path: root/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_6.ipynb

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  {
   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "#CHAPTER 11 : MULTISTAGE TRANSISTOR AMPLIFIERS"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "##Example 11.1 : Page number 285"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from math import log10\n",
      "#Variable declaration\n",
      "#(i)\n",
      "A_v=30;                 #Voltage gain\n",
      "\n",
      "#(ii)\n",
      "A_p=100;                #Power gain\n",
      "\n",
      "\n",
      "#Calculations\n",
      "#(i)\n",
      "A_v_dB=20*log10(A_v);          #Voltage gain, dB\n",
      "A_p_dB=10*log10(A_p);          #Power gain, dB\n",
      "\n",
      "#Results\n",
      "print(\"(i) Voltage gain in dB=%.2fdB\"%A_v_dB);\n",
      "print(\"(ii) Power gain in dB=%ddB\"%A_p_dB);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i) Voltage gain in dB=29.54dB\n",
        "(ii) Power gain in dB=20dB\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.2 : Page number 285-286\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#(i)\n",
      "A_p_dB=40.0;                #Power gain in dB\n",
      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
      "A_p=10**A_p_b;              #Power gain in number\n",
      "\n",
      "#Result\n",
      "print(\"(i) Power gain in number=%d\"%A_p);\n",
      "\n",
      "#(ii)\n",
      "A_p_dB=43.0;                #Power gain in dB\n",
      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
      "A_p=round(10**A_p_b,-4);              #Power gain in number\n",
      "\n",
      "#Result\n",
      "print(\"(ii) Power gain in number=%d\"%A_p);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i) Power gain in number=10000\n",
        "(ii) Power gain in number=20000\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.3 : Page number 286\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from math import log10\n",
      "\n",
      "#Variable declaration\n",
      "Av_1=100.0;                     #Voltage gain of stage 1\n",
      "Av_2=200.0;                     #Voltage gain of stage 2\n",
      "Av_3=400.0;                     #Voltage gain of stage 3\n",
      "\n",
      "#Calculations\n",
      "Av_1_dB=20*log10(Av_1);                     #Voltage gain of stage 1, dB\n",
      "Av_2_dB=20*log10(Av_2);                     #Voltage gain of stage 2, dB\n",
      "Av_3_dB=20*log10(Av_3);                     #Voltage gain of stage 3, dB\n",
      "\n",
      "Av_T=Av_1_dB+Av_2_dB+Av_3_dB;            #Total voltage gain\n",
      "\n",
      "#Result\n",
      "print(\"The total voltage gain=%ddB\"%Av_T);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The total voltage gain=138dB\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.4 : Page number 286\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log10\n",
      "\n",
      "#Variable declaration\n",
      "A_p_absolute=30.0;                          #Absolute gain of each stage\n",
      "number_of_stages=5.0;                       #number of stages\n",
      "negative_feedback=10.0;                     #negative feedback, dB\n",
      "\n",
      "#Calculations\n",
      "#(i)\n",
      "A_p_dB=round(10*log10(A_p_absolute),2);              #Power gain of one stage. dB\n",
      "A_p_T=number_of_stages * A_p_dB;            #Total power gain, dB\n",
      "\n",
      "#(ii)\n",
      "A_p_resultant=A_p_T-negative_feedback;      #Resultant power gain with negative feedback, dB\n",
      "\n",
      "#Results\n",
      "print(\"The total power gain = %.2fdB.\"%A_p_T);\n",
      "print(\"The resultant power gain with negative feedback = %.2fdB.\"%A_p_resultant);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The total power gain = 73.85dB.\n",
        "The resultant power gain with negative feedback = 63.85dB.\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.5 : Page number 286\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log10\n",
      "\n",
      "#Variable declaration\n",
      "P_out_2kHz=1.5;                 #Output power at 2 kHz, W\n",
      "P_out_20kHz=0.3;                #Output power at 20 kHz, W\n",
      "P_in=10.0;                      #Input power, mW\n",
      "\n",
      "#Calculations\n",
      "A_p_dB_2kHz=10*log10(P_out_2kHz*1000/P_in);         #dB power gain at 2 kHz\n",
      "A_p_dB_20kHz=10*log10(P_out_20kHz*1000/P_in);       #dB power gain at 20 kHz\n",
