{"id":23845,"date":"2020-02-22T10:51:25","date_gmt":"2020-02-22T05:21:25","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=23845"},"modified":"2021-07-02T15:24:40","modified_gmt":"2021-07-02T09:54:40","slug":"2nd-puc-electronics-question-bank-chapter-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-electronics-question-bank-chapter-3\/","title":{"rendered":"2nd PUC Electronics Question Bank Chapter 3 Transistor Amplifiers"},"content":{"rendered":"

Students can Download 2nd PUC Electronics Chapter 3 Transistor Amplifiers 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 3 Transistor Amplifiers<\/h2>\n

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

Question 1.
\nWhat is an amplifier?
\nAnswer:
\nAmplifier is a circuit which increases the strenght of weak input signal without any change in its shape.<\/p>\n

Question 2.
\nWhat is faithful amplification?
\nAnswer:
\nFaithful amplification is the process of increasing the strength of the signal without any change in its shape.<\/p>\n

Question 3.
\nWhat are the functions of input and output coupling capacitors?
\nAnswer:
\nThe coupling capacitors block dC and allow ac signal.<\/p>\n

Question 4.
\nWhat is the phase difference between input and output in CB amplifier?
\nAnswer:
\n0\u00b0<\/p>\n

\"2nd<\/p>\n

Question 5.
\nWhat is the phase difference between input and output in CE amplifier?
\nAnswer:
\n180o<\/sup><\/p>\n

Question 6.
\nWhat is the phase difference between input and output in CC amplifier?
\nAnswer:
\n0C<\/sup><\/p>\n

Question 7.
\nWhich of the three types of amplifiers is best suited for low input impedance and high output impedance?
\nAnswer:
\nCB amplifier.<\/p>\n

Question 8.
\nWhat is operating point?
\nAnswer:
\nThe zero signal collector current and collector to emitter voltage is called operating point.<\/p>\n

Question 9.
\nWhy does base collector junction get heated more than the emitter base junction in a common emitter amplifier?
\nAnswer:
\nThe collector current is more than base current and collector base junction has more resistance as it is reverse biased. Hence collector base junction gets heated more.<\/p>\n

Question 10.
\nWhich transistor amplifier produces maximum current gain, voltage gain and power gain?
\nAnswer:
\nCommon emitter amplifier.<\/p>\n

\"2nd<\/p>\n

Question 11.
\nWhy is a CC amplifier called as emitter follower?
\nAnswer:
\nCC amplifier is called emitter follower because the emitter voltage follows the base voltage.<\/p>\n

Question 12.
\nWhy does a transformer coupled amplifier give poor frequency response?
\nAnswer:
\nThe disproportionate amplification of frequencies in complete signals the reason for poor frequency response.<\/p>\n

Question 13.
\nWhy is transformer coupled amplifier used in the final stage of multistage amplifier?
\nAnswer:
\nFor impedance matching and maximum output.<\/p>\n

Question 14.
\nWhat is the use of a bypass capacitor in CE amplifier?
\nAnswer:
\nThe emitter by pass capacitor bypasses the alternating current at the emitter to the ground by provinding easy alternate path.<\/p>\n

Question 15.
\nWhat is the use of RE<\/sub> in CE amplifier?
\nAnswer:
\nThe resistor RE<\/sub> provides the stabilisation of the operating point.<\/p>\n

Question 16.
\nWhat happens to base current when emitter base junction is reverse biased in CB
\namplifier?
\nAnswer:
\nBase current decreases.<\/p>\n

Question 17.
\nWhat is buffer amplifier?
\nAnswer:
\nBuffer amplifier is an amplifier with unit gain which is used to deliver the maximum power to the load.<\/p>\n

Question 18.
\nWhat is meant by frequency response of an amplifier?
\nAnswer:
\nThe variation of the gain of an amplifier with the frequency of the input signal is called frequency response of the amplifier.<\/p>\n

\"2nd<\/p>\n

Question 19.
\nWhat is a power amplifier?
\nAnswer:
\nPower amplifiers are used to boost the power level of the input signal.<\/p>\n

Question 20.
\nWhat type of amplifier is used to drive the loud speaker system of a record player?
\nAnswer:
\nCC amplifier.<\/p>\n

Question 21.
\nHow do you classify the power amplifier based on the position of operating point?
\nAnswer:<\/p>\n

    \n
  • Class A<\/li>\n
  • Class B<\/li>\n
  • Class AB<\/li>\n
  • Class C<\/li>\n<\/ul>\n

    Question 22.
    \nWrite the expression for the efficiency of power amplifier?
    \nAnswer:
    \nEfficient,\\({ n }=\\frac { { { P }_{ out({ a }{ c }) }\\quad } }{ { P }_{ { in }({ d }{ c }) } }\\)<\/p>\n

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

    Question 1.
    \nMention the biasing conditions for normal operation of a transistor amplifier?
    \nAnswer:<\/p>\n

      \n
    • The operating point should be at the centre of load line.<\/li>\n
    • Collector current should be kept constant against temperature variations.<\/li>\n
    • The operating point should not change when transistor is replaced by another of same type.<\/li>\n<\/ul>\n

