{"id":23757,"date":"2020-02-21T12:54:51","date_gmt":"2020-02-21T07:24:51","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=23757"},"modified":"2021-07-02T15:24:44","modified_gmt":"2021-07-02T09:54:44","slug":"2nd-puc-electronics-question-bank-chapter-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-electronics-question-bank-chapter-2\/","title":{"rendered":"2nd PUC Electronics Question Bank Chapter 2 Transistor Biasing"},"content":{"rendered":"

Students can Download 2nd PUC Electronics Chapter 2 Transistor Biasing Questions and Answers, Notes Pdf, 2nd PUC Electronics Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Electronics Question Bank Chapter 2 Transistor Biasing<\/h2>\n

2nd PUC Electronics Transistor Biasing One Mark Questions and Answers<\/h3>\n

Question 1.
\nWhat is biasing of a transistor?
\nAnswer:
\nTranistor biasing is the application of suitable DC voltages across the transistor terminals.<\/p>\n

Question 2.
\nWhy is the transistor called as current controlled device?
\nAnswer:
\nThe output current of a transistor depends upon the input current. Hence, the transistor is called as current controlled device.<\/p>\n

Question 3.
\nWhat is an unbiased transistor?
\nAnswer:
\nA transistor with its terminals not connected to any source is called an unbiased transistor.<\/p>\n

\"2nd<\/p>\n

Question 4.
\nDefine quiescent point.
\nAnswer:
\nThe zero signal collector current and collector to emitter voltage is called quiescent point.<\/p>\n

Question 5.
\nWhat is meant by dc load line?
\nAnswer:
\nDc load line is a straight line drawn on the output characteristics representing output DC voltage and corresponding direct current under no signal conditions.<\/p>\n

Question 6.
\nWhat is thermal runaway?
\nAnswer:
\nThe self destruction of an unbiased transistor due to increase in temparature and hence increase in leakage current is called thermal runaway.<\/p>\n

Question 7.
\nDefine stabilisation of a transistor.
\nAnswer:
\nStabilisation of a transistor is the process of making. Q point independent of transistor parameters and temparature variations.<\/p>\n

Question 8.
\nMention any one factor on which stabilisation depends.
\nAnswer:
\nstabilisation depends upon \u03b2.<\/p>\n

Question 9.
\nName any one parameter of the transistor which is temparature dependent.
\nAnswer:
\nLeakage current<\/p>\n

Question 10.
\nWhat is leakage current?
\nAnswer:
\nLeakage current is the current through a device due to the motion of minority charge carriers under reverse biased condition.<\/p>\n

\"2nd<\/p>\n

Question 11.
\nMention any one leakage current in a transistor.
\nAnswer:
\nCollector to emitter current with the base open ICEO<\/sub>.<\/p>\n

Question 12.
\nName anyone biasing circuit.
\nAnswer:
\nFixed bias.<\/p>\n

Question 13.
\nName the biasing circuit which gives excellent stabilisation.
\nAnswer:
\nVoltage divider biasing circuit.<\/p>\n

Question 14.
\nWhat is a heat sink?
\nAnswer:
\nHeat sink is a device which absorbs the unnecessary heat generated in the transistor and radiates it to surroundings and protects it.<\/p>\n

Question 15.
\nWrite the relation between ICE0<\/sub> and ICB0<\/sub>.
\nAnswer:
\nICEO<\/sub>\u00a0= (I+p)ICBO<\/sub>.<\/p>\n

2nd PUC Electronics Transistor Biasing Two Marks and Three Marks Questions and Answers<\/h3>\n

Question 1.
\nExplain the biasing of a transistor.
\nAnswer:
\nThe application of suitable DC voltage across the transistor terminals is called biasing of the transistor.<\/p>\n

The reasons for biasing of a transistor are:<\/p>\n

    \n
  1. To make the operating point to be at the centre of the load line.<\/li>\n
  2. To make the collector current independent of temperature variations.<\/li>\n
  3. To maintain the operating point constant even though transistor is replaced by another of same type.<\/li>\n<\/ol>\n

    \"2nd<\/p>\n

    Question 2.
    \nDraw the DC load line and mark the endpoints.
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 3.
    \nMention any two biasing circuits.
    \nAnswer:<\/p>\n

      \n
    • Fixed bias<\/li>\n
    • Collector to base feedback.<\/li>\n<\/ul>\n

      Question 4.
      \nWhat are the quiescent point and load line?
      \nAnswer:
      \nOperating point is the zero signal values of collector Ic<\/sub> and collector to emitter voltage VCE <\/sub>of a transistor. DC load line is a straight line drawn on the output characteristics representing output DC voltage and corresponding direct current under no signal conditions.<\/p>\n

      Question 5.
      \nWhy is the voltage divider bias circuit preferred in the amplification process?
      \nAnswer:
      \nThe operating point does not depend upon the value of \u03b2 of the transistor. Hence the operating point does not change its position due to rise in temperature or replacing a transistor of different p value. Hence voltage divider bias provides excellent stabilisation and is preferred to other biasing methods.<\/p>\n

