path: root/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter6.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "# Chapter 6: General Linear Applications"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.1"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Low-freq cutoff is 10.61 kHz\n",
      "\n",
      " High-freq cutoff is 90.91 kHz\n",
      "\n",
      " Bandwidth is 80.3 kHz\n"
     ]
    }
   ],
   "source": [
    "#Example 6.1\n",
    "#In the circuit of figure 6-3(a) Rin=50 Ohm,Ci=0.1 uF,R1=100 Ohm, Rf=1 KOhm\n",
    "#Rl=10 kOhm and supply voltages +15, -15 V.\n",
    "#Determine the bandwidth of the amplifier.\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "R1=100\n",
    "Rf=1*10**3\n",
    "Rin=50\n",
    "Rl=10*10**3\n",
    "Ci=0.1*10**-6    #Capacitance b/w 2 stages being coupled \n",
    "RiF=R1           #ac input resistance of the second stage\n",
    "Ro=Rin           #ac output resistance of the 1st stage\n",
    "UGB=10**6        #Unity gain bandwidth\n",
    "\n",
    "\n",
    "#calculation\n",
    "fl=1/(2*math.pi*Ci*(RiF+Ro)) #Low-freq cutoff\n",
    "K=Rf/(R1+Rf)\n",
    "Af=-Rf/R1                    #closed loop voltage gain\n",
    "fh=UGB*K/abs(Af)             #High-freq cutoff\n",
    "BW=fh-fl                     #Bandwidth\n",
    "\n",
    "#result\n",
    "print \"\\n Low-freq cutoff is\",round(fl/10**3,2),\"kHz\"\n",
    "print \"\\n High-freq cutoff is\",round(fh/10**3,2),\"kHz\"\n",
    "print\"\\n Bandwidth is\",round(BW/10**3,2),\"kHz\" \n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Low-freq cutoff is 31.8 kHz\n",
      "High-freq cutoff is 90.91 kHz\n",
      "Bandwidth is 90.88 kHz\n",
      "The ideal maximum output voltage swing is 15 Volts\n"
     ]
    }
   ],
   "source": [
    "\n",
    "#Example 6.2\n",
    "#For the noninverting amplifier of figure 6-4(c), Rin=50 Ohm,Ci=C1=0.1 uF\n",
    "#R1=R2=R3=100 kOhm,Rf=1 MOhm,Vcc=15 V.Determine\n",
    "#a)the bandwidth of the amplifier\n",
    "#b)the maximum output voltage swing\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "R1=100*10**3\n",
    "R2=100*10**3\n",
    "R3=100*10**3\n",
    "Rf=1*10**6\n",
    "Rin=50\n",
    "Ci=0.1*10**-6      #Capacitance b/w 2 stages being coupled\n",
    "Ro=Rin             #ac output resistance of the 1st stage\n",
    "Vcc=15\n",
    "UGB=10**6          #Unity gain bandwidth\n",
    "Rif=R2*R3/(R2+R3)  #since Ri*(1+A*B)>>R2 or R3\n",
    "\n",
    "#calculation\n",
    "fl=1/(2*math.pi*Ci*(Rif+Ro)) #Low-freq cutoff\n",
    "K=Rf/(R1+Rf)\n",
    "Af=-Rf/R1                    #closed loop voltage gain\n",
    "fh=UGB*K/abs(Af)             #High-freq cutoff\n",
    "BW=fh-fl                     #Bandwidth\n",
    "\n",
    "#result\n",
    "print \"Low-freq cutoff is\",round(fl,2),\"kHz\"\n",
    "print \"High-freq cutoff is\",round(fh/10**3,2),\"kHz\"\n",
    "print\"Bandwidth is\",round(BW/10**3,2),\"kHz\"\n",
    "print \"The ideal maximum output voltage swing is\",Vcc,\"Volts\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.3"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Inductance is 9.9 mH\n",
      "Figure of merit of the coil is 33.2\n",
      "Parallel resistance of the tank circuit is 32.98 kHz\n",
      "Feedback resistance is 1.03 kHz\n"
     ]
    }
   ],
   "source": [
    "#Example 6.3\n",
    "#The circuit of figure 6-5(a) is to provide a gain of 10 at a peak frequency of\n",
    "#16 kHz. Determine the value of all its components.\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "fp=16*10**3   #Peak frequency\n",
    "Af=10         #Gain at peak frequency\n",