      Question 2.
      \nWhy do we prefer to express the gain in decibel?
      \nAnswer:<\/p>\n

        \n
      • Loudness heard by human ear is according to the logarithm of the intensity of sound.<\/li>\n
      • In multistage amplifiers, total gain in decibel is equal to the sum of the decibel gains of individual stages.<\/li>\n<\/ul>\n

        \"2nd<\/p>\n

        Question 3.
        \nA CC amplifier does not produce any voltage gain. Explain.
        \nAnswer:
        \nIn CC amplifier, voltage gain =
        \n\"2nd<\/p>\n

        Question 4.
        \nMention alternate names for CC, CB and CE amplifiers?
        \nAnswer:
        \nCC amplifier = Emitter follower, Buffer amplifier.
        \nCB amplifier = Radio frequency amplifier.
        \nCE amplifier = Audio frequency amplifier.<\/p>\n

        Question 5.
        \nHow CE amplifier can produce 180\u00b0 phase shift?
        \nAnswer:
        \nOutput voltage VCE<\/sub> = VCC <\/sub>– IC<\/sub>RC<\/sub>. During positive half cycle forward bias increases. Hence IC <\/sub>increases and as Icc<\/sub> is constant, VCE<\/sub>\u00a0decreases.<\/p>\n

        Question 6.
        \nWrite the steps involved in drawing AC equivalent circuit af an amplifier.
        \nAnswer:<\/p>\n

          \n
        • Replace DC sources by short circuit.<\/li>\n
        • Replace capacitors by short circuit.<\/li>\n<\/ul>\n

          Question 7.
          \nDraw AC equivalent circuit diagram of a single stage CE amplifier.
          \nAnswer:
          \n\"2nd<\/p>\n

          Question 8.
          \nDraw DC equivalent circuit diagram of single stage CE amplifier.
          \nAnswer:
          \n\"2nd<\/p>\n

          Question 9.
          \nWrite the steps involved in drawing DC equivalent circuit of an amplifier.
          \nAnswer:<\/p>\n

            \n
          • Reduce all the AC sources to zero.<\/li>\n
          • open all the capacitors.<\/li>\n<\/ul>\n

            \"2nd<\/p>\n

            Question 10.
            \nWhy do we consider frequencies above 70.7% of the maximum gain for bandwidth?
            \nAnswer:
            \nIf the gain falls to 70.7% of maximum gain, the ear can not detect the change in the intensity of sound. If the gain falls below 70.7%, the ear will hear distortion. Hence the range of frequencies above 70.7% of the maximum gain is called as bandwidth.<\/p>\n

            Question 11.
            \nWhy do we avoid RC and transformer coupling for very high and very low frequencies?
            \nAnswer:
            \nSince reactance (XL<\/sub> and Xc<\/sub>) depend on frequency, RC and transformer coupling is avoided for very high and low frequencies.<\/p>\n

            Question 12.
            \nExplain the functions of the coupling and bypass capacitors?
            \nAnswer:
            \nCoupling capacitors are used to block dc and allow only ac to pass through. The emitter bypass capacitor bypasses the alternating current at the emitter to the ground by providing easy alternate path.<\/p>\n

            Question 13.
            \nDraw the frequency response of CE amplifier.
            \nAnswer:
            \n\"2nd<\/p>\n

            Question 14.
            \nMention two characteristics of a CC amplifier.
            \nAnswer:<\/p>\n

              \n
            • Voltage gain is nearly equal to 1.<\/li>\n
            • Input and output signals are in phase.<\/li>\n<\/ul>\n

              Question 15.
              \nWhy does the gain of an amplifier decrease at very low and very high frequencies?
              \nAnswer:
              \nThe capacitive reactance of coupling and by pass capacitors is large at low frequencies. As this reactance with is in series with source resistance, the input signal is attenuated and the gain is less.<\/p>\n

              \"2nd<\/p>\n

              At high frequencies, the reactance offered by junction capacitance of transistor and wiring capacitance is less. As this is in paralled with output resistance, gain decreases with increase in frequency.<\/p>\n

              Question 16.
              \nDraw the circuit of a CB amplifier?
              \nAnswer:
              \n\"2nd<\/p>\n

              Question 17.
              \nWhy docs a transformer-coupled amplifier produce humming noise?
              \nAnswer:
              \nMagnetostriction (extension and contraction of magnetic sheet steel when magnetised) is the reason for noise in transformer-coupled amplifier.<\/p>\n

              Question 18.
              \nWhat are the disadvantages of a direct-coupled amplifier?
              \nAnswer:<\/p>\n

                \n
              • High frequency signals cannot be amplified.<\/li>\n
              • It has poor temperature stability.<\/li>\n<\/ul>\n

                Question 19.
                \nWhat are the advantages of an RC coupled amplifier?
                \nAnswer:<\/p>\n

                  \n
                • Distortion is minimum.<\/li>\n
                • Overall gain is high.<\/li>\n
                • It has good frequency response.<\/li>\n
                • It costs less.<\/li>\n<\/ul>\n

                  Question 20.
                  \nWrite any two applications of CC amplifier?
                  \nAnswer:<\/p>\n

