumentation amplifier is\n",
    "R1 = 25 # # K ohm\n",
    "Av = (R4/R3)*(1+(2*R2/R1))#\n",
    "print 'the minimum differential voltage gain of the instrumentation amplifier is = %0.2f'%Av,' '#\n",
    "\n",
    "print ' the range of the differential voltage gain of the instrumentation amplifier is '#\n",
    "print '                    5 <= Av <= 101 '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.11 Pg 196"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 36,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The variable resistance R1 varies is  401 ohm <= R1 <= 100.2 K-ohm \n"
     ]
    }
   ],
   "source": [
    "# To design an instrumentation amplifier\n",
    "# 4 <= Av <= 1000 #  gain\n",
    "Ad = 2 #\n",
    "Res = 100 #  # K ohm\n",
    "\n",
    "# we cosider the variable resistance is R1 , the maximum and the minimum range of variable resistance \n",
    "# R1min = R1 #             \n",
    "# R1max = R1+100 #\n",
    "\n",
    "# the gain of difference amplifier \n",
    "#A3 = Ad = Vo/(Vo2-Vo1) = (R4/R3)\n",
    "\n",
    "# the maximum range of differential voltage gain Avmax = 1000 when R1min = R1\n",
    "#Avmax = R4/R3*(1+(2*R2/R1min))#\n",
    "\n",
    "# by solvin we get following equation\n",
    "#  499*R1-2*R2=0                       equation 1\n",
    "\n",
    "# the maximum range of differential voltage gain Avmin =4 when R1max = R1+100 K ohm\n",
    "# Avmin = (R4/R3)*(1+(2R2/R1max))#\n",
    "\n",
    "# by solving above equation we get\n",
    "# R1 -2 R2 = -200 K ohm                equation 2\n",
    "\n",
    "#by solving equation 1 and 2 we get\n",
    "R1 = 401 # #  ohm\n",
    "R2 = 100.2 # # ohm\n",
    "print 'The variable resistance R1 varies is  401 ohm <= R1 <= 100.2 K-ohm ' #"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.12 Pg 198"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 41,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " The time constant of the given filter is RC = 0.20 msec \n"
     ]
    }
   ],
   "source": [
    " # Determine the time constant of the integrator\n",
    "Vo = 10 #\n",
    "t = 2*10**-3 #\n",
    "VI = -1 #  # at t =0 #\n",
    "\n",
    "# The output voltage of an integrator is define as\n",
    "RC = t/10*1e3 # ms\n",
    "print ' The time constant of the given filter is RC = %0.2f'%RC,'msec '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.13 Pg 198"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 48,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " The time constant of the given filter is RC = 0.1 msec \n",
      "The capacitor value is = 0.1 F\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "# Determine the time constant of the integrator\n",
    "Vo = 20 #\n",
    "t = 1*10**-3 #\n",
    "VI = -1 #  # at t =0 #\n",
    "\n",
    "# The output voltage of an integrator is define as\n",
    "RC = t/10 #\n",
    "print ' The time constant of the given filter is RC = %0.1f'%(RC*1000),'msec '#\n",
    "\n",
    "R = 1*10**3 #  # we assume \n",
    "C = RC/R*10**6 #\n",
    "print 'The capacitor value is = %0.1f'%C,'F'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.14 Pg 199"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 49,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The value of resistance RIa is = 90.00  K ohm \n",
      "The value of resistance RIb is = 22.50  K ohm \n",
      "The value of resistance RIc is = 18.00  K ohm \n",
      "The value of resistance RId is = 15.00  K ohm \n"
     ]
    }
   ],
   "source": [
    " # to design a summing amplifier\n",
    "\n",
    "# the output of the summing amplifier is given by\n",
    "#Vo = -R2*((VIa/RIa)+(VIb/RIb)+(VIc/RIc)+(VId/RId))#    equation 1\n",
    "\n",
    "# the equation given is\n",
