path: root/Electronic_Devices_and_Circuits_by_D._C._Kulshreshtha/Chapter06.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter06:Transistor Biasing and Stabilization"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E1 - Pg 191"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the Q point\n",
      "#given\n",
      "B=50.;            #dc beta\n",
      "Rc=2.2*10.**3.;#ohm      #resistor connected to collector\n",
      "Rb=270.*10.**3.;#ohm      #resistor connected to base\n",
      "Vcc=9.;#V            #Voltage supply across the collector resistor\n",
      "Vbe=0.7;#V          #base to emitter voltage\n",
      "Ib=(Vcc-Vbe)/Rb;         #Base current\n",
      "Ic=B*Ib;         #Colletor current\n",
      "Ics=Vcc/Rc;       #Colletor saturation current\n",
      "\n",
      "#Actual Ic is the smaller of the above two values\n",
      "Vce=Vcc-Ic*Rc;\n",
      "print '%s %.1f %s %.1f %s' %(\"The Q point is =\",Vce,'V',Ic*1000,'mA');\n",
      "\n",
      "#Note--In book Vce = 5.7 V because of approaximation\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Q point is = 5.6 V 1.5 mA\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E2 - Pg 192"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the Q point\n",
      "#given\n",
      "B=150.;            #dc beta\n",
      "Rc=1.*10.**3.;#ohm      #resistor connected to collector\n",
      "Rb=100.*10.**3.;#ohm      #resistor connected to base\n",
      "Vcc=10.;#V            #Voltage supply across the collector resistor\n",
      "Vbe=0.7;#V          #base to emitter voltage\n",
      "Ib=(Vcc-Vbe)/Rb;         #Base current\n",
      "Ic=B*Ib;         #Colletor current\n",
      "Ics=Vcc/Rc;       #Colletor saturation current\n",
      "\n",
      "#Actual Ic is the smaller of the above two values i.e. Ic(sat) and since the transistor is in saturation mode therefore Vce will become 0\n",
      "\n",
      "Vce=0;\n",
      "print '%s %.f %s %.f %s' %(\"The Q point is =\",Vce,\"V\",Ics*1000,\"mA\");\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Q point is = 0 V 10 mA\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E3 - Pg 192"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine Rb and percentage change in collector current due to temperature rise\n",
      "#given\n",
      "\n",
      "#Calculating the base resistance\n",
      "B=20.;            #dc beta\n",
      "Rc=1.*10.**3.;#ohm      #resistor connected to collector\n",
      "Ic=1.*10.**-3.;#A       #collector current\n",
      "Vcc=6.;#V            #Voltage supply across the collector resistor\n",
      "Vbe=0.3;#V       #for germanium\n",
      "Icbo=2.*10.**-6.;#A       #collector  to base leakage current\n",
      "\n",
      "Ib=(Ic-(1.+B)*Icbo)/B;\n",
      "Rb=(Vcc-Vbe)/Ib;\n",
      "\n",
      "print '%s %.f %s' %(\"The value of resistor Ib is =\",120,'kohm');\n",
      "\n",
      "Rb=120.*10.**3.;#ohm approax\n",
      "\n",
      "#Now when temperature rise\n",
      "Icbo=10.*10.**-6.;#A #collector  to base leakage current\n",
      "B=25.;#dc beta\n",
      "Ic1=B*Ib+(B+1)*Icbo;# #changed collector current\n",
      "perc=(Ic1-Ic)*100./Ic;#percentage increase\n",
      "print '%s %.f %s' %(\"The percentage change in collector current is =\",perc,\"percent\");\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of resistor Ib is = 120 kohm\n",
        "The percentage change in collector current is = 46 percent\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E4 - Pg 193"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the Q point at two different B\n",
      "#given\n",
      "\n",
      "#At B=50\n",
      "\n",
      "B=50.;            #dc beta\n",
      "Rc=2.*10.**3.;#ohm      #resistor connected to collector\n",
      "Rb=300.*10.**3.;#ohm      #resistor connected to base\n",
      "Vcc=9.;#V            #Voltage supply across the collector resistor\n",
      "Ib=Vcc/Rb;         #Base current\n",
      "Ic=B*Ib;         #Colletor current\n",
      "Ics=Vcc/Rc;       #Colletor saturation current\n",
      "\n",
      "#Actual Ic is the smaller of the above two values\n",
      "\n",
      "Vce=Vcc-Ic*Rc;\n",
      "print '%s %.2f %s %.1f %s' %(\"The Q point (At B=50) =\",Vce,\"V\",Ic*1000,\"mA\");\n",
      "\n",