                    \n
                  • CC amplifier is used as buffer amplifier to deliver maximum power to the load.<\/li>\n
                  • It is used as current amplifier<\/li>\n<\/ul>\n

                    Question 21.
                    \nWhat is the necessity of cascading?
                    \nAnswer:
                    \nCascading of amplifiers is required to get high overall gain.<\/p>\n

                    \"2nd<\/p>\n

                    Question 22.
                    \nWhat are the functions of power amplifier?
                    \nAnswer:
                    \nPower amplifiers drive the loud speakers in audio systems, radio and TV receivers.<\/p>\n

                    Question 23.
                    \nWrite a note on cross over distortion in a power amplifier?
                    \nAnswer:
                    \nThe transistor does not conduct until the input signal voltage exceeds 0.7V for silicon and 3V for Germanium. Due to this, there is time interval between positive and negative cycles of input signal. The distortion when neither of transistors is conducting in push pull amplifier is called cross over distortion.<\/p>\n

                    Question 24.
                    \nDistinguish between power amplifier and voltage amplifier?
                    \nAnswer:
                    \nPower amplifiers are used to increase the power level of the input signal. Voltage amplifiers are the amplifiers that amplify small input signals whose amplitudes are very much smaller than DC operating point value.<\/p>\n

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

                    Question 1.
                    \nWith a circuit diagram, explain how a transistor can amplify a signal OR Explain the principle of amplification.
                    \nAnswer:
                    \n\"2nd
                    \nDuring positive half cycle of ac input signal, the forward bias on the base emitter junction increases. Hence iB<\/sub> and ic<\/sub> increase and voltage drop across Rc<\/sub> increases. As VCE<\/sub> = Vcc<\/sub> – ic<\/sub>Rc<\/sub> we get negative half cycle amplified at the output.<\/p>\n

                    During negative half cycle of ac input signal, forward bias on the base emitter junction decreases. Due to this iB<\/sub> and ic<\/sub> decrease and voltage drop across Rc<\/sub> decreases.
                    \nAS VCE<\/sub> = Vcc<\/sub> – ic<\/sub>Rc<\/sub> We get amplified positive half cycle at the output.<\/p>\n

                    \"2nd<\/p>\n

                    Question 2.
                    \nWhat is cascaded amplifier? Why is it necessary?
                    \nAnswer:
                    \nCascaded amplifier is number of amplifiers connected in series. Cascading of amplifiers is required to get high gain.<\/p>\n

                    Question 3.
                    \nWith circuit diagram, explain working of CB amplifier?
                    \nAnswer:
                    \n\"2nd
                    \nThe voltage divider biasing is used biasing CB amplifier. Emitter base junction is forward biased by voltage across R, and the collector base junction is reverse biased by the voltage across Rr<\/sub> Coupling capacitors block DC and allow AC to pass through. By pass capacitor CB <\/sub>by passes AC.<\/p>\n

                    During positive half cycle of input signal, forward bias on the base emitter junction decreases as it is reverse biased,Hence Ic<\/sub> and IC<\/sub>RC<\/sub> is decreased.The output voltage VCB<\/sub> is increased because VCB<\/sub> = Vcc<\/sub> – IC<\/sub>RC<\/sub>. Thus positive half cycle is obtained at the output.<\/p>\n

                    During negative half cycle of the input signal, forward bias on base emitter junction is increased. Hence IE<\/sub>, and Ic<\/sub> increase and IC<\/sub>,RC<\/sub> also increases. As VCB<\/sub> = Vcc<\/sub> – Ic<\/sub>Rr<\/sub>, VCB<\/sub> decreases as LRr <\/sub>increases. Thus negative half cycle of output is obtained. The input and output signals are in phase.<\/p>\n

                    Question 4.
                    \nWith circuit diagram, explain the working of CE amplifier?
                    \nAnswer:
                    \nThe ac signal is applied between base and emitter terminals and output is taken at the collector. R1<\/sub>, R2<\/sub> provide biasing to the transistor. R1<\/sub> provides stabilisation and CF<\/sub> provides easy alternate path for the ac to flow, coupling capacitors C1<\/sub> and C2<\/sub> block DC and allow ac signals.
                    \n\"2nd<\/p>\n

                    During the positive half cycle of input signal, forward bias emitter base junction increases. Hence base current IB<\/sub>, IE<\/sub> and Ic<\/sub> increase. This makes IC<\/sub>RC<\/sub> increase. Output voltage VCP<\/sub> is given by VCE<\/sub> = Vcc<\/sub> – IC<\/sub>RC<\/sub>. As IC<\/sub>RC<\/sub> increases, VCE<\/sub> decreases and negative half cycle at the output is obtained.<\/p>\n

                    During the negative half cycle of ac input signal, forward bias on emitter base junction
                    \ndecreases. Hence IB<\/sub>, Ic<\/sub> and IE<\/sub> decrease, decreasing IC<\/sub>RC<\/sub>.<\/p>\n

                    The output voltage VCE<\/sub> = Vcc<\/sub> – IC<\/sub>RC<\/sub>. As IC<\/sub>RC<\/sub> decreases, VCE<\/sub> increases and we get positive half cycle at the output. Thus there is a phase difference of 180\u00b0 between the input and voltage.<\/p>\n