      "Fall_in_gain=A_p_dB_2kHz-A_p_dB_20kHz;              #Fall in gain from 2kHz to 20kHz\n",
      "\n",
      "#Results\n",
      "print(\"The fall in gain from 2kHz to 20kHz=%.2fdB\"%Fall_in_gain);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The fall in gain from 2kHz to 20kHz=6.99dB\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.6 : Page number 287\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log10\n",
      "\n",
      "#Variable declaration\n",
      "A_v=15.0;                   #Voltage gain, dB\n",
      "V_1=0.8;                    #Input signal voltage, V\n",
      "\n",
      "#Calculations\n",
      "#Since, Av(in decibel)=20*log10(V_2/V_1),\n",
      "V_2=V_1*(10**(A_v/20));                 #Output voltage, V\n",
      "\n",
      "#Results\n",
      "print(\"The output voltage= %.1fV.\"%V_2);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The output voltage= 4.5V.\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.7 : Page number 287\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "A_0_dB=70.0;                   #Open circuit voltage gain, dB\n",
      "A_v_dB=67.0;                   #Voltage gain, dB\n",
      "R_out=1.5;                     #Output resistance, kilo ohm\n",
      "\n",
      "#Calculations\n",
      "#Since, A_0_dB-A_v_dB=20*log10(A_0/A_v)\n",
      "ratio_A0_Av=round(10**((A_0_dB-A_v_dB)/20),2);                  #Ratio of open-circuit voltage gain to normal voltage gain\n",
      "\n",
      "#Since, A_v/A_0 = RL/(R_out+RL)\n",
      "RL=R_out/(ratio_A0_Av-1);                               #Load resistor, kilo ohm\n",
      "\n",
      "\n",
      "#Results\n",
      "print(\"The load resistance=%.2f kilo ohm.\"%RL);\n",
      "\n",
      "#Note: The value of load resistor is calculated to be 3.6585 kilo ohm and approximated to 3.66. But, in the text it has been approximated to 3.65.\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The load resistance=3.66 kilo ohm.\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.8 : Page number 287\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "RL=1.0;                 #Load resistance,  kilo ohm\n",
      "A_v=40.0;               #Voltage gain, dB\n",
      "V_in=10.0;              #Input signal voltage, mV\n",
      "\n",
      "#Calcultaions\n",
      "#(i)\n",
      "#Since, A_v=20*log10(V_out/V_in)\n",
      "V_out=V_in*(10**(A_v/20))/1000;              #Output voltage, V\n",
      "\n",
      "#(ii)\n",
      "P_L=(V_out**2/RL);                            #The load power, mW\n",
      "\n",
      "\n",
      "#Results\n",
      "print(\"(i)The output voltage is %dV.\"%V_out);\n",
      "print(\"(ii)The load poweris %dmW.\"%P_L);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i)The output voltage is 1V.\n",
        "(ii)The load poweris 1mW.\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.9 : Page number 287-288\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from math import log10\n",
      "\n",
      "#Variable declaration\n",
      "P_2=40.0;                     #Output power, W\n",
      "R=10.0;                        #Resistance of speaker, ohm\n",
      "\n",
      "\n",
      "#Calculations\n",
      "#(i)\n",
      "A_p_dB=25.0;                                   #Power gain, dB\n",
      "#Since, A_p_dB=10*log10(P_2/P_1)\n",
      "P_1=(P_2/10**(A_p_dB/10))*1000;                #Input power, mW\n",
      "\n",
      "\n",
      "#(ii)\n",
      "A_v_dB=40.0;                                    #Voltage gain, dB\n",
      "\n",
      "#Since, P=(V**2)/R,\n",
      "V_2=(P_2*R)**0.5;                               #Output voltage, V\n",
      "\n",
      "#Since, A_v_dB=20*log10(V_2/V_1)\n",
      "V_1=(V_2/10**(A_v_dB/20))*1000;                #Input voltage, mV\n",
      "\n",
      "\n",
      "#Results\n",
      "\n",
      "print(\"(i)The input power=%.1fmW.\"%P_1);\n",
      "print(\"(ii)The input voltage=%dmV.\"%V_1);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i)The input power=126.5mW.\n",
        "(ii)The input voltage=200mV.\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.10 : Page  number 288\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "A_v_max=2000.0;                 #Maximum voltage gain\n",
      "f_max=2.0;                      #Frequency at which maximum voltage gain occurs,kHz\n",
      "A_v=1414.0;                     #Voltage gain at 50 Hz and 10kHz\n",
      "f1=50;                          #Lower frequency at which gain is 1414, Hz\n",
