{
 "metadata": {
  "name": ""
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 18 : Power Circuits and Systems"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.1a, Page No 780"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#initialisation of variables\n",
      "\n",
      "Vo=25.0     #in V\n",
      "ro=10.0     #in ohm\n",
      "\n",
      "print('select a silicon reference diode')\n",
      "print('two IN7555 diodes are provided')\n",
      "Rz = 12.0        #in ohm\n",
      "Vo=25.0        #output voltage in V\n",
      "Vr = 7.5 + 7.5   #because two diodes are used\n",
      "Iz = 20.0    #in mA\n",
      "Ie2=10.0     #in mA\n",
      "Ic2 = Ie2\n",
      "Icmax=30.0   #in mA\n",
      "Vcemax=45.0  #in V\n",
      "hFE2=220.0 \n",
      "hfe2=200.0 \n",
      "hie2=800.0    #in ohm\n",
      "Id=10.0       #in mA\n",
      "Il = 1000.0   #in mA\n",
      "Vi = 50.0     #in V\n",
      "dVi = 10.0    #change in input voltage\n",
      "dIl = 1.0     #change in load current\n",
      "\n",
      "#Calculations\n",
      "\n",
      "#For D1 and D2 operate\n",
      "Iz = Id + Id\n",
      "Rd = (Vo-Vr)/Id\n",
      "print(\"The value of Rd is %.2f K \" %Rd)\n",
      "Ib2 = (1000.0*Ic2)/hFE2\n",
      "print(\"The value of Ib2 is %.2f microA \" %Ib2)\n",
      "#Since we require I1>Ib2 , we select\n",
      "I1=10*(10**-3)#in A\n",
      "Vbe = 0.7#in V\n",
      "V2 = Vbe + Vr\n",
      "print(\"The value of V2 is %.2f v \" %V2)\n",
      "R1 = (Vo-V2)/I1\n",
      "R2 = V2/I1\n",
      "print(\"The value of R1 is %.2f ohm \" %R1)\n",
      "print(\"The value of R2 is %.2f ohm \" %R2)\n",
      "#We are selecting Texas Instruments 2N1722 silicon power transistor, so following parameters are required\n",
      "print('We are selecting Texas Instruments 2N1722 silicon power transistor')\n",
      "Ic1 = 1#in A\n",
      "hFE1=125.0\n",
      "hfe1=100.0\n",
      "hie1=20.0\n",
      "\n",
      "\n",
      "#Results\n",
      "Ib1 =(1000*I1 + Il + Id)/hFE1\n",
      "print(\"The current through resistor R3 is %.2f mA \" %(Ib1+Ic2))\n",
      "I=Ib1 + Ic2\n",
      "\n",
      "R3 = (Vi - (Vbe + Vo))/I\n",
      "print(\"The value of R3 is %.2f K \" %R3)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "select a silicon reference diode\n",
        "two IN7555 diodes are provided\n",
        "The value of Rd is 1.00 K \n",
        "The value of Ib2 is 45.45 microA \n",
        "The value of V2 is 15.70 v \n",
        "The value of R1 is 930.00 ohm \n",
        "The value of R2 is 1570.00 ohm \n",
        "We are selecting Texas Instruments 2N1722 silicon power transistor\n",
        "The current through resistor R3 is 18.16 mA \n",
        "The value of R3 is 1.34 K \n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.1b, Page No 780"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#initialisation of variables\n",
      "Vo=25.0    #in V\n",
      "ro=10.0    #in ohm\n",
      "\n",
      "Rz = 12.0 #in ohm\n",
      "Vo=25.0   #output voltage in V\n",
      "Vr = 7.5 + 7.5#because two diodes are used\n",
      "Iz = 20.0 #in mA\n",
      "Ie2=10.0 #in mA\n",
      "Ic2 = Ie2\n",
      "Icmax=30.0 #in mA\n",
      "Vcemax=45.0 #in V\n",