      Question 6.
      \nWhat is leakage current? Mention different types of leakage currents.
      \nAnswer:
      \nLeakage current is current through a device due to the motion of minority charge carriers under reverse bias conditions.<\/p>\n

      Different leakage currents are:<\/p>\n

        \n
      • Collector to base leakage current with emitter open (ICBO<\/sub>).<\/li>\n
      • Collector to emitter leakage current with base open (1CEO<\/sub>).<\/li>\n<\/ul>\n

        \"2nd<\/p>\n

        Question 7.
        \nWhat are advantages of voltage divider bias?
        \nAnswer:<\/p>\n

          \n
        • Q point gets stabilised very well as the stabilication factor does not
          \ndepend upon p.<\/li>\n
        • It provides better amplification when used in amplifiers.<\/li>\n<\/ul>\n

          Question 8.
          \nWhat is the significance of load line?
          \nAnswer:
          \nLoad line is a straight line drawn on output characteristics representing output DC voltage and corresponding output direct current under no signal condition. Load line contains every possible operating point for the circuit.<\/p>\n

          Question 9.
          \nExplain the collector to base leakage current and collector to emitter leakage current.
          \nAnswer:
          \n1. The current flowing through the collector when the emitter terminal is open is called collector to base leakage current with emitter open ICB0<\/sub>. This leakage current doubles for every 10\u00b0c rise in temperature. This leakage current also increases when the reverse bias voltage is increased.<\/p>\n

          2. The current flowing through the collector when base terminal is open is called collector to emitter leakage current with base open.This current increases with increase in temperature. This leakage current increases when the reverse bias voltage is increased.<\/p>\n

          Question 10.
          \nWhat is thermal runaway? Explain.
          \nAnswer:
          \nThe self destruction of an unbiased transistor due to increase in temperature and hense increase in leakage current.<\/p>\n

          \"2nd\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ‘<\/p>\n

          Question 11.
          \nOn what factors does stabilisation of a transistor amplifier depend upon?
          \nAnswer:
          \nThe value of p and temperature.<\/p>\n

          Question 12.
          \nWhat is meant by heat sink? Mention its importance.
          \nAnswer:
          \nHeat sink is a device which absorbs unwanted heat generated in the transistor and radiates to the surroundings. A copper conductor with different shapes are connected to the device to take out heat from the device. With heat sink, power rating of the device increases.<\/p>\n

          Question 13.
          \nWhat is unbiased transistor? Explain the need for biasing the transistor.
          \nAnswer:
          \nUnbiased transistor is a transistor with its terminals not connected to any source. Biasing a transistor is applying a suitable DC voltage across the transistor terminals to operate the transistor in the desired region.<\/p>\n

          Need for biasing a transistor are:<\/p>\n

            \n
          • To make operating point to be at the centre of load line for faithful amplification.<\/li>\n
          • To make collector current independent of temperature variations.<\/li>\n
          • To make operating point stabilised even when one transistor is replaced by another.<\/li>\n<\/ul>\n

            Question 14.
            \nDescribe the steps to draw dc load line for a transistor.
            \nAnswer:
            \nConsider the equation VCE<\/sub> = Vcc<\/sub>-Ic<\/sub> x Rc<\/sub>)
            \nTo plot the point on x-axis, put I =0 in equ (1)
            \nVCE<\/sub> = Vcc
            \n<\/sub> To plot the point on y-axis, put VCE<\/sub>=0, in equ (1)
            \neqn(1) = ICsat<\/sub> \\(=\\frac{V_{C C}}{R_{C}}\\)<\/p>\n

            The load lines is drawn joining these end points
            \n\"2nd<\/p>\n

            Question 15.
            \nWhat is voltage divider bias? Mention its advantages.
            \nAnswer:
            \nVoltage divider bias is the method of biasing a transistor. The voltage drop across forword biases the base emitter junction. This causes the base current and hence collector current to flow in zero signal condition and establishes operating point.<\/p>\n

            Advantages:<\/p>\n

              \n
            • It offers excellent stabilisation and hence Q point does not shift.<\/li>\n
            • when used in amplifiers, it provides better amplification.<\/li>\n<\/ul>\n

              \"2nd<\/p>\n

              Question 16.
              \nExplain the steps required to draw the dc load line for voltage divider bias.
              \nAnswer:
              \n\"2nd
              \nApplying KVL to the circuit,
              \nVcc<\/sub>= Ic<\/sub>Re +<\/sub> Vce +<\/sub> Ic<\/sub>Re
              \n<\/sub>Vcc<\/sub> =Ic<\/sub>(Rc<\/sub>+Re<\/sub>)+ Vce<\/sub>.
              \nVcc<\/sub>=Vcc <\/sub>– Ic<\/sub>(Rc<\/sub>+Re<\/sub>) ……………….. (1)
              \nTo plot the point on x axis, put Ic<\/sub> =0.
              \nEqu (1) becomes,\u00a0 VCE<\/sub> = VCC
              \n<\/sub>To plot the point on y axis, put VCE<\/sub>=0.
              \nEqu (1) becomes,
              \n\\(\\mathrm{I}_{c}=\\frac{\\mathrm{V}_{c c}}{\\mathrm{R}_{c}+\\mathrm{R}_{\\varepsilon}}\\)
              \nDC load line is drawn from the end points and is plotted as shown below:
              \n\"2nd<\/p>\n