      "#At B=150\n",
      "\n",
      "B1=150.;            #dc beta\n",
      "Ic1=B*Ib;         #Colletor current\n",
      "Ics1=Vcc/Rc;       #Colletor saturation current\n",
      "\n",
      "#Actual Ic is the smaller of the above two values i.e. Ic(sat) and since the transistor is in saturation mode therefore Vce will become 0\n",
      "\n",
      "Vce=0;\n",
      "print '%s %.f %s %.1f %s' %(\"\\nThe Q point (At B=150) is =\",Vce,\"V\",Ics*1000,\"mA\");\n",
      "\n",
      "print '%s %.f' %(\"\\nThe factor at which collector current increases =\",Ics1/Ic);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Q point (At B=50) = 6.00 V 1.5 mA\n",
        "\n",
        "The Q point (At B=150) is = 0 V 4.5 mA\n",
        "\n",
        "The factor at which collector current increases = 3\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E5 - Pg 196"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine Q point in collector to base bias circuit\n",
      "#given\n",
      "B=100.;            #dc beta\n",
      "Rc=500.;#ohm      #resistor connected to collector\n",
      "Rb=500.*10.**3.;#ohm      #resistor connected to base\n",
      "Vcc=10.;#V            #Voltage supply across the collector resistor\n",
      "Ib=Vcc/(Rb+B*Rc);         #Base current\n",
      "Ic=B*Ib;         #Colletor current\n",
      "Ics=Vcc/Rc;       #Colletor saturation current\n",
      "\n",
      "#Actual Ic is the smaller of the above two values\n",
      "\n",
      "Vce=Vcc-(Ic+Ib)*Rc;\n",
      "print '%s %.1f %s %.1f %s' %(\"The Q point is =\",Vce,\"V\",Ic*1000,\"mA\");\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Q point is = 9.1 V 1.8 mA\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E6 - Pg 196"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate the collector current and change in it if B is changed by three times of previous B\n",
      "#given\n",
      "B=50.;            #dc beta\n",
      "Rc=2.*10.**3.;#ohm      #resistor connected to collector\n",
      "Rb=300.*10.**3.;#ohm      #resistor connected to base\n",
      "Vcc=9.;#V            #Voltage supply across the collector as it is PNP so taking positive\n",
      "Ib=Vcc/(Rb+B*Rc);         #Base current\n",
      "Ic=B*Ib;         #Colletor current\n",
      "print '%s %.3f %s' %(\"Collector current (B=50)=\",Ic*1000,\"mA\\n\");\n",
      "#Now B=150\n",
      "B=3.*B;      #three times of previous B\n",
      "Ib1=Vcc/(Rb+B*Rc);         #Base current\n",
      "Ic1=B*Ib1;         #Colletor current\n",
      "print '%s %.2f %s' %(\"Collector current (B=150)=\",Ic1*1000,\"mA\\n\");\n",
      "print '%s %.f' %(\"The factor at which collector current increases =\",Ic1/Ic);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Collector current (B=50)= 1.125 mA\n",
        "\n",
        "Collector current (B=150)= 2.25 mA\n",
        "\n",
        "The factor at which collector current increases = 2\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E7 - Pg 199"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate the value of all three current Ie and Ic and Ib\n",
      "#given\n",
      "B=90.;            #dc beta\n",
      "Rc=1.*10.**3.;#ohm      #resistor connected to collector\n",
      "Rb=500.*10.**3.;#ohm      #resistor connected to base\n",
      "Re=500.;#ohm      #resistor connected to emitter\n",
      "Vcc=9.;#V            #Voltage supply across the collector resistor\n",
      "Ib=Vcc/(Rb+B*Re);         #Base current\n",
      "Ic=B*Ib;         #Colletor current\n",
      "Ie=Ic+Ib;        #Emitter current\n",
      "print '%s %.1f %s %s %.3f %s %s %.3f %s' %(\"Base current =\",Ib*10**6,\"uA\\n\",\"\\nCollector current =\",Ic*10**3,\"mA\\n\",\"\\nEmitter current =\",Ie*10**3,\"mA\");\n",
      " "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Base current = 16.5 uA\n",
        " \n",
        "Collector current = 1.486 mA\n",
        " \n",
        "Emitter current = 1.503 mA\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E8 - Pg 199"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate max and min value of emitter current\n",
      "#given\n",
      "\n",
      "#At B=50\n",
      "\n",
      "B=50.;            #dc beta\n",
      "Rc=75.;#ohm      #resistor connected to collector\n",