      "hFE2=220.0 \n",
      "hfe2=200.00 \n",
      "hie2=800.0 #in ohm\n",
      "Id=10.0 #in mA\n",
      "Il = 1000.0 #in mA\n",
      "Vi = 50.0 #in V\n",
      "dVi = 10.0  #change in input voltage\n",
      "dIl = 1.0  #change in load current\n",
      "\n",
      "#Calculations\n",
      "\n",
      "#For D1 and D2 operate\n",
      "Iz = Id + Id\n",
      "Rd = (Vo-Vr)/Id\n",
      "Ib2 = (1000*Ic2)/hFE2\n",
      "\n",
      "#Since we require I1>Ib2 , we select\n",
      "I1=10*(10**-3)#in A\n",
      "Vbe = 0.7#in V\n",
      "\n",
      "V2 = Vbe + Vr\n",
      "\n",
      "R1 = (Vo-V2)/I1\n",
      "R2 = V2/I1\n",
      "\n",
      "#We are selecting Texas Instruments 2N1722 silicon power transistor, so following parameters are required\n",
      "Ic1 = 1#in A\n",
      "hFE1=125.0\n",
      "hfe1=100.0\n",
      "hie1=20.0 \n",
      "\n",
      "Ib1 =(1000.0*I1 + Il + Id)/hFE1\n",
      "#The current through resistor R3\n",
      "I=Ib1 + Ic2\n",
      "\n",
      "R3 = (Vi - (Vbe + Vo))/I\n",
      "Gm = hfe2*(R2/(R2+R1))*(1/((R1*R2/(R1+R2))+hie2+(1+hfe2)*Rz)) \n",
      "Sv = (10**-3)/(Gm*R3)\n",
      "\n",
      "#Results\n",
      "\n",
      "print(\"The value of Sv is %.2f \" %Sv)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of Sv is 0.02 \n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.1c Page No 780"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#initialisation of variables\n",
      "Vo=25.0 #in V\n",
      "ro=10.0  #in ohm\n",
      "Rz = 12.0  #in ohm\n",
      "Vo=25.0  #output voltage in V\n",
      "Vr = 7.5 + 7.5#because two diodes are used\n",
      "Iz = 20.0 #in mA\n",
      "Ie2=10.0 #in mA\n",
      "Ic2 = Ie2\n",
      "Icmax=30.0 #in mA\n",
      "Vcemax=45.0 #in V\n",
      "hFE2=220.0\n",
      "hfe2=200.0\n",
      "hie2=800.0  #in ohm\n",
      "Id=10.0  #in mA\n",
      "Il = 1000.0 #in mA\n",
      "Vi = 50.0 #in V\n",
      "dVi = 10.0 #change in input voltage\n",
      "dIl = 1.0  #change in load current\n",
      "\n",
      "#Calculations\n",
      "#For D1 and D2 operate\n",
      "Iz = Id + Id\n",
      "Rd = (Vo-Vr)/Id\n",
      "\n",
      "Ib2 = (1000.0*Ic2)/hFE2\n",
      "\n",
      "#Since we require I1>Ib2 , we select\n",
      "I1=10*(10**-3)#in A\n",
      "Vbe = 0.7#in V\n",
      "\n",
      "V2 = Vbe + Vr\n",
      "\n",
      "R1 = (Vo-V2)/I1\n",
      "R2 = V2/I1\n",
      "\n",
      "#We are selecting Texas Instruments 2N1722 silicon power transistor, so following parameters are required\n",
      "Ic1 = 1#in A\n",
      "hFE1=125.0\n",
      "hfe1=100.0\n",
      "hie1=20.0\n",
      "\n",
      "Ib1 =(1000*I1 + Il + Id)/hFE1\n",
      "#The current through resistor R3 is\n",
      "I=Ib1 + Ic2\n",
      "\n",
      "R3 = (Vi - (Vbe + Vo))/I\n",
      "\n",
      "Gm = hfe2*(R2/(R2+R1))*(1/((R1*R2/(R1+R2))+hie2+(1+hfe2)*Rz)) \n",
      "\n",
      "#Results\n",
      "print(\"The value of Gm is %.2f \" %Gm)\n",
      "Ro = (ro + (((1000*R3) + hie1)/(1+hfe1)))/(1 + (Gm*((1000*R3) + ro)))\n",
      "print(\"The output impedence is = %.2f K \" %Ro)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of Gm is 0.03 \n",