                    \"2nd<\/p>\n

                    Question 5.
                    \nMention types of coupling of amplifier with block diagrams.
                    \nAnswer:
                    \n\"2nd<\/p>\n

                    Question 6.
                    \nCompare different types of coupling of amplifiers?
                    \nAnswer:<\/p>\n\n\n\n\n\n\n\n\n
                    Parameter<\/td>\nRC coupling<\/td>\nTransformer coupling<\/td>\nDirect coupling<\/td>\n<\/tr>\n
                    Frequency
                    \nResponse<\/td>\n                    		Excellent<\/td>\nPoor<\/td>\nBest<\/td>\n<\/tr>\n
                    Cost<\/td>\nLess<\/td>\nmore<\/td>\nleast<\/td>\n<\/tr>\n
                    Weight<\/td>\nLess<\/td>\nmore<\/td>\nleast<\/td>\n<\/tr>\n
                    Impedance
                    \nmatching<\/td>\n                    		Not good<\/td>\nExcellent<\/td>\nGood<\/td>\n<\/tr>\n
                    uses<\/td>\nFor voltage amplification<\/td>\nFor power amplification<\/td>\nFor amplifying very low frequencies.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                    Question 7.
                    \nDefine Class A, Class B, Class C power amplifiers.
                    \nAnswer:<\/p>\n

                      \n
                    • Class A power amplifier: The Q point is located exactly at the centre of the load line. The current flows for full cycle of input signal.<\/li>\n
                    • Class B power amplifier: The Q point is located in between X-axis and the centre of load line. The output current flows for more than half cycle and less than full cycle of the input signal.<\/li>\n
                    • Class C power amplifier: The Q point is located beyond cut off region below the X-axis. The output current flows for less than half cycle of input signal.<\/li>\n<\/ul>\n

                      Question 8.
                      \nWhich of the power amplifiers amongst Class A, Class B or Class C produce maximum distortion? Which is more suitable for faithful amplification? Where is Q point fixed in the case of class AB amplifier?
                      \nAnswer:
                      \nClass A is more suitable for faithful amplification. Class C power amplifier produces maximum. distortion. In Class AB amplifier, Q point is located in between X-axis and the centre of load line.<\/span><\/p>\n

                      Question 9.
                      \nWhich of the Class A, Class B or Class C power amplifiers have maximum efficiency? Mention approximate power efficiency of these amplifiers.
                      \nAnswer:
                      \nClass C power amplifier has maximum efficiency.<\/p>\n\n\n\n\n\n\n\n
                      Class<\/strong><\/td>\nEfficiency<\/strong><\/td>\n<\/tr>\n
                      A<\/td>\n25 to 30%<\/td>\n<\/tr>\n
                      B<\/td>\n70 to 80%<\/td>\n<\/tr>\n
                      C<\/td>\nHigher than 80%<\/td>\n<\/tr>\n
                      AB<\/td>\n50-70%<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                      Question 10.
                      \nWrite one application each for CB, CC, and CE amplifiers.
                      \nAnswer:<\/p>\n\n\n\n\n\n\n
                      Amplifier<\/strong><\/td>\nApplication<\/strong><\/td>\n<\/tr>\n
                      CB<\/td>\nRF amplification<\/td>\n<\/tr>\n
                      CE<\/td>\nAF amplification<\/td>\n<\/tr>\n
                      CC<\/td>\nImpedance matching<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                      \"2nd<\/p>\n

                      Question 11.
                      \nExplain the working of a CC amplifier?
                      \nAnswer:
                      \nThe circuit of a single stage common collector amplifier using npn transistor is shown below:
                      \n\"2nd<\/p>\n

                      The ac is applied between base and collector terminals collector circuit. The collector of the transistor is at ac ground as Rc<\/sub> = 0 Vcc<\/sub>, R1<\/sub>,R2<\/sub>,RE<\/sub> provide voltage divider DC bias with constant operating point in the active region. The output is taken across the emitter resistor RE<\/sub>. C, and C2<\/sub> are the coupling Capacitors which block DC and allow ac signal.<\/p>\n

                      During positive half cycle of input signal, forward bias on emitter base junction increases. Hence the base current emitter current and IE<\/sub>RE<\/sub> increases which is output voltage. Hence we get positive half cycle at the output.<\/p>\n

                      During negative half cycle, forward bias between emitter base junction decreases. Hence, IB<\/sub>, IE<\/sub> and Ie<\/sub>Re<\/sub> decrease which is the output voltage. Thus negative half cycle is obtained. The input and out signals are in phase. CC amplifier is also called emitter follower because emitter voltage follows the base voltage.
                      \nVo<\/sub>= IE<\/sub>RE<\/sub> = Vin<\/sub> -VBP
                      \n<\/sub>Vo <\/sub>= Vin<\/sub> -VBP\u00a0 <\/sub>is very small
                      \n\u2234 Vo <\/sub>\u2248Vin<\/sub>
                      \nVoltage gain
                      \n\\(A_{V}=\\frac{V_{o}}{V_{i n}}=1\\)
                      \nCC amplifier is used as impedance matching device.<\/p>\n