      "Re=100.;#ohm      #resistor connected to emitter\n",
      "Rb=10.*10.**3.;#ohm      #resistor connected to base\n",
      "Vcc=6.;#V            #Voltage supply across the collector resistor\n",
      "Vbe=0.3;#V       #for germanium\n",
      "Ib=(Vcc-Vbe)/(Rb+(1.+B)*Re);         #Base current\n",
      "Ie=(1.+B)*Ib;\n",
      "Vce=Vcc-(Rc+Re)*Ie\n",
      "print '%s %.2f %s' %(\"Minimum emitter current =\",Ie*10**3,\"mA\\n\");\n",
      "print '%s %.2f %s' %(\"The collector to emitter volatge =\",Vce,\"V\\n\");\n",
      "\n",
      "#At B=300 \n",
      "\n",
      "B1=300.; #dc beta\n",
      "Ib1=(Vcc-Vbe)/(Rb+(1.+B1)*Re);#Base current\n",
      "Ie1=(1.+B1)*Ib1;\n",
      "Vce1=Vcc-(Rc+Re)*Ie1\n",
      "#Here Vce1= -1.4874 V but can never have negative voltage because Ie1 is wrong as it cant be more than saturation value therefore\n",
      "Ie1=Vcc/(Rc+Re);\n",
      "\n",
      "#And Vce=0 V\n",
      "\n",
      "Vce1=0;#V\n",
      "print '%s %.2f %s' %(\"Maximum emitter current =\",Ie1*10**3,\"mA\\n\");\n",
      "print '%s %.f %s' %(\"The collector to emitter volatge(saturation) =\",Vce1,\"V\");\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Minimum emitter current = 19.25 mA\n",
        "\n",
        "The collector to emitter volatge = 2.63 V\n",
        "\n",
        "Maximum emitter current = 34.29 mA\n",
        "\n",
        "The collector to emitter volatge(saturation) = 0 V\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E9 - Pg 200"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the value of base resistance\n",
      "#given\n",
      "\n",
      "B=100.;            #dc beta\n",
      "Rc=200.;#ohm      #resistor connected to collector\n",
      "Re=500.;#ohm      #resistor connected to emitter\n",
      "Vcc=9.;#V         #Voltage supply across the collector as it is PNP so taking positive\n",
      "Vce=4.5;#V       #Collector to emitter voltage\n",
      "Ic=(Vcc-Vce)/(Rc+Re);\n",
      "Ib=Ic/B;\n",
      "Rb=(Vcc-B*Re*Ib)/Ib;\n",
      "print '%s %.f %s' %(\"The value of base resistance is =\",Rb/1000,\"kohm\");\n",
      " "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of base resistance is = 90 kohm\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E10 - Pg 200"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the collector current at two different B\n",
      "#given\n",
      "\n",
      "#At B=50\n",
      "\n",
      "B=50.;#dc beta\n",
      "Rc=2.;#ohm #resistor connected to collector\n",
      "Re=1000.;#ohm #resistor connected to emitter\n",
      "Rb=300.*10.**3.;#ohm #resistor connected to base\n",
      "Vcc=9.;#V #Voltage supply across the collector resistor\n",
      "Ib=Vcc/(Rb+B*Re);         #Base current\n",
      "Ic=B*Ib;         #Colletor current\n",
      "print '%s %.2f %s' %(\"The collector current at (B=50)=\",Ic*1000,\"mA\\n\");\n",
      "\n",
      "#At B=150\n",
      "\n",
      "B1=150.;#dc beta\n",
      "Ib1=Vcc/(Rb+B1*Re);         #Base current\n",
      "Ic1=B1*Ib1;         #Colletor current\n",
      "print '%s %.1f %s' %(\"The collector current at (B=150)=\",Ic1*1000,\"mA\\n\");\n",
      "print '%s %.1f' %(\"The factor at which collector current increases=\",Ic1/Ic);\n",
      "\n",
      "#IN BOOK Ic(AT B=50)= 1.25 mA and Ic1/Ic=2.4 DUE TO APPROAXIMATION\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The collector current at (B=50)= 1.29 mA\n",
        "\n",
        "The collector current at (B=150)= 3.0 mA\n",
        "\n",
        "The factor at which collector current increases= 2.3\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E11 - Pg 205"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate Q point in voltage divider\n",
      "#given\n",
      "B=100.;            #dc beta\n",
      "Rc=2.*10.**3.;#ohm      #resistor connected to collector\n",
      "R1=10.*10.**3.;#ohm      #voltage divider resistor 1\n",
      "R2=1.*10.**3.;#ohm      #voltage divider resistor 2\n",
      "Re=200.;#ohm      #resistor connected to emitter\n",
      "Vcc=10.;#V            #Voltage supply across the collector resistor\n",
      "Vbe=0.3;#V          #base to emitter voltage\n",
      "I=Vcc/(R1+R2);       #current through voltage divider\n",
      "Vb=I*R2;        #voltage at base\n",
      "Ve=Vb-Vbe;\n",