        "The output impedence is = 0.51 K \n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.1d, Page No 780"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#initialisation of variables\n",
      "Vo=25.0 #in V\n",
      "ro=10.0 #in ohm\n",
      "\n",
      "Rz = 12.0  #in ohm\n",
      "Vo=25.0  #output voltage in V\n",
      "Vr = 7.5 + 7.5#because two diodes are used\n",
      "Iz = 20.0  #in mA\n",
      "Ie2=10.0  #in mA\n",
      "Ic2 = Ie2\n",
      "Icmax=30.0 #in mA\n",
      "Vcemax=45.0 #in V\n",
      "hFE2=220.0 \n",
      "hfe2=200.0 \n",
      "hie2=800.0 #in ohm\n",
      "Id=10.0 #in mA\n",
      "Il = 1000.0 #in mA\n",
      "Vi = 50.0 #in V\n",
      "dVi = 10.0  #change in input voltage\n",
      "dIl = 1.0 #change in load current\n",
      "\n",
      "#Calculations\n",
      "\n",
      "#For D1 and D2 operate\n",
      "Iz = Id + Id\n",
      "Rd = (Vo-Vr)/Id\n",
      "\n",
      "Ib2 = (1000.0*Ic2)/hFE2\n",
      "\n",
      "#Since we require I1>Ib2 , we select\n",
      "I1=10.0*(10**-3)#in A\n",
      "Vbe = 0.7#in V\n",
      "\n",
      "V2 = Vbe + Vr\n",
      "\n",
      "R1 = (Vo-V2)/I1\n",
      "R2 = V2/I1\n",
      "\n",
      "#We are selecting Texas Instruments 2N1722 silicon power transistor, so following parameters are required\n",
      "Ic1 = 1#in A\n",
      "hFE1=125.0\n",
      "hfe1=100.0\n",
      "hie1=20.0\n",
      "\n",
      "Ib1 =(1000.0*I1 + Il + Id)/hFE1\n",
      "#The current through resistor R3 is\n",
      "I=Ib1 + Ic2\n",
      "\n",
      "R3 = (Vi - (Vbe + Vo))/I\n",
      "\n",
      "Gm = hfe2*(R2/(R2+R1))*(1/((R1*R2/(R1+R2))+hie2+(1+hfe2)*Rz)) \n",
      "Sv = (10**-3)/(Gm*R3)\n",
      "\n",
      "Ro = (ro + (((1000*R3) + hie1)/(1+hfe1)))/(1 + (Gm*((1000*R3) + ro)))\n",
      "\n",
      "dVo = (Sv*dVi)+(Ro*dIl)\n",
      "\n",
      "#Results\n",
      "print(\"Change in output voltage = %.2f v \" %dVo)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Change in output voltage = 0.74 v \n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.2a Page No 792"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from scipy import integrate\n",
      "#initialisation of variables\n",
      "Vs=230.0  #in V\n",
      "Rl=200.0  #in ohm\n",
      "#Trigger is adjusted so that conduction starts after 60degree of start of cycle\n",
      "#Instantaneous Current il = (230*2^0.5*sin(a))/200\n",
      "\n",
      "#Calculations\n",
      "# to find rms value\n",
      "xo = math.pi/3.0 #lower limit of integration\n",
      "x1 = math.pi     #upper limit of integration\n",
      "def f(x):\n",
      "    return((230.0*(2.0**0.5)*math.sin(x))/200.0)**2.0\n",
      "\n",
      "\n",
      "X = integrate.quad(f,xo,x1)\n",
      "Irms = (X[0]/(2*math.pi))**0.5\n",
      "\n",
      "#Results\n",
      "print(\"The value of Irms is %.2f A \" %Irms)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of Irms is 0.73 A \n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.2b Page No 792"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from scipy import integrate\n",
      "#initialisation of variables\n",
      "Vs=230.0 #in V\n",
      "Rl=200.0 #in ohm\n",
      "#Trigger is adjusted so that conduction starts after 60degree of start of cycle\n",