                      Question 12.
                      \nExplain the working of direct coupled amplifier.
                      \nAnswer:
                      \n\"2nd
                      \nThe direct coupled amplifier using two npn transistors is as shown in the diagram. In this method, output of first stage is directly coupled to the input of second stage. Resistors R1<\/sub> and R2<\/sub> form voltage divider circuit. Any signal at the base of the first transistor is amplified A, times and appears at the collector of first stage.<\/p>\n

                      \"2nd<\/p>\n

                      It is then applied to the base of transistor in the second stage. Hence it is amplified A2<\/sub> times. The gain of the direct coupled amplifier is A = A x A . Direct-coupled amplifiers are used to amplify extremely low frequency signals.
                      \n\"2nd
                      \nFrequency Response of Direct coupled transistor amplifier.<\/p>\n

                      Question 13.
                      \nExplain the working of RC coupled amplifier.
                      \nAnswer:
                      \n\"2nd
                      \nRC coupling is most widely used method of coupling signal from one stage to next stage in multistage amplifiers. The two stage RC coupled amplifier is as shown in the diagram. The signal developed across the collector resistor Rc<\/sub> of the first stage is capacitively coupled to the input of second stage through the coupling capacitor Cc<\/sub>.<\/p>\n

                      RC coupled transistor amplifier is widely used as it provides excellent audio fidelity over wide range of frequency.<\/p>\n

                      R1<\/sub>R2<\/sub>,Rc<\/sub>,RE<\/sub> provide biasing resistors. RE<\/sub> provides stabilisation and CE<\/sub> is used for bypassing ac across Rg.<\/p>\n

                      Working:<\/p>\n

                      When an AC signal is applied to the input of first stage, it gets amplified and appears at its output with a phase reversal of 180\u00b0. Through the coupling capacitor Cc<\/sub>, the amplified signal is fed to the input of second stage.<\/p>\n

                      Cc<\/sub> blocks DC and allows AC to pass through. The second stage further amplifies the signal, further providing a phase reversal of 180\u00b0 with respect to its input. The input and output signal are in phase as the signal is reversed twice. Overall gain of RC coupled amplifier is the product of gain of individual stages.
                      \n\"2nd
                      \nAt low frequencies, reactance of coupling capacitor is high. Hence it allows only a small part of the signal to pass from one stage to next stage. Hence gain is low at low frequencies. At high frequencies, the coupling capacitor Cc<\/sub> offers low reactance and acts like a short. Hence loading effect of next stage increases and the gain decreases.
                      \nIn mid frequency range, as frequency increases, the capacitive reactance of coupling capacitor decreases, which tries to increase gain. As capacitive reactance decreases, due to loading effect of next stage, the gain reduces. Hence, voltage gain is constant.<\/p>\n

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

                      Question 1.
                      \nClassify amplifiers on different criteria.
                      \nAnswer:<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
                      Criteria<\/strong><\/td>\nClassification.<\/strong><\/td>\n<\/tr>\n
                      1. Input signal<\/td>\ni. Small signal amplifier, large amplifier<\/td>\n<\/tr>\n
                      2. Nature of output<\/td>\ni. Voltage amplifier.
                      \nii. Power amplifier.<\/td>\n<\/tr>\n
                      3. Frequency range<\/td>\ni.\u00a0\u00a0\u00a0\u00a0 AF amplifier
                      \nii.\u00a0\u00a0 RF amplifier
                      \niii.\u00a0 IF amplifier
                      \niv. VHF amplifier
                      \nv.\u00a0\u00a0 UHF amplifier
                      \nvi.\u00a0 SHF amplifier<\/td>\n<\/tr>\n
                      4. Operating point<\/td>\nClass A Class B Class C Class AB<\/td>\n<\/tr>\n
                      5. Configuration.<\/td>\nCB amplifier CE amplifier CC amplifier<\/td>\n<\/tr>\n
                      6. Coupling.<\/td>\nRC coupled Direct coupled Transformer coupled<\/td>\n<\/tr>\n
                      7. Stages.<\/td>\nSingle stage. Multistage<\/td>\n<\/tr>\n
                      8. Bandwidth<\/td>\nWideband amplifiers. Narrow band amplifiers.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                      \"2nd<\/p>\n

                      Question 2.
                      \nCompare on the CB, CE and CC amplifiers.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
                      Parameter<\/strong><\/td>\nTypes of Amplifiers.<\/strong><\/td>\n<\/td>\n<\/td>\n<\/tr>\n
                      <\/td>\nCE<\/td>\nCB<\/td>\nCC<\/td>\n<\/tr>\n
                      Current gain<\/td>\nHigh (P)<\/td>\nLess than 1<\/td>\nHighest (1+ p)<\/td>\n<\/tr>\n
                      Voltage gain<\/td>\nHigh<\/td>\nModerate<\/td>\nLow<\/td>\n<\/tr>\n
                      Power gain<\/td>\nHighest<\/td>\nModerate<\/td>\nModerate<\/td>\n<\/tr>\n
                      Phase shift<\/td>\n180\u00b0<\/td>\n0\u00b0<\/td>\n0-<\/td>\n<\/tr>\n
                      Input impedance<\/td>\nModerate<\/td>\nLow<\/td>\nHigh<\/td>\n<\/tr>\n
                      Output impedance<\/td>\nModerate<\/td>\nHigh<\/td>\nLow<\/td>\n<\/tr>\n
                      Bandwidth<\/td>\nNarrow<\/td>\nWide<\/td>\nWide<\/td>\n<\/tr>\n
                      Applications<\/td>\nUsed as AF amplifies<\/td>\nUsed as RF amplifies<\/td>\nimpedance
                      \nmatching<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                      Question 3.
                      \nGive the graphical representation of amplification for small signal by choosing Q point near to cut off, active and saturation region.
                      \nAnswer:
                      \n1. When the Q point is at the centre of the load line:
                      \n\"2nd<\/p>\n