      "f2=10;                          #Upper frequency at which gain is 1414, kHz\n",
      "\n",
      "#Calculation\n",
      "#Since, bandwidth is the range of frequency  over which gain is greater than or equal to 70.7% of maximum gain\n",
      "if((A_v/A_v_max)*100 ==70.7):                               \n",
      "    print(\"(i)The bandwidth is from %dHz to %dkHz.\"%(f1,f2));\n",
      "    print(\"(ii)The lower cut-off frequency=%dHz.\"%f1);\n",
      "    print(\"(iii)The upper cut-off frequency=%dkHz.\"%f2);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i)The bandwidth is from 50Hz to 10kHz.\n",
        "(ii)The lower cut-off frequency=50Hz.\n",
        "(iii)The upper cut-off frequency=10kHz.\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.11 : Page number 291\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "A_v=60.0;               #Voltage gain of single stage amplifier\n",
      "R_C=500.0;              #Collector load, ohm\n",
      "R_in=1.0;               #Input impedance, kilo ohm\n",
      "\n",
      "\n",
      "#Calculation\n",
      "#Since, there is no loading , second stage gain remains at A_v\n",
      "#But, due to loading effect of input impedance of second stage, gain of first stage decreases\n",
      "A_v_2=A_v;                                                    #Voltage gain of second stage\n",
      "R_AC=round((R_C*R_in*1000)/(R_C+R_in*1000),0);              #Effective load of first stage, ohm\n",
      "A_v_1=A_v*R_AC/R_C;                                         #Gain of first stage\n",
      "A_v_T=A_v_1*A_v_2;                                            #Total gain\n",
      "\n",
      "\n",
      "#Results\n",
      "print(\"The total gain=%d.\"%A_v_T);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The total gain=2397.\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.12 : Page number 291-292\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "Rin=1.0;                        #Input resistance, kilo ohm\n",
      "beta=100.0;                     #base current amplification factor\n",
      "RC=2.0;                         #Collector load, kilo ohm\n",
      "\n",
      "\n",
      "#Calculation\n",
      "#(i)\n",
      "R_AC=(RC*Rin)/(RC+Rin);               #Effective load on first stage, kilo ohm\n",
      "A_v_1=round(beta*(R_AC/Rin),0);                #Voltage gain of first stage\n",
      "\n",
      "#(ii)\n",
      "A_v_2=round(beta*RC/Rin,0);                      #Voltage gain of second stage\n",
      "\n",
      "#(iii)\n",
      "A_v_T=A_v_1*A_v_2;                      #Total voltage gain\n",
      "\n",
      "\n",
      "#Results\n",
      "print(\"(i)The voltage gain of first stage =%d.\"%A_v_1);\n",
      "print(\"(ii)The voltage gain of second stage =%d.\"%A_v_2);\n",
      "print(\"(iii)The total voltage gain =%d.\"%A_v_T);\n",
      "\n",
      "#Note: The approximation inthe text for A_v_1=66.66 is taken as 66 but here it has been taken 67 and therefore the total voltage is 13400 instead of 13200.\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i)The voltage gain of first stage =67.\n",
        "(ii)The voltage gain of second stage =200.\n",
        "(iii)The total voltage gain =13400.\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.13 : Page number 292\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "RC=10.0;                    #Collector load of single stage amplifier, kilo ohm\n",
      "Rin=1.0;                    #Input resistance, kilo ohm\n",
      "beta=100.0;                 #base current amplification factor\n",
      "RL=100.0;                   #Load resistor, ohm\n",
      "\n",
      "\n",
      "#Calculations\n",
      "R_AC=round((RC*1000)*RL/(RC*1000+RL),-1);                 #Effective collector load,\n",
      "A_v=beta*R_AC/(Rin*1000);                              #Voltage gain\n",
      "\n",
      "#Results\n",
      "print(\"The voltage gain=%d.\"%A_v);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The voltage gain=10.\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.14 : Page number 292-293\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "VCC=20.0;                       #Collector suppply voltage, V\n",
      "R1=10.0;                        #resistor R1, kilo ohm\n",
      "R2=2.2;                         #resistor R2, kilo ohm\n",
      "R3=10.0;                        #resistor R3, kilo ohm\n",
      "R4=2.2;                         #resistor R4, kilo ohm\n",