    "C=0.01*10**-6 #Assume\n",
    "R=30          #Assume the value of internal resistance of the inductor\n",
    "R1=100        #Assume the value of internal resisrance of the coil\n",
    "\n",
    "#calculation\n",
    "L=1/(((2*math.pi*fp)**2)*10**-8) #Simplifying fp=1/(2*pi*sqrt(L*C))\n",
    "Xl=2*math.pi*fp*L                #Inductive reactance\n",
    "Qcoil=Xl/R                       #Figure of merit of the coil\n",
    "Rp=((Qcoil)**2)*R                #Parallel resistance of the tank circuit\n",
    "Rf=-Rp/(1-(Rp/(Af*R1)))          #Simplifying Af=(Rf||Rp)/R1\n",
    "\n",
    "\n",
    "#result\n",
    "print \"Inductance is\",round(L*10**3,1),\"mH\"\n",
    "print \"Figure of merit of the coil is\",round(Qcoil,1)\n",
    "print \"Parallel resistance of the tank circuit is\",round(Rp/10**3,2),\"kHz\"\n",
    "print \"Feedback resistance is\",round(Rf/10**3,2),\"kHz\"\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.4"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Output voltage is -2.0 Volts\n"
     ]
    }
   ],
   "source": [
    "#Example 6.4\n",
    "#In the circuit of figure 6-6 Va=1V, Vb=2V, Vc=3V, Ra=Rb=Rc=3 kOhm,Rf=1 kOhm\n",
    "#Supply voltages are 15V and -15V. Assuming that the opamp is initially nulled,\n",
    "#determine the output voltage Vo\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "Va=1          #Input voltage in Volts\n",
    "Vb=2          #Input voltage in Volts\n",
    "Vc=3          #Input voltage in Volts\n",
    "Ra=3*10**3    #Resistance in ohms\n",
    "Rb=3*10**3    #Resistance in ohms\n",
    "Rc=3*10**3    #Resistance in ohms\n",
    "Rf=1*10**3    #Resistance in ohms\n",
    "\n",
    "\n",
    "#calculation\n",
    "Vo=-((Rf/Ra)*Va+(Rf/Rb)*Vb+(Rf/Rc)*Vc) #Output voltage\n",
    "\n",
    "\n",
    "#result\n",
    "print \"Output voltage is\",Vo,\"Volts\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.5"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Voltage at non-inverting terminal is 1.0 Volts\n",
      "Output voltage is 3.0 Volts\n"
     ]
    }
   ],
   "source": [
    "\n",
    "#Example 6.5\n",
    "#In the circuit of figure 6-7 Va=2V, Vb=-3V, Vc=4V, R=R1=1 kOhm,Rf=2 kOhm\n",
    "#Supply voltages are 15V and -15V. Assuming that the opamp is initially nulled,\n",
    "#determine the output voltage Vo and voltage V1 at the noninverting terminal.\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "Va=2             #Input voltage in volts\n",
    "Vb=-3            #Input voltage in volts\n",
    "Vc=4             #Input voltage in volts\n",
    "R1=1*10**3       #Resistance in ohms      \n",
    "Rf=2*10**3       #Resistance in ohms\n",
    "\n",
    "\n",
    "#calculation\n",
    "V1=(Va+Vb+Vc)/3  #Voltage at non-inverting terminal\n",
    "Vo=(1+Rf/R1)*V1  #Output voltage\n",
    "\n",
    "#result\n",
    "print \"Voltage at non-inverting terminal is\",V1,\"Volts\"\n",
    "print \"Output voltage is\",Vo,\"Volts\"\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.6"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Output voltage is 4 Volts\n"
     ]
    }
   ],
   "source": [
    "#Example 6.6\n",
    "#In the circuit of figure 6-9 Va=2V, Vb=3V,Vc=4V,Vc=4V,Vd=5V,R=1 kOhm\n",
    "#Supply voltages are 15V and -15V. Assuming that the opamp is initially nulled,\n",
    "#Determine the output voltage Vo\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "Va=2             #Input voltage in volts\n",
    "Vb=3             #Input voltage in volts\n",
    "Vc=4             #Input voltage in volts\n",
    "Vd=5             #Input voltage in volts\n",