    "#Vo = -(3*VIa+12*VIb+15*VIc+18*VId)#               equation 2\n",
    "\n",
    "# comparing equation 1 and 2\n",
    "#R2/RIa = 3 #\n",
    "#R2/RIb = 12 #\n",
    "#R2/RIc = 15 #\n",
    "#R2/RId = 18 # \n",
    "\n",
    "# the feedback resistance R2= 270 K ohm \n",
    "R2 = 270 #  # K ohm\n",
    "RIa = R2/3 #\n",
    "print 'The value of resistance RIa is = %0.2f'%RIa,' K ohm '#\n",
    "\n",
    "RIb = R2/12 #\n",
    "print 'The value of resistance RIb is = %0.2f'%RIb,' K ohm '#\n",
    "\n",
    "RIc = R2/15 #\n",
    "print 'The value of resistance RIc is = %0.2f'%RIc,' K ohm '#\n",
    "\n",
    "RId = R2/18 #\n",
    "print 'The value of resistance RId is = %0.2f'%RId,' K ohm '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.15 Pg 200"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 50,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The output of first op-amp A1 is = -275*sin wt mV \n",
      "The output of second op-amp A2 is = 275*sin wt mV \n",
      "The output of third op-amp A3 is = 825*sin wt mV \n",
      "current through the resistor R1 and R2 is = 5 sin wt uA \n",
      "current through the non-inverting terminal resistor R3 and R4 = 5.5 sin wt uA \n",
      "current through the inverting terminal resistor R3 and R4 = 22 sin wt uA \n"
     ]
    }
   ],
   "source": [
    "# for the instrumentation amplifier find Vo1 , Vo2 , Vo \n",
    "# Vi1 = -25 sin wt  #  # mV\n",
    "# Vi2 = 25 sin wt # # mV\n",
    "R1 = 10*10**3 #\n",
    "R2 = 20*10**3 #\n",
    "R3 = 20*10**3 #\n",
    "R4 = 10*10**3 #\n",
    "\n",
    "# the output of first op-amp A1 is given by\n",
    "# Vo1 = (1+(R2/R1))*Vi1-(R2/R1)*Vi2 #\n",
    "#by solving above equation we get\n",
    "print 'The output of first op-amp A1 is = -275*sin wt mV '#\n",
    "\n",
    "# the output of second op-amp A2 is given by\n",
    "# Vo2 = (1+(R2/R1))*Vi2-(R2/R1)*Vi1 #\n",
    "#by solving above equation we get\n",
    "print 'The output of second op-amp A2 is = 275*sin wt mV '#\n",
    "\n",
    "# the output of third op-amp A3 is given by\n",
    "# Vo = (R4/R3)-(1+(2R2/R1)*(Vi2-Vi1) #\n",
    "#by solving above equation we get\n",
    "print 'The output of third op-amp A3 is = 825*sin wt mV '#\n",
    "\n",
    "# current through the resistor R1 and R2 is\n",
    "#i = (Vi1-Vi2)/R1 #\n",
    "print 'current through the resistor R1 and R2 is = 5 sin wt uA '#\n",
    "\n",
    "# current through the non-inverting terminal resistor R3 and R4 \n",
    "#i3 = Vo2/(R3+R4)#\n",
    "print 'current through the non-inverting terminal resistor R3 and R4 = 5.5 sin wt uA '#\n",
    "\n",
    "# current through the inverting terminal resistor R3 and R4 \n",
    "#i2 = Vo1-(R3/(R3+R4))*Vo2/R3 #\n",
    "print 'current through the inverting terminal resistor R3 and R4 = 22 sin wt uA '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.16 Pg 202"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 52,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The input resistance Rin is = 0.00999  ohm \n",
      "The value of Resistance Rs is = 1.0990  K ohm \n"
     ]
    }
   ],
   "source": [
    " # for the a current to voltage converter show a) Rin = (Rf/1+Aop)  b) Rf = 10 K ohm Aop = 1000 \n",
    "\n",
    "#a) The input resistance given as\n",
    "#Rin = (Rf)/(1+Aop) #\n",
    "\n",
    "# The input resistance of the circuit can be written as\n",
    "#Rin = (V1/i!)#\n",
    "\n",
    "#  the feedback current of the given circuit is defined as\n",
    "#i1 =(V1-Vo)/RF #\n",
    "\n",
    "# the feedback resistance RF is \n",
    "#RF =(V1-Vo)/i1 #\n",
    "\n",
    "# The output voltage Vo is\n",
    "#Vo = -Aop*V1 #\n",
    "\n",
    "#by using this output feedback currenty i1 can be reformed as\n",