      "RC_1=3.6;                         #Collector resistor of first stage, kilo ohm\n",
      "RC_2=4.0;                         #Collector resistor of second stage, kilo ohm\n",
      "RE_1=900.0;                     #Emitter resistor of first stage, ohm\n",
      "RE_2=1.0;                       #Emitter resistor of second stage, kilo ohm\n",
      "\n",
      "\n",
      "#Calculations\n",
      "#Biasing potential for the second stage is the voltage across R4 resistor,\n",
      "#so, by voltage divider rule:\n",
      "VB=VCC*R4/(R3+R4);                #Biasing potential for second stage,(Voltage across R4), V\n",
      "\n",
      "print(\"The biasing voltage for the second stage=%.1fV.\"%VB);\n",
      "\n",
      "#If coupling capacitor C_c is replaced by a wire, RC_1 and R3 become parallel\n",
      "Req=round((RC_1*R3)/(RC_1+R3),2);        #Equivalent resistance of R3 parallel with RC_1, kilo ohm\n",
      "VB=VCC*R4/(Req+R4);             #Biasing voltage if coupling capacitor is replaced by a wire, V\n",
      "\n",
      "print(\"The biasing voltage after replacing coupling capacitor by wire=%.2fV.\"%VB);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The biasing voltage for the second stage=3.6V.\n",
        "The biasing voltage after replacing coupling capacitor by wire=9.07V.\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.15 : Page number 293-294\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "#Function for calculating parallel resistance\n",
      "def pr(r1,r2):\n",
      "    return (r1*r2)/(r1+r2);\n",
      "\n",
      "#Variable declaration\n",
      "VCC=15.0;                           #Collector supply voltage, V\n",
      "R1=22.0;                            #Resistor R1, kilo ohm\n",
      "R2=3.3;                            #Resistor R2, kilo ohm\n",
      "R3=5.0;                            #Resistor R3, kilo ohm\n",
      "R4=1.0;                            #Resistor R4, kilo ohm\n",
      "R5=15.0;                            #Resistor R5, kilo ohm\n",
      "R6=2.5;                            #Resistor R6, kilo ohm\n",
      "R7=5.0;                            #Resistor R7, kilo ohm\n",
      "R8=1.0;                            #Resistor R8, kilo ohm\n",
      "beta=200;                           #Base current amplification factor\n",
      "RL=10.0;                            #Load resistor, kilo ohm\n",
      "V_BE=0.7;                           #Base-emitter voltage, V\n",
      "\n",
      "\n",
      "#Calculations\n",
      "#for 2nd stage\n",
      "V_R6=round(VCC*R6/(R5+R6),2);                #Voltage across R6, V (voltage divider rule)\n",
      "V_R8=round(V_R6-V_BE,2);                     #Voltage across R8, V\n",
      "IE_2=round(V_R8/R8,2);                         #Emitter current through R8, mA (OHM's LAW)\n",
      "re_2nd_stage=round(25/IE_2,1);                 #a.c emitter resistance for 2nd stage, ohm\n",
      "\n",
      "#For 1st stage\n",
      "V_R2=round(VCC*R2/(R1+R2),2);                     #Voltage across R2, V (voltage divider rule)\n",
      "V_R4=round(V_R2-V_BE,2);                          #Voltage across R4, V\n",
      "IE_1=round(V_R4/R4,2);                            #Emitter current through R4, mA (OHM's LAW)\n",
      "re_1st_stage=round(25/IE_1,1);                    #a.c emitter resistance for 1st stage, ohm\n",
      "\n",
      "#(i)\n",
      "Zin_base_2nd_stage=round((beta*re_2nd_stage)/1000,2);               #input resistance of transistor base of 2nd stage, kilo ohm\n",
      "Zin=round(pr(pr(R5,R6),Zin_base_2nd_stage),2);                      #Input impedance of the 2nd stage, kilo ohm\n",
      "R_AC_1st_stage=round(pr(R3,Zin),2);                                          #Effective collector load for 1st stage, kilo ohm\n",
      "A_v_1=round(R_AC_1st_stage*1000/re_1st_stage,0);                                  #voltage gain of 1st stage\n",
      "\n",
      "#(ii)\n",
      "R_AC_2nd_stage=round(pr(R7,RL),2);                          #Effective collector load for 2nd stage, kilo ohm\n",
      "A_v_2=round(R_AC_2nd_stage*1000/re_2nd_stage,1);                  #voltage gain of 2nd stage\n",
      "\n",
      "#(iii)\n",
      "A_v_overall=A_v_1*A_v_2;                            #overall voltage gain\n",
      "\n",
      "\n",
      "#results\n",
      "print(\"(i)The voltage gain of 1st stage=%.0f.\"%A_v_1);\n",
      "print(\"(i)The voltage gain of 2nd stage=%.1f.\"%A_v_2);\n",