    "R=1*10**3        #Resistance in ohms\n",
    "\n",
    "#calculation\n",
    "Vo=-Va-Vb+Vc+Vd  #Output voltage\n",
    "\n",
    "#result\n",
    "print \"Output voltage is\",Vo,\"Volts\"\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.7"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Output voltage at 0 degree is 1.47 Volts\n",
      "Output voltage at 100 degree is -4.41 Volts\n"
     ]
    }
   ],
   "source": [
    "\n",
    "#Example 6.7\n",
    "#In the circuit of figure 6-12 R1=1 kOhm, Rf=4.7 kOhm, Ra=Rb=Rc=100 kOhm.\n",
    "#Vdc=5V and Supply voltages are 15V and -15V.\n",
    "#The transducer is a thermistor with the following specifications.\n",
    "#Rt=100 kOhm at a reference temperature of 25 degree celcius, temperature\n",
    "#coefficient of resistance =-1 kOhm/ degree celcius or 1%/degree celcius.\n",
    "#Determine the output voltage Vo at o degree C and 100 degree C.\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "R1=1*10**3         #Resistance in ohms\n",
    "Rf=4.7*10**3       #Resistance in ohms\n",
    "Ra=100*10**3       #Resistance in ohms\n",
    "Rb=100*10**3       #Resistance in ohms\n",
    "Rc=100*10**3       #Resistance in ohms\n",
    "Vdc=5              #dc voltage in Volts\n",
    "Rt=100*10**3       #Resistance of a thermistor\n",
    "temp_coeff=1*10**3\n",
    "R=Ra               #Ra=Rb=Rc=R\n",
    "\n",
    "#calculation\n",
    "delta_R=-temp_coeff*(0-25)      #Change in resistance\n",
    "Vo1=((Rf*delta_R)/(R1*4*R))*Vdc #Output voltage at degrees\n",
    "delta_R=-temp_coeff*(100-25)    #Change in resistance\n",
    "Vo2=((Rf*delta_R)/(R1*4*R))*Vdc #Output voltage at 100 degrees\n",
    "\n",
    "#result\n",
    "print \"Output voltage at 0 degree is\",round(Vo1,2),\"Volts\"\n",
    "print \"Output voltage at 100 degree is\",round(Vo2,2),\"Volts\"\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.8"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Change in resistance is 0.1 ohm\n"
     ]
    }
   ],
   "source": [
    "\n",
    "#Example 6.8\n",
    "#The circuit of Figure 6-12 is used as an analog weight scale with the following\n",
    "#specifications. The gain of the differential instrumentation amplifier = -100.\n",
    "#Assume that Vdc= +10 V and the opamp supply voltages = +/- 10 V. The unstrained\n",
    "#resistance of each of the four elements of the strain gage is 100 ohm Vo= 1 V.\n",
    "#When a certain weight is placed on the scale platform,the output voltage Vo=1 V.\n",
    "#Assuming that the output is initially 0,determine the change in the resistance\n",
    "#of each strain-gage element.\n",
    "\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "A=-100                      #Gain of the differential instrumentation amplifier\n",
    "Ra=100\n",
    "Rb=100\n",
    "Rc=100\n",
    "Vdc=10\n",
    "Vo=1\n",
    "R=Ra                        #Ra=Rb=Rc=R\n",
    "\n",
    "#calculation\n",
    "delta_R=(Vo*R)/(Vdc*abs(A)) #Change in resistance\n",
    "\n",
    "#result\n",
    "print \"Change in resistance is\",delta_R,\"ohm\"\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.9"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Feedback resisrance is 1.8 kOhm\n",
      "Gain of the differential amplifier is 38.0\n"
     ]
    }
   ],
   "source": [
    "\n",
    "#Example 6.9\n",
    "#The differential input and output amplifier of figure 6-14(a) is used as a\n",
    "#pre-amplifier and requires a differential output of atleast 3.7 V. Determine\n",
    "#the gain of the circuit if the differential input Vin=100 mV.\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "Vo=3.7                  #differential output voltage in Volts\n",