      "#Instantaneous Current il = (230*2^0.5*sin(a))/200\n",
      "#It is noted that between 0 to pi/3 SCR voltage equals line voltage and between pi/3 to pi it is zer and for the rest it is equal to line voltage\n",
      "#Vl = 230*2^0.5*sin(x)\n",
      "#To find instantaneous power\n",
      "\n",
      "x0=math.pi/3#lower limit of integral\n",
      "x1=math.pi#upper limit of integral\n",
      "def f(x):\n",
      "    return((230*230*2*(math.sin(x)**2))/200)\n",
      "\n",
      "X = integrate.quad(f,xo,x1)\n",
      "P = X[0]/(2*3.14)\n",
      "print(\"The value of P is %.2f W \" %P)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of P is 106.45 W \n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.2c Page No 792"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#initialisation of variables\n",
      "\n",
      "Vs=230.0 #in V\n",
      "Rl=200.0 #in ohm\n",
      "#Trigger is adjusted so that conduction starts after 60degree of start of cycle\n",
      "#Instantaneous Current il = (230*2**0.5*sin(a))/200\n",
      "\n",
      "#To find Vrms\n",
      "\n",
      "#Calculations\n",
      "#It is noted that between 0 to pi/3 SCR voltage equals line voltage and between pi/3 to pi it is zer and for the rest it is equal to line voltage\n",
      "xo=0      #lower limit of first integral\n",
      "x1=math.pi/3.0  #upper limit of first integral\n",
      "x2=math.pi      #lower limit of second integral\n",
      "x3=2.0*(math.pi)    #upper limit of second integral\n",
      "\n",
      "def f(x):\n",
      "    return((230*(2**0.5)*math.sin(x))**2)\n",
      "\t\n",
      "def g(x):\n",
      "    return((230*(2**0.5)*math.sin(x))**2)\n",
      "\n",
      "X1 = integrate.quad(f,xo,x1)\n",
      "X2 = integrate.quad(g,x2,x3)\n",
      "Vrms = ((X1[0]+X2[0])/(2*math.pi))**0.5\n",
      "\n",
      "#Results\n",
      "print(\"The value of Vrms is %.2f v \" %Vrms)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of Vrms is 177.82 v \n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.3, Page No 794"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#initialisation of variables\n",
      "\n",
      "#Caption:SCR Relaxation Oscillator Phase control Circuit\n",
      "#Given Data\n",
      "C=0.1#in microF\n",
      "V=60.0 #in V\n",
      "Vb=32.0 #in V\n",
      "Vh=10.0 #holding voltage in V\n",
      "Ih=100.0 #in microA\n",
      "c=45.0 #conductance angle in degree\n",
      "cd = 360.0 - c#angle in which capacitor will get charged\n",
      "td = (cd/360.0)*(1.0/60)#in ms\n",
      "\n",
      "#Calculations\n",
      "#if the anode voltage is positive,the SCR will fire when vc=32V\n",
      "vc=32#in V\n",
      "#let time constant = t = R*C\n",
      "#vc-Vh = (V-Vh)(1-exp(-td/t))\n",
      "t = -td/math.log(1-((vc-Vh)/(V-Vh)))\n",
      "\n",
      "#Results\n",
      "\n",
      "print(\"The value of time constant = is %.2f sec \" %t)\n",
      "R = t/C#Resistance in K\n",
      "print(\"The value of R is %.2f K \" %(R*1000))\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of time constant = is 0.03 sec \n",
        "The value of R is 251.52 K \n"
       ]
      }
     ],
     "prompt_number": 8
    }
   ],
   "metadata": {}
  }
 ]
}