      "print(\"(i)The overall voltage gain =%d.\"%A_v_overall);\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i)The voltage gain of 1st stage=53.\n",
        "(i)The voltage gain of 2nd stage=191.4.\n",
        "(i)The overall voltage gain =10144.\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.16 : Page number 297\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "Primary_impedance=1000.0;                       #Primary impedance, ohm\n",
      "Load_impedance=10.0;                            #Load impedance, ohm\n",
      "\n",
      "#Calculation\n",
      "#since,for maximum power transfer primary impedance should be equal to output impedance\n",
      "#and, impedance of secondary should be equal to load impedance\n",
      "#therfore, primary_impedance/load_impedance=square of(primary to secondary turn ratio)\n",
      "n=(Primary_impedance/Load_impedance)**0.5;           #Primary to secondary turn ratio\n",
      "\n",
      "\n",
      "#Result\n",
      "print('The primary to secondary turn ratio for maximum power transfer=%d.'%n);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The primary to secondary turn ratio for maximum power transfer=10.\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.17 : Page number 297\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "RL=16.0;                    #Load resistor, ohm\n",
      "R_p=10.0;                   #Output impedance of primary, kilo ohm\n",
      "Vp=10.0;                    #Terminal voltage of the source, V\n",
      "\n",
      "#Calculation\n",
      "#Since, for maximum power transfer, the impedance of the primary should be equal to output impedance of the source\n",
      "n=(R_p*1000/RL)**0.5;                        #Primary to secondary turns ratio\n",
      "\n",
      "#Since, power in a transformer remains constant,\n",
      "#ratio of primary to secondary voltageis equal to primary to secondary turns ratio\n",
      "Vs=Vp/n;                                #Voltage across the external load, V\n",
      "\n",
      "\n",
      "#Result\n",
      "print(\"The primary to secondary turns ratio=%d.\"%n);\n",
      "print(\"The voltage across the external load=%.1fV.\"%Vs);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The primary to secondary turns ratio=25.\n",
        "The voltage across the external load=0.4V.\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.18 : Page number 297-298\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "Rp=300.0;                   #D.C resistance of primary, ohm\n",
      "RL=3.0;                     #Load resistance, ohm\n",
      "R_out=3.0;                  #Ouput resistance of the transistor, kilo ohm               \n",
      "\n",
      "#Calculation\n",
      "#when no signal is applied, only Rp is seen to be the load.\n",
      "#But, when a.c signal is applied, RL in secondary reflects as RL*(squre of turns ratio).\n",
      "#Therefore, load is seen to be Rp in series with the reflected RL in primary.\n",
      "#i.e, R_out=Rp+(n**2 * RL), where n is the turns ratio\n",
      "n=((R_out*1000-Rp)/RL)**0.5;                     #turns ratio\n",
      "\n",
      "#Result\n",
      "print(\"Turns ratio for maximum power transfer=%d.\"%n);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Turns ratio for maximum power transfer=30.\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.19 : Page number 298"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import pi\n",
      "\n",
      "#Variable declaration\n",
      "f=200.0;                        #Frequency, Hz\n",
      "Z_out=10.0;                     #Output impedance of the transistor, kilo ohm\n",
      "Z_in=2.5;                       #Input impedance of the next stage, kilo ohm\n",
      "\n",
      "#Calculation\n",
      "#For perfect impedance matching,\n",
      "#Z_out should be equal to primary impedance\n",
      "#Z_out=2*pi*f*(primary inductance)\n",
      "Lp=(Z_out*1000)/(2*pi*f);                   #Primary inductance, H\n",
      "\n",
      "#for the secondary side,\n",
      "#Z_in should be equal to impedance of secondary\n",
      "Ls=(Z_in*1000)/(2*pi*f);                    #Secondary inductance, H\n",
      "\n",
      "\n",
      "#result\n",
      "print(\"The primary inductance=%.0fH.\"%Lp);\n",
      "print(\"The secondary inductance=%.0fH.\"%Ls);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The primary inductance=8H.\n",