    "Vin=100*10**-3          #differential input voltage in Volts\n",
    "R1=100                  #Assume\n",
    "Rf=0.5*((Vo*R1)/Vin-1)  #Feedback resisrance\n",
    "\n",
    "#calculation\n",
    "A=(1+2*Rf/R1)           #Gain of the differential amplifier\n",
    "\n",
    "#result\n",
    "print \"Feedback resisrance is\",round(Rf/10**3,1),\"kOhm\"\n",
    "print \"Gain of the differential amplifier is\",round(A,0)\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.10"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Minimum input voltage is 1.1 Volts\n",
      "Maximum input voltage is 7.48 Volts\n"
     ]
    }
   ],
   "source": [
    "#Example 6.10\n",
    "#In the figure 6-17, for the indicated values of resistors, determine the full\n",
    "#scale range for the input voltage.\n",
    "\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "R1min=1*10**3\n",
    "R1max=6.8*10**3\n",
    "io=1*10**-3              #Meter current for full-wave rectification\n",
    "\n",
    "\n",
    "#calculation\n",
    "vin_min=1.1*R1min*io     #Minimum input voltage\n",
    "vin_max=1.1*R1max*io     #Maximum input voltage\n",
    "\n",
    "#result\n",
    "print \"Minimum input voltage is\",vin_min,\"Volts\"\n",
    "print \"Maximum input voltage is\",vin_max,\"Volts\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.11"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Current through diode is 5.0 mA\n",
      "Voltage drop across diode is 0.7 Volts\n"
     ]
    }
   ],
   "source": [
    "\n",
    "#Example 6.11\n",
    "#The circuit of figure 6-18,when the switch is in position 1, Vin=0.5 V and\n",
    "#Vo=1.2 V. Determine the current through the diode and the voltage drop across\n",
    "#it.Assume that the opamp is initially nulled.\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "Vin=0.5        #Input voltage\n",
    "Vo=1.2         #Output voltage\n",
    "R1=100     \n",
    "\n",
    "\n",
    "#calculation\n",
    "Io=Vin/R1      #Current through diode\n",
    "Vd=Vo-Vin      #Voltage drop across diode\n",
    "\n",
    "#result\n",
    "print \"Current through diode is\",Io*10**3,\"mA\"\n",
    "print \"Voltage drop across diode is\",Vd,\"Volts\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.12"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Load Current is 0.5 mA\n",
      "Output voltage is 2 Volts\n"
     ]
    }
   ],
   "source": [
    "\n",
    "#Example 6.12\n",
    "#The circuit of figure 6-19,Vin=5 V, R=1 Kilo Ohm and V1=1 V. Find\n",
    "#a) the load current.\n",
    "#b) the output voltage Vo.\n",
    "#Assume that the op-amp is initially nulled.\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "Vin=5          #Input voltage in Volts\n",
    "V1=1           #Voltage in Volts\n",
    "R1=10*10**3    #Resistance in ohms\n",
    "\n",
    "\n",
    "#calculation\n",
    "I1=Vin/R1      #Load current\n",
    "Vo=2*V1        #Output voltage\n",
    "\n",
    "\n",
    "#result\n",
    "print \"Load Current is\",I1*10**3,\"mA\"\n",
    "print \"Output voltage is\",Vo,\"Volts\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.13"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Minimum output voltage is 0.0 Volts\n",
      "Maximum output voltage is 5.38 Volts\n"
     ]
    }
   ],
   "source": [
    "#Example 6.13\n",
    "#The circuit of figure 6-20, Vref=2V, R1=1 kilo Ohm. Rf=2.7 kilo Ohm. Assuming\n",
    "#that the opamp is initially nulled, determine the range for the output voltage\n",