    "#i1 = (V1-(-Aop*V1))/RF #\n",
    "\n",
    "#i1 = V1*(1+Aop)/RF #\n",
    "\n",
    "# Then Rin Becomes \n",
    "#Rin =Rf/(1+Aop)#\n",
    "\n",
    "Rf =10*10**3 #\n",
    "Aop = 1000 #\n",
    "\n",
    "# the input current and output voltage of the circuit are defined as\n",
    "#i1 =(Rs)/(Rs+Rin) #\n",
    "# Vo = -(Aop*(RF/1+Aop))*i1 #\n",
    "\n",
    "#the input resistance Rin is \n",
    "Rin =(Rf/(1+Aop)) #\n",
    "\n",
    "# subsituting the value of RF Aop Rin and Vo we get \n",
    "RF = 10 #\n",
    "Rin = RF/(1+Aop)\n",
    "print 'The input resistance Rin is = %0.5f'%Rin,' ohm '#\n",
    "\n",
    "Aop = 1000 #\n",
    "#(1000/1001)*(Rs/(Rs*0.00999))> 0.99 #\n",
    "# by solving above equation we get \n",
    "Rs = 1.099  #  # K ohm \n",
    "print 'The value of Resistance Rs is = %0.4f'%Rs,' K ohm '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.17 Pg 204"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 54,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "for Aop = 10**4 closed loop gain is = 0.9999  \n",
      "for Aop = 10**3 closed loop gain is = 0.9990  \n",
      "for Aop = 10**2 closed loop gain is = 0.9901  \n",
      "for Aop = 10**1 closed loop gain is = 0.9091  \n"
     ]
    }
   ],
   "source": [
    " # determine the closed loop gain\n",
    "\n",
    "# the output of the voltage follower is given as\n",
    "#Vo = Aop(V1-Vo)#\n",
    "\n",
    "# the closed loop gain of the voltage follower \n",
    "#A = 1/(1+(1/Aop))#\n",
    " \n",
    "# for Aop = 10**4 closed loop gain\n",
    "Aop = 10**4 #\n",
    "A = 1/(1+(1/Aop))#\n",
    "print 'for Aop = 10**4 closed loop gain is = %0.4f'%A,' '#\n",
    "\n",
    "# for Aop = 10**3 closed loop gain\n",
    "Aop = 10**3 #\n",
    "A = 1/(1+(1/Aop))#\n",
    "print 'for Aop = 10**3 closed loop gain is = %0.4f'%A,' '#\n",
    "\n",
    "# for Aop = 10**2 closed loop gain\n",
    "Aop = 10**2 #\n",
    "A = 1/(1+(1/Aop))#\n",
    "print 'for Aop = 10**2 closed loop gain is = %0.4f'%A,' '#\n",
    "\n",
    "# for Aop = 10**1 closed loop gain\n",
    "Aop = 10**1 #\n",
    "A = 1/(1+(1/Aop))#\n",
    "print 'for Aop = 10**1 closed loop gain is = %0.4f'%A,' '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.18 Pg 205"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 56,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "for the frequency f = 10 Hz  the output is = 106.10  V \n",
      "for the frequency f = 1000 Hz  the output is = 1.06  V \n",
      "for the frequency f = 10000 Hz  the output is = 0.106  V \n"
     ]
    }
   ],
   "source": [
    "from math import pi\n",
    "# To determine the output voltage of integrator\n",
    "Vin = 1 #\n",
    "R = 150*10**3 ##  ohm\n",
    "C = 1*10**-9 # # F\n",
    "\n",
    "# the output voltage of an integrator is given as\n",
    "#Vo = (fc/f)*Vin #\n",
    "\n",
    "#fc = 1/(2*pi*R*C)#\n",
    "\n",
    "#Vo = (1/(2*pi*R*C*f))*Vin#\n",
    "\n",
    "#for the frequency f = 10 Hz  the output is\n",
    "f = 10 # # Hz\n",
    "Vo = (1/(2*pi*R*C*f))*Vin#\n",
    "print 'for the frequency f = 10 Hz  the output is = %0.2f'%Vo,' V '#\n",
    "\n",
    "#for the frequency f = 1000 Hz  the output is\n",
    "f = 1000 # # Hz\n",
    "Vo = (1/(2*pi*R*C*f))*Vin#\n",
    "print 'for the frequency f = 1000 Hz  the output is = %0.2f'%Vo,' V '#\n",
    "\n",
    "#for the frequency f = 10000 Hz  the output is\n",
    "f = 10000 # # Hz\n",
    "Vo = (1/(2*pi*R*C*f))*Vin#\n",
    "print 'for the frequency f = 10000 Hz  the output is = %0.3f'%Vo,' V '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.19 Pg 206"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 63,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The cutoff frequency of the integrator is = 13.263 kHz\n",