        "The secondary inductance=2H.\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.20 : Page number 299\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "Lp=8.0;                     #Primary inductance, H\n",
      "Ls=2.0;                     #Secondary inductance, H\n",
      "K=10**-5;                   #Inductance to turns ratio, constant\n",
      "\n",
      "#Calculations\n",
      "Np=(Lp/K)**0.5;                  #Primary turns\n",
      "Ns=(Ls/K)**0.5;                  #Secondary turns\n",
      "\n",
      "#Result\n",
      "print(\"The primary turns=%.0f.\"%Np);\n",
      "print(\"The secondary turns=%.0f.\"%Ns);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The primary turns=894.\n",
        "The secondary turns=447.\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11.21 : Page number 300-301\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "#Variable declaration\n",
      "VCC=12.0;                   #Collector supply voltage, V\n",
      "R1=100.0;                   #Resistor R1, kilo ohm\n",
      "R2=22.0;                    #Resistor R2, kilo ohm\n",
      "R3=22.0;                    #Resistor R3, kilo ohm\n",
      "R4=4.7;                     #Resistor R4, kilo ohm\n",
      "R5=10.0;                    #Resistor R5, kilo ohm\n",
      "R6=10.0;                    #Resistor R6, kilo ohm\n",
      "beta=125;                   #Base current amplification factor\n",
      "V_BE=0.7;                   #Base-emitter voltage, V\n",
      "\n",
      "\n",
      "#Calculation\n",
      "#(i) D.C voltages\n",
      "#For 1st stage:\n",
      "V_B1=VCC*R2/(R1+R2);                    #Voltage at the base of 1st transistor, V (Voltage across R2, using voltage divider rule)\n",
      "V_E1=V_B1-V_BE;                         #Emitter voltage of the 1st transistor, V\n",
      "I_E1=round(V_E1/R4,2);                           #Emitter current of 1st transistor, mA (OHM's LAW)\n",
      "I_C1=I_E1;                              #Collector current of 1st transistor, mA(approximately equals to emitter current)\n",
      "V_C1=VCC-I_C1*R3;                       #Collector voltage of 1st transistor, V\n",
      "\n",
      "#For 2nd stage:\n",
      "V_B2=V_C1;                              #Voltage at the base of 2nd transistor, V (equals to collector voltage of 1st transistor)\n",
      "V_E2=V_C1-V_BE;                         #Emitter voltage of the 2nd transistor, V\n",
      "I_E2=V_E2/R6;                           #Emitter current of 2nd transistor, mA (OHM's LAW)\n",
      "I_C2=I_E2;                              #Collector current 2nd transistor, mA(approximately equals to emitter current)\n",
      "V_C2=VCC-I_C2*R5;                       #Collector voltage of 2nd transistor, V\n",
      "\n",
      "print(\"(i) D.C voltages\");\n",
      "print(\"First stage: VB1=%.2fV , VE1=%.2fV and VC1=%.2fV\"%(V_B1,V_E1,V_C1));\n",
      "print(\"First stage: VB2=%.2fV , VE2=%.2fV and VC2=%.2fV\"%(V_B2,V_E2,V_C2));\n",
      "\n",
      "#(ii)Voltage gain\n",
      "#First stage\n",
      "re_1=25/I_E1;                                   #a.c emitter resistance of 1st transistor, ohm\n",
      "re_2=25/I_E2;                                   #a.c emitter resistance of 2nd transistor, ohm\n",
      "Zin_2nd_stage=beta*re_2/1000;                   #Input impedance of 2nd stage, kilo ohm\n",
      "R_AC=R3*Zin_2nd_stage/(R3+Zin_2nd_stage);       #Total a.c collector load, kilo ohm\n",
      "A_v1=round(R_AC*1000/re_1,0);                                 #Voltage gain of first stage\n",
      "\n",
      "print(\"The voltage gain of first stage=%d.\"%A_v1);\n",
      "\n",
      "#Second stage\n",
      "R_AC=R5;                                        #Total a.c collector load for 2nd stage, kilo ohm(Due to no loading effect, equal to R5)\n",
      "A_v2=round(R5*1000/re_2,0);                                   #Voltage gain of 2nd stage\n",
      "\n",
      "print(\"The voltage gain of second stage=%d.\"%A_v2);\n",
      "\n",
      "A_vT=A_v1*A_v2;                                 #Overall voltage gain\n",
      "\n",
      "print(\"Overall voltage gain=%d.\"%A_vT);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i) D.C voltages\n",
        "First stage: VB1=2.16V , VE1=1.46V and VC1=5.18V\n",
        "First stage: VB2=5.18V , VE2=4.48V and VC2=7.52V\n",
        "The voltage gain of first stage=66.\n",
        "The voltage gain of second stage=179.\n",
        "Overall voltage gain=11814.\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}