              Question 17.
              \nName the different biasing circuits. Which biasing circuit is better recommeded to get faithful amplification?
              \nAnswer:
              \nDifferent biasing circuits are:<\/p>\n

                \n
              • Fixed bias<\/li>\n
              • collector to base bias.<\/li>\n
              • Emitter feedback bias.<\/li>\n
              • Voltage divider bias.
                \nvoltage divider biasing circuit is the best method to get faithful amplification.<\/li>\n<\/ul>\n

                2nd PUC Electronics Transistor Biasing Five Marks Questions and Answers<\/h3>\n

                Question 1.
                \nDerive the expressions for the Q point in voltage divider bias.
                \nAnswer:
                \n\"2nd<\/p>\n

                Voltage across R2<\/sub> is
                \n\\(\\mathrm{V}_{2}=\\frac{\\mathrm{V}_{C C} \\mathrm{R}_{2}}{\\mathrm{R}_{1}+\\mathrm{R}_{2}}\\)
                \nApplying KVL to the base circuit , We get
                \nV2 <\/sub>= VBE <\/sub>+ V2
                \n<\/sub>V2\u00a0<\/sub>= VBE <\/sub>+ IE <\/sub>RE
                \n<\/sub>\"2nd
                \n<\/sub>Equations 1 and 2 define operating point or Q point.<\/p>\n

                \"2nd<\/p>\n

                Question 2.
                \nHow does stabilisation takes place in voltage divider bias? Explain
                \nAnswer:
                \n\"2nd
                \nFrom voltage divider theorem, voltage across the resistor R2<\/sub> is \\(\\mathrm{Y}=\\mathrm{A} \\overline{\\mathrm{C}} \\overline{\\mathrm{D}}+\\overline{\\mathrm{B}} \\mathrm{D}+\\overline{\\mathrm{A}} \\mathrm{D}+\\mathrm{BC} \\overline{\\mathrm{D}}\\)
                \nApply KVL to base emitter loop
                \nV2<\/sub> =VBE +<\/sub> VE
                \n<\/sub>V2<\/sub> = VBE<\/sub> = IE<\/sub> RE<\/sub>
                \nVBE<\/sub> \u00a0= V2<\/sub> – IE<\/sub> RE<\/sub><\/p>\n

                Whenever Ic<\/sub> increases, due to increase in \u03b2 or temperature, the current IE<\/sub> also increases (IE<\/sub>=IC<\/sub>). Hence IE<\/sub> RE <\/sub>increases and VBE<\/sub> \u00a0decreases due to which base current IB <\/sub>decreases. Decrease in L decreases Ic<\/sub> and thus the operating point gets stabilised.<\/p>\n

                Question 3.
                \nFind the end points of the load line for the circuit given below:
                \nAnswer:
                \n\"2nd
                \nApplying KVL to collector circuit, We get,
                \nVcc =Ic<\/sub>Rc + Vce
                \n16 =Ic <\/sub>x 4 x 103<\/sup>+Vce<\/sub> …………. (1)
                \nPut Ic<\/sub> = 0 in equation (1), we get,
                \nVCE<\/sub> ( cut off)
                \n= Vcc
                \n= 16 V
                \nThis is cut off point on the x-axis.
                \nPut VCE<\/sub> = 0 in equation (1), we get,
                \n16 = IC x 4 x 103<\/sup>+ 0, IC = 16 x 10-3<\/sup> \/4 = 4mA.
                \nThis is saturation point on y axis
                \n\"2nd<\/p>\n

                Question\u00a0 4.
                \nApplying KVL to collector circuit, we have
                \nAnswer:
                \n\"2nd<\/p>\n

                Applying KVL to collector circuit, We get,
                \nVcc<\/sub> = ic<\/sub>Re<\/sub>+VCE
                \n<\/sub>40 = Ic <\/sub>x 10 x 103<\/sup>+Vce <\/sub>…………(1)
                \nPut Ic<\/sub> =0,
                \n40 = VCE<\/sub> (cut off)
                \nThis is cut off point on x-axis.
                \n\"2nd<\/p>\n

                \"2nd<\/p>\n

                Question 5.
                \nDraw the load line for the following circuit:
                \nAnswer:
                \n\"2nd
                \n\"2nd<\/p>\n

                Question 6.
                \nFind the operating point of the circuit given below:
                \nAnswer:
                \n\"2nd
                \n\"2nd
                \n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

                Students can Download 2nd PUC Electronics Chapter 2 Transistor Biasing Questions and Answers, Notes Pdf, 2nd PUC Electronics Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Electronics Question Bank Chapter 2 Transistor Biasing 2nd PUC Electronics Transistor Biasing One Mark …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/23757"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=23757"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/23757\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=23757"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=23757"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=23757"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}