      "The gain of the integrator is = 3.18\n"
     ]
    }
   ],
   "source": [
    "from math import sqrt\n",
    "# To determine the magnitude gain of the integrator\n",
    "Vin = 1 #\n",
    "f = 50*10**3 #\n",
    "Rf = 120*10**3 #\n",
    "R = 10*10**3 #\n",
    "C = 0.1*10**-9 #\n",
    "\n",
    "# the magnitude gain of the integrator is given by\n",
    "#A = (Rf/R)/(sqrt(1+(f/fc)**2))#\n",
    "\n",
    "# the cutoff frequency of the integrator \n",
    "fc = 1/(2*pi*Rf*C)/1e3\n",
    "print 'The cutoff frequency of the integrator is = %0.3f'%fc,'kHz'#\n",
    "\n",
    "\n",
    "A = (Rf/R)/(sqrt(1+(f/fc)**2))*1e3#\n",
    "print 'The gain of the integrator is = %0.2f'%A"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.20 Pg 207"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 67,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the break frequency fa is = 31.83 kHz \n",
      "the break frequency fb is = 21.22 kHz \n",
      "The gain of the differentiator is = 0.6667  \n"
     ]
    }
   ],
   "source": [
    " # To determine the magnitude gain of the differentiator\n",
    "Vin = 1 #\n",
    "f = 50*10**3 #\n",
    "R = 75*10**3 #\n",
    "R1 = 50*10**3 #\n",
    "C = 0.1*10**-9 #\n",
    "\n",
    "# the magnitude gain of the differentiator is given by\n",
    "#A = (f/fa)/(sqrt(1+(f/fb)**2))#\n",
    "\n",
    "# the break frequency fa is defined as\n",
    "fa = 1/(2*pi*R1*C) / 1e3\n",
    "print 'the break frequency fa is = %0.2f'%fa,'kHz '#\n",
    "\n",
    "# the break frequency fb is defined as\n",
    "fb = 1/(2*pi*R*C) /1e3\n",
    "print 'the break frequency fb is = %0.2f'%fb,'kHz '#\n",
    "\n",
    "\n",
    "A = (f/fa)/(sqrt(1+(f/fb)**2))#\n",
    "print 'The gain of the differentiator is = %0.4f'%A,' '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.21 Pg 209"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 70,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The input voltage of an op-amp is = -40 mV\n"
     ]
    }
   ],
   "source": [
    " # to determine the input voltage of an op-amp\n",
    "Vo = 2 # # V\n",
    "R1 = 20*10**3 # # ohm\n",
    "R2 = 1*10**6 #  # ohm\n",
    "\n",
    "# the input voltage of an op-amp\n",
    "Vin = -(R1/R2)*Vo *1000 # mV\n",
    "print 'The input voltage of an op-amp is = %0.f'%Vin,'mV'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.22 Pg 209"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 71,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the output voltage of follower Vo1 is = 2.00   V\n",
      "The output voltage of an inverting amplifier is = -20.00  V \n"
     ]
    }
   ],
   "source": [
    "# To determine the output voltage\n",
    "Vin = 2 #\n",
    "R2 = 20*10**3 #\n",
    "R1 = 2*10**3 #\n",
    "\n",
    "# the output voltage of follower Vo1 is\n",
    "Vo1 = Vin #\n",
    "print 'the output voltage of follower Vo1 is = %0.2f'%Vo1,'  V'#\n",
    "# the output voltage of an inverting amplifier\n",
    "Vo = -(R2/R1)*Vo1 #\n",
    "print 'The output voltage of an inverting amplifier is = %0.2f'%Vo,' V '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.23 Pg 210"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 72,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The output of the inverting amplifier is = -15.00  V\n",
      "The output of the non-inverting amplifier is = 20.00  V\n"
     ]
    }
   ],
   "source": [
    "# to determine the output voltage of an op-amp\n",
    "Vin = 5 # # V\n",
    "R1 = 25*10**3 # # ohm\n",
    "R2 = 75*10**3 #  # ohm\n",
    "\n",
    "# in this problem op-amp A1 perform the voltage follower and op-amp A2 perform inverting amplifier and op-amp A3 perform non-inverting amplifier\n",
    "\n",