                      2. When Q point is near to saturation.
                      \n\"2nd<\/p>\n

                      3. When Q point is nearer to cut-off.
                      \n\"2nd<\/p>\n

                      Question 4.
                      \nDerive the expressions for the current gain, voltage gain, input impedance, output impedance and power gain for CE amplifier with re<\/sub>1<\/sup> model.
                      \nAnswer:
                      \nAC equivalent circuit of CE amplifier using re<\/sub>1<\/sup>\u00a0model,
                      \n\"2nd
                      \nExpression for current gain:
                      \nCurrent gain is the ratio of output current to input current.
                      \nCurrent gain, = \\(A_{t}=\\frac{1}{i_{s}}=\\beta\\)<\/p>\n

                      \"2nd<\/p>\n

                      Expression for input impedance:
                      \nInput impedance is the ratio of input voltage to the input current.
                      \n\\(Z_{i n}=\\frac{V_{i n}}{I_{i n}}\\)<\/p>\n

                      This is total input impedance includes biasing resistors and impedance looking into base of the transistor.<\/p>\n

                      Zin <\/sub>= R1 <\/sub>|| R2<\/sub> || Zin <\/sub>(base)<\/p>\n

                      Impedance looking directly into the base is
                      \n\"2nd<\/p>\n

                      Expression for Output impedance: Output impedance is the ratio of output voltage to output current.\u00a0 Output impedance \\(Z_{o}=\\frac{V_{o}}{i_{o}}=\\frac{R_{C} i_{C}}{i_{C}}=R_{C}\\)<\/p>\n

                      If RL<\/sub> is the load resistance, then output impedance is equal to Zo<\/sub> = Rc<\/sub> || RL<\/sub><\/p>\n

                      Expression for voltage gain:
                      \nVoltage gain of an amplifier is the ratio of output voltage to input voltage.
                      \n\"2nd<\/p>\n

                      Expression for the power gain:
                      \nPower gain of an amplifier is the ratio of output power to input power.
                      \n\\(\\mathrm{A}_{P}=\\frac{\\mathrm{P}_{o}}{\\mathrm{P}_{i}}\\)
                      \nPower gain is also equal to the product of voltage gain and current gain.<\/p>\n

                      Question 5.
                      \nDraw the circuit diagram and explain with graph the frequency response of CE amplifier.
                      \nAnswer:
                      \nAt low frequencies, coupling and by pass capacitors offer large reactance. As this reactance is in series with source resistance, it attenuates the input signal. As a result, gain decreases with decrease in frequency.
                      \n\"2nd<\/p>\n

                      \"2nd<\/p>\n

                      At higher frequencies, the reactance offered by junction capacitance of the transistor and wiring capacitance becomes very low. As this is in parallel with output resistance, the gain decreases with increase in frequency.<\/p>\n

                      In mid-frequency range, the resistance offered by the coupling and by pass capacitors become small and negligible. Also reactance offered by junction and wiring capacitance is quite large and negligible. Hence the gain remains a constant.<\/p>\n

                      Question 6.
                      \nWrite a note on applications of CB, CE, CC amplifiers?
                      \nAnswer:
                      \nCB amplifier: It is used to match low input impedance circuit with the high impedance load. As it is suitable for amplifying high frequencies, it is used as voltage amplifier in RF circuits. CE amplifier: It is used in applications where high voltage gain and power gain are required. Its important applications are as<\/p>\n

                        \n
                      • RF amplifier.<\/li>\n
                      • AF amplifier.<\/li>\n
                      • Signal generator.<\/li>\n<\/ul>\n

                        CC amplifier: It is mainly used as buffer amplifier to match high impedance circuit with Sow impedance load for maximum power transmission. It is used as power amplifier and current amplifier.<\/p>\n

                        \"2nd<\/p>\n

                        Question 7.
                        \nDraw the circuit diagram and explain the working of CE amplier. Also mention any two applications.
                        \nAnswer:
                        \n\"2nd
                        \nThe power supply Vcc<\/sub>, resistors R1<\/sub>, R2<\/sub>, Rc<\/sub> and RE<\/sub> provide required DC bias for the transistor, C1<\/sub> and C2<\/sub> are blocking capacitors which allow ac and block dc. The bypass capacitor CE<\/sub> bypass es ac at the emitter to ground.<\/p>\n