      "Ie=Ve/Re;\n",
      "Ic=Ie        #approaximating Ib is nearly equal to 0\n",
      "Vc=Vcc-Ic*Rc;\n",
      "Vce=(Vc)-Ve; \n",
      "print '%s %.1f %s %.f %s' %(\"The Q point is =\",Vce,\"V\",Ic*1000,\"mA\");\n",
      "\n",
      "Ibc=I/20.;     #critical value of base current\n",
      "Ib=Ic/B;      #actual base current\n",
      "\n",
      "#Since Ib < Ibc, hence assumption is alright\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Q point is = 3.3 V 3 mA\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E12 - Pg 207"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "#Solve the voltage divider accurately by applying thevenin's theorem\n",
      "#given\n",
      "B=100.;            #dc beta\n",
      "Rc=2.*10.**3.;#ohm      #resistor connected to collector\n",
      "R1=10.;#ohm      #voltage divider resistor 1\n",
      "R2=1.;#ohm      #voltage divider resistor 2\n",
      "Re=200.;#ohm      #resistor connected to emitter\n",
      "Vcc=10.;#V            #Voltage supply across the collector resistor\n",
      "Vbe=0.3;#V          #base to emitter voltage\n",
      "\n",
      "Vth=Vcc*R2/(R1+R2);\n",
      "Rth=R1*R2/(R1+R2);\n",
      "\n",
      "print '%s %.1f %s' %(\"\\nThevenin equivalent voltage Vth =\",Vth,\"V\");\n",
      "print '%s %.1f %s' %(\"\\nThevenin equivalent resistance Rth =\",Rth,\"kohm\");\n",
      "\n",
      "Ib=(Vth-Vbe)/(Rth+(1.+B)*Re);\n",
      "Ic=B*Ib;\n",
      "Ie=Ic+Ib;\n",
      "Vce=Vcc-Ic*Rc-Ie*Re; \n",
      "print '%s %.4f %s' %(\"\\nThe accurate value of Ic =\",Ic*10**3,\"mA\");\n",
      "print '%s %.5f %s' %(\"\\nThe accurate value of Vce =\",Vce,\"V\");\n",
      "Icp=3.*10.**-3.; # Current calculated by voltage divider in previous example\n",
      "Vcep=3.4; # Voltage calculated by voltage divider in previous example\n",
      "Err_Ic=(Ic-Icp)*100./Ic;\n",
      "Err_Vce=(Vce-Vcep)*100./Vce;\n",
      "print '%s %.1f %s' %(\"\\nError in Ic =\",Err_Ic,\"percent\\n\");\n",
      "print '%s %.1f %s' %(\"\\nError in Vce =\",Err_Vce,\"percent\");\n",
      "\n",
      "# The errors and The accurate values are different \n",
      "# because of the approaximation in Vth and Rth in book\n",
      "\n",
      "# In Book Ic = 2.8436 mA and Vce = 3.73839 V\n",
      "# Error in Ic = -5.5% \n",
      "# Error in Vce = +9% \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "Thevenin equivalent voltage Vth = 0.9 V\n",
        "\n",
        "Thevenin equivalent resistance Rth = 0.9 kohm\n",
        "\n",
        "The accurate value of Ic = 3.0152 mA\n",
        "\n",
        "The accurate value of Vce = 3.36060 V\n",
        "\n",
        "Error in Ic = 0.5 percent\n",
        "\n",
        "\n",
        "Error in Vce = -1.2 percent\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E13 - Pg 209"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the Q point for the emitter bias circuit\n",
      "#given\n",
      "B=100.;               #dc beta\n",
      "Rc=5.*10.**3.;#ohm      #resistor connected to collector\n",
      "Rb=10.*10.**3.;#ohm      #resistor connected to base\n",
      "Re=10.*10.**3.;#ohm      #resistor connected to emitter \n",
      "Vcc=12.;#V #Voltage supply across the collector resistor\n",
      "Vee=15;#V #supply at emitter\n",
      "Ie=Vee/Re;\n",
      "Ic=Ie;\n",
      "Vce=Vcc-Ic*Rc;\n",
      "print '%s %.1f %s %.1f %s' %(\"The Q point is =\",Vce,\"V\",Ic*1000,\"mA\");\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Q point is = 4.5 V 1.5 mA\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E14 - Pg 211"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate Vgs and Rs\n",
      "#given\n",
      "import math\n",
      "Vp=2.;#V\n",
      "Idss=1.75*10.**-3.;#A      #drain current at Vgs=0\n",
      "Vdd=24.;#V             #drain to supply source\n",
      "Id=1.*10.**-3.;#A       #drain current\n",
      "Vgs=(-Vp)*(1-math.sqrt(Id/Idss));\n",
      "Rs=abs(Vgs)/Id;\n",
      "print '%s %.3f %s' %(\"Vgs =\",Vgs,\"V\\n\");\n",
      "print '%s %.f %s' %(\"Rs =\",Rs,\"ohm\");\n",
      " "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Vgs = -0.488 V\n",
        "\n",
        "Rs = 488 ohm\n"
       ]
      }
     ],
     "prompt_number": 14
    }
   ],
   "metadata": {}
  }
 ]
}