    "#Vo.\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "#Variable declaration\n",
    "R1=1*10**3       #Resistance in ohms\n",
    "Rf=2.7*10**3     #Resistance in ohms\n",
    "Vref=2           #Voltage in Volts\n",
    "Io=0             #Since all the binary inputs D0 to D7 are logic zero\n",
    "\n",
    "\n",
    "#calculation\n",
    "Vo_min=Io*Rf                                         #Minimum output voltage\n",
    "Io=(Vref/R1)*(1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256)\n",
    "Vo_max=Io*Rf                                         #Maximum output voltage\n",
    "\n",
    "#result\n",
    "print \"Minimum output voltage is\",Vo_min,\"Volts\"\n",
    "print \"Maximum output voltage is\",round(Vo_max,2),\"Volts\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.14"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Minimum output voltage at darkness is -0.15 Volts\n",
      "Maximum output voltage at illumination is -10.0 Volts\n"
     ]
    }
   ],
   "source": [
    "#Example 6.14\n",
    "#The circuit of figure 6-21, Vdc=5 V and Rf=3 kilo Ohm. Determine the change in\n",
    "#the output voltage if the photocell is exposed to light of 0.61 lux from a dark\n",
    "#condition.Assume that the opamp is initially nulled.\n",
    "\n",
    "from __future__ import division     #to perform decimal division\n",
    "import math\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "Rf=3*10**3\n",
    "Vdc=5\n",
    "Rt1=100*10**3        #Resistance at darkness in ohms\n",
    "Rt2=1.5*10**3        #Resistance at Illumination in ohms\n",
    "\n",
    "#calculation\n",
    "Vomin=-(Vdc/Rt1)*Rf  #Min output voltage at darkness\n",
    "Vomax=-(Vdc/Rt2)*Rf  #Max output voltage at Illumination\n",
    "\n",
    "#result\n",
    "print \"Minimum output voltage at darkness is\",Vomin,\"Volts\"\n",
    "print \"Maximum output voltage at illumination is\",round(Vomax,2),\"Volts\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 6.15"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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      "text/plain": [
       "<matplotlib.figure.Figure at 0x7f75432e8a90>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "\n",
    "\n",
    "#Example 6.15\n",
    "#In the figure 6-23, R1Cf=1 second, and the input is a step(dc) voltage, as\n",
    "#shown in figure 6-26(a). Determine the output voltage and sketch it.\n",
    "#Assume that the opamp is initially nulled.\n",
    "\n",
    "%matplotlib inline\n",
    "import math\n",
    "import scipy\n",
    "from matplotlib.pyplot import ylabel, xlabel, title, plot, show\n",
    "import scipy.integrate\n",
    "\n",
    "\n",
    "#Variable declaration\n",
    "Vin=2   #Input voltage in Volts\n",
    "VoO=0   #Output offset voltage\n",
    "\n",
    "\n",
    "#calculation\n",
    "\n",
    "def integrnd(x) :\n",
    " return 2\n",
    "val1, err = scipy.integrate.quad(integrnd, 0, 1)\n",
    "val2, err = scipy.integrate.quad(integrnd, 1, 2)\n",
    "val3, err = scipy.integrate.quad(integrnd, 2, 3)\n",
    "val4, err = scipy.integrate.quad(integrnd, 3, 4)\n",
    "\n",
    "a=-val1\n",
    "b=a+-val2\n",
    "c=b+-val3\n",
    "d=c+-val4\n",
    "\n",
    "import matplotlib.pyplot as plt\n",
    "x=[0,1,2,3,4]\n",
    "y=[VoO,a,b,c,d]\n",
    "plt.plot(x,y)\n",
    "title('Output voltage')\n",
    "xlabel('time')\n",
    "ylabel('Voutput')\n",
    "plt.show()\n"
   ]
  }
 ],
 "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": "2.7.6"
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 },
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