                        During positive half cycle at the input signal, forward bias on the emitter base junction increases. Hence the currents ib and ic increase. As a result, ic R increases and V decreases because VCE<\/sub> = VCC<\/sub> -icRc
                        \n<\/sub>
                        \nThus we get negative half cycle at the output. During negative half cycle of input signal, forward bias on the emitter junction decreases. Hence ib, ic decrease and icRc also decreases. As VCB<\/sub> = VCC <\/sub>– icRc<\/sub> , VCE<\/sub>\u00a0increases.
                        \nThus, we get positive half cycle at the output.
                        \nThus a phase shift of 180\u00b0 exists between the input and the output.
                        \nApplications: It is used as AF amplifier, RF amplifier and signal generator.<\/p>\n

                        Question 8.
                        \nDraw the circuit diagram and explain the working of a CB amplifier. Also mention any two applications.
                        \nAnswer:
                        \n\"2nd<\/p>\n

                        The weak signal to be amplified is applied to emitter-base junction and output signal is taken across collector-base junction.<\/p>\n

                        The DC power supply, R1<\/sub>, R2<\/sub>, RE<\/sub>, Rc<\/sub> provide the required biasing for the transistor. Capacitor C1<\/sub>, C2<\/sub> block dc and allow ac to pass through. Capacitor CB by passes ac to the ground. During positive half cycle of input signal, the forward bias between base-emitter junction decreases, as VBE<\/sub> is already negative with respect to the ground.<\/p>\n

                        Hence iE<\/sub> and ic<\/sub>. decrease and ic<\/sub> Rc<\/sub> also decreases.
                        \nVCB <\/sub>=Vc<\/sub>c<\/sub> – ic <\/sub>Rc\u00a0 <\/sub>. As ic <\/sub>Rc <\/sub>decreases, VCB<\/sub>increases. Hence we get amplified positive half cycle. During negative half cycle of the input signal, forward bias is increased because VBE<\/sub>is negative w.r.t. the ground.<\/p>\n

                        Hence iC , <\/sub>iE <\/sub>are increased and ic <\/sub>Rc<\/sub> is increased. As VCB <\/sub>=Vc<\/sub>c<\/sub> – ic <\/sub>Rc\u00a0 <\/sub>\u00a0when ic <\/sub>Rc <\/sub>increases, output voltage VCB<\/sub> decreases.<\/p>\n

                        Thus, the positive going input signal produces positive going output signal and negative going input signal produces negative going output signal. Hence in CB amplifier, the input and output signals are in phase.<\/p>\n

                        \"2nd<\/p>\n

                        Question 9.
                        \nDraw circuit diagram and explain the working of CC amplifier. Also mention any two applications.
                        \nAnswer:
                        \n\"2nd<\/p>\n

                        The stable operating point in the active region is obtained by R1<\/sub> R2<\/sub>, RE<\/sub>. The input is applied between base and emitter and the output is taken across the emitter resistance RE <\/sub>During the positive half cycle of the input signal, forward bias on emitter base junction increases.<\/p>\n

                        Hence iB<\/sub>, iE<\/sub> increase and output voltage iE<\/sub>RE<\/sub> increases. Thus, during positive half cycle of the input, we get positive half cycle at the output.\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

                        During the negative half cycle of the input signal, iB<\/sub> and iE<\/sub> decrease and hence output voltage iE<\/sub>RE<\/sub> decreases. Thus output and the input are in phase and the voltage gain is 1 as input signal is equal to output signal.<\/p>\n

                        CC amplifiers are used in<\/p>\n

                          \n
                        • Power amplifiers<\/li>\n
                        • Current amplifiers<\/li>\n
                        • Impedance matching devices.<\/li>\n<\/ul>\n

                          \"2nd<\/p>\n

                          Question 10.
                          \nExplain with the circuit diagram, the working of RC coupled CE amplifier. Also explain the frequency response.
                          \nAnswer:
                          \n\"2nd
                          \nThe circuit diagram shows the two stage RC Coupled transistor amplifier using voltage divider bias in CE configuration. The connection to second stage from the first stage is done through the coupling capacitor Cc<\/sub>.C1<\/sub> couples input signal to first stage and C2<\/sub> couples the signal from first stage to the second stage.<\/p>\n

                          R1<\/sub>,R2<\/sub> provide the required biasing and RE<\/sub> provides the stabilization. CE<\/sub> bypasses ac across the resistor Rc<\/sub>.<\/p>\n

                          When the ac input signal is applied to the input of the first stage, it gets amplified and appears at its output with a phase difference of 180\u00b0 This amplified signal is fed to the input of the second stage through the coupling capacitor C2<\/sub>. The second stage further amplifies the signal and produces a phase reversal of 180\u00b0 with respect to its input signal. The signal is inverted twice and the input and output of two stage CE amplifier are in phase.<\/p>\n

                          Frequency response:
                          \n\"2nd
                          \nAt low frequencies, the coupling capacitor Cc<\/sub> offers high reactance Xc<\/sub>. Hence it allows only a small part of the signal to pass from one stage to the next stage. Thus gain is less at low frequencies.<\/p>\n

                          At high frequencies, the coupling capacitor Cc offers a low reactance and acts like a switch. Hence , due to loading effect of the next stage increases and the gain decreases.<\/p>\n