    "# the output voltage of follower op-amp A1\n",
    "Vo1 = Vin #\n",
    "\n",
    "# the output of the inverting amplifier A2\n",
    "Vo2 = -((R2/R1)*Vo1) #\n",
    "print 'The output of the inverting amplifier is = %0.2f'%Vo2,' V'#\n",
    "\n",
    "# the output of the non-inverting amplifier A3\n",
    "Vo =(1+(R2/R1))*Vo1 #\n",
    "print 'The output of the non-inverting amplifier is = %0.2f'%Vo,' V'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.24 Pg 211"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 73,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The output voltage of an inverting amplifier is = 27.50  V \n",
      "the output voltage of follower Vo1 is = 2.50   V\n",
      "the output of the inverting summing amplifier is = 101.25  V \n"
     ]
    }
   ],
   "source": [
    " # To determine the output voltage\n",
    "Vin = 2.5 #\n",
    "Rf = 100*10**3 #\n",
    "R1 = 10*10**3 #\n",
    "RI1 = 25*10**3 #\n",
    "RI2 = 10*10**3 #\n",
    "R2 = 100*10**3 #\n",
    "\n",
    "# the output voltage of an inverting amplifier\n",
    "Vo1 = (1+(R2/R1))*Vin # #\n",
    "print 'The output voltage of an inverting amplifier is = %0.2f'%Vo1,' V '#\n",
    "\n",
    "# the output voltage of follower Vo2 is\n",
    "Vo2 = Vin #\n",
    "print 'the output voltage of follower Vo1 is = %0.2f'%Vo2,'  V'#\n",
    "\n",
    "# the output of the inverting summing amplifier\n",
    "R2 = 75*10**3 #\n",
    "Vo = R2*((Vo1/RI1)+(Vo2/RI2))#\n",
    "print 'the output of the inverting summing amplifier is = %0.2f'%Vo,' V '#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.25 Pg 212"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 74,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the total gain of the circuit is = 36.00  \n",
      "The  output voltage of the op-amp is = 90.00  V\n"
     ]
    }
   ],
   "source": [
    "# To calculate the output voltage\n",
    "Vin = 2.5 # \n",
    "R1 = 10*10**3 #\n",
    "R2 = 10*10**3 #\n",
    "R3 = 10*10**3 #\n",
    "Rf = 30*10**3 #\n",
    "\n",
    "# the total gain of the circuit \n",
    "#Av =A1v*A2v*A3v #\n",
    "Av = (1+(Rf/R1))*(-Rf/R2)*(-Rf/R3)#\n",
    "print 'the total gain of the circuit is = %0.2f'%Av,' '#\n",
    "\n",
    "# The  output voltage of the op-amp \n",
    "Vo = Av*Vin #\n",
    "print 'The  output voltage of the op-amp is = %0.2f'%Vo,' V'#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.26 Pg 212"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 76,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The output of op-amp A1 is = -10.00 V1\n",
      "The output of op-amp A2 is  Vo = 40V1 - 2V2 \n",
      "The output is equal to the difference between 40V1 and 2V2 \n"
     ]
    }
   ],
   "source": [
    "# to calculate the output voltage of op-amp circuit\n",
    "Rf = 100*10**3 # # ohm\n",
    "R1 = 10*10**3 # # ohm\n",
    "R2 = 25*10**3 # # ohm\n",
    "R3 = 50*10**3 # # ohm\n",
    "\n",
    "# the output of op-amp A1 is\n",
    "# VA1 = (-Rf/R1)*V1 #\n",
    "VA1 = (-Rf/R1)#\n",
    "print 'The output of op-amp A1 is = %0.2f'%VA1,'V1' #       # *V1 because the output is come from 1 op-amp\n",
    "\n",
    "# the output of op-amp A2 is\n",
    "# Vo = -Rf*((VA1/R2)+(V2/R3))#\n",
    "#Vo = -100*(-0.4*V1+0.02V2)#\n",
    "print 'The output of op-amp A2 is  Vo = 40V1 - 2V2 '#   \n",
    "\n",
    "print 'The output is equal to the difference between 40V1 and 2V2 '#   "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.27 Pg 213"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 77,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the upper crossover voltage of schmitt trigger is = 1.00  V\n",
      "the lower crossover voltage of schmitt trigger is = -1.00  V\n",
      "the hysteresis width HW of schmitt trigger is = 2.00  V\n"
     ]
    }
   ],
   "source": [
    " # to determine the hysteresis width of a schmitt trigger\n",
    "R1 = 25*10**3 # # ohm\n",
    "R2 = 75*10**3 # # ohm\n",
    "VTH = 4 # # V\n",
    "VTL = -4 # # V\n",
    "\n",
    "# the upper crossover voltage of schmitt trigger is defined as\n",
    "VU = (R1/(R1+R2))*VTH#\n",
    "print 'the upper crossover voltage of schmitt trigger is = %0.2f'%VU,' V'\n",
    "\n",
    "# the lower crossover voltage of schmitt trigger is defined as\n",
    "VL = (R1/(R1+R2))*VTL#\n",
    "print 'the lower crossover voltage of schmitt trigger is = %0.2f'%VL,' V'\n",
    "\n",
    "# the hysteresis width of schmitt trigger is\n",
    "HW = VU-VL #\n",
    "print 'the hysteresis width HW of schmitt trigger is = %0.2f'%HW,' V'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.28 Pg 214"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 78,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the upper crossover voltage of schmitt trigger is = 1.43  V\n",
      "the lower crossover voltage of schmitt trigger is = -1.43  V\n",
      "the hysteresis width HW of schmitt trigger is = 2.86  V\n"
     ]
    }
   ],
   "source": [
    " # to determine the hysteresis width of a schmitt trigger\n",
    "R1 = 15*10**3 # # ohm\n",
    "R2 = 90*10**3 # # ohm\n",
    "VTH = 10 # # V\n",
    "VTL = -10 # # V\n",
    "\n",
    "# the upper crossover voltage of schmitt trigger is defined as\n",
    "VU = (R1/(R1+R2))*VTH#\n",
    "print 'the upper crossover voltage of schmitt trigger is = %0.2f'%VU,' V'\n",
    "\n",
    "# the lower crossover voltage of schmitt trigger is defined as\n",
    "VL = (R1/(R1+R2))*VTL#\n",
    "print 'the lower crossover voltage of schmitt trigger is = %0.2f'%VL,' V'\n",
    "\n",
    "# the hysteresis width of schmitt trigger is\n",
    "HW = VU-VL #`\n",
    "print 'the hysteresis width HW of schmitt trigger is = %0.2f'%HW,' V'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.29 Pg 214"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 80,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the value of resistance R1 is = 5.00 kohm\n"
     ]
    }
   ],
   "source": [
    "# to determine the resistance R1 when low and high saturated output states are given\n",
    "R2 = 20*10**3 # # ohm\n",
    "VH = 2 # # V      crossover voltage\n",
    "VL = -2 #  # V     crossover voltage\n",
    "VOH = 10 # # V    saturated output states\n",
    "VOL = -10 # # V    saturated output states\n",
    "\n",
    "# the upper crossover voltage of schmitt trigger is defined as\n",
    "# V = (R1/(R1+R2))*VOH#\n",
    "# solving above equation we get \n",
    "# 2R1+2R2 = 10R1 #\n",
    "R1 = (2*R2)/8/1000 # kohm\n",
    "print 'the value of resistance R1 is = %0.2f'%R1,'kohm'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6.30 Pg 215"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 82,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the value of resistance R2 is = 30.00 kohm\n"
     ]
    }
   ],
   "source": [
    "# to determine the value of resistance R1 and R2 when low and high saturated output states are given\n",
    "VH = 3 # # V      crossover voltage\n",
    "VL = -3 #  # V     crossover voltage\n",
    "VOH = 12 # # V    saturated output states\n",
    "VOL = -12 # # V    saturated output states\n",
    "\n",
    "# the upper crossover voltage of schmitt trigger is defined as\n",
    "# V = (R1/(R1+R2))*VOH#\n",
    "# solving above equation we get \n",
    "# 3R1+3R2 = 12R1 #\n",
    "\n",
    "# 3*R1 = R2 #\n",
    "R1 = 10*10**3 # # ohm  we assume\n",
    "R2 = 3*R1 / 1e3#\n",
    "print 'the value of resistance R2 is = %0.2f'%R2,'kohm'"
   ]
  }
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