                          In mid-frequency range, the effect of coupling capacitor is such that it maintains a constant voltage gain. When the frequency increases, the capacitive reactance Cc<\/sub> decreases which tries to increases the gain. As the capacitive reactance decreases, the loading effect of next stage reduces the gain. The voltage gain remains a constant as the two factors cancel each other.<\/p>\n

                          \"2nd<\/p>\n

                          Question 12.
                          \nWith input and output waveforms, explain class-A, Class B, Class C and Class AB power amplifiers.
                          \nAnswer:
                          \nThe four types of power amplifiers are:<\/p>\n

                          i. Class A: In class A operation of an amplifier, the Q point is located exactly at the centre of linear amplifying region of the characteristic. The current in the output circuit flows for full cycle of the input signal.
                          \n\"2nd<\/p>\n

                          ii. Class B: In class B operation, the Q point is located at the cut off region. The output current will flow only for half the cycle of the input signal.
                          \n\"2nd<\/p>\n

                          iii. Class AB: Q point is located between X-axis and the centre of the load line. The output current flow for more than half cycle and less than full cycle of the input signal.
                          \n\"2nd<\/p>\n

                          iv. Class C: The Q point is located beyond the cut-off region, below the X-axis. The output current flows for less than half of input signal.<\/p>\n

                          Question 13.
                          \nWith a circuit diagram, explain the working of class B push pull amplifier.
                          \nAnswer:
                          \n\"2nd
                          \nClass B push-pull amplifier is as shown in the diagram. T1 <\/sub>and T2<\/sub> are two centre tapped transformers and Q1<\/sub>, Q2<\/sub> are two identical transistors. Transformer T1<\/sub> produces signal voltages V1<\/sub> and V2<\/sub> which are 180\u00b0 out of phase with each other.<\/p>\n

                          These two signals are applied to the Polarities reverse during negative half cycle of input voltage. Q2<\/sub>then is ON and Q1<\/sub> is OFF. Q2<\/sub>amplifies the signal and the alternate half cycle appears across the loud speaker.<\/p>\n

                          two transistors. Transformer T2<\/sub> couples AC output signal from collector to loud speaker. The two emitters are connected to centre tap of transformer T1<\/sub> secondary and Vcc<\/sub> to the centre tap of T2<\/sub> secondary.<\/p>\n

                          During positive half cycle of input voltage, secondary winding of T1<\/sub> has voltages V, and V2<\/sub>. Transistor Q1<\/sub>, conducts and Q2<\/sub> is cut off. The collector current through Q1<\/sub> produces an amplified and inverted voltage which applied to loud speaker through a transformer.<\/p>\n

                          \"2nd<\/p>\n

                          Question 14.
                          \nExplain with the circuit diagram, the action of class C collector tuned power amplifier
                          \nAnswer:
                          \n\"2nd
                          \nIt has parallel tuned circuit as collector load. The value of capacitance and inductance of tuned collector are selected such that it resonates with the same frequency, which is equal to frequency to be amplified.<\/p>\n

                          The high frequency signal to be amplified is given to the input of the amplifier. The resonant frequency of parallel tuned circuit is made equal to frequency of signal by changing value of capacitor. Then the tuned circuit offers very high impedance. If the input signal has several frequencies, amplifier will select and amplify those frequencies which are close to resonant frequency. In this way, the amplifier selects and amplifies the desired frequency.<\/p>\n

                          Question 15.
                          \nCompare four types of power amplifiers.
                          \nAnswer:<\/p>\n\n\n\n\n\n\n\n
                          Class<\/strong><\/td>\nA<\/strong><\/td>\nB<\/strong><\/td>\nC<\/strong><\/td>\nAB<\/strong><\/td>\n<\/tr>\n
                          Conduction
                          \nangle<\/td>\n                          		360\u00b0<\/td>\n180\u00b0<\/td>\nLess than 90\u00b0<\/td>\n180\u00b0 to 360\u00b0<\/td>\n<\/tr>\n
                          Position of Q point<\/td>\nCentre of load line.<\/td>\non x-axis (cut off)<\/td>\nBelow x-axis<\/td>\nBetween x-axis and centre of load line<\/td>\n<\/tr>\n
                          Overall
                          \nefficiency<\/td>\n                          		25%<\/td>\n70-80%<\/td>\nHigher than 80%<\/td>\n50-70%<\/td>\n<\/tr>\n
                          Signal
                          \ndistortion<\/td>\n                          		None<\/td>\nAt x-axis cross over point<\/td>\nLarge
                          \nAmounts<\/td>\n                          		Small amounts.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                          Problems with Solutions<\/span><\/p>\n

                          Question 1.
                          \nThe microphone converts audio frequency sound into electrical signal of 5mV and is connected to a CB amplifier having input impedance of 25\u03a9 and output impedance of 500 kQ and load resistance 5k\u03a9. Find the output voltage.
                          \nAnswer:
                          \n\"2nd
                          \n\\(\\mathrm{I}_{i}=\\frac{\\mathrm{V}_{i}}{\\mathrm{Z}_{i}}=\\frac{5}{25 \\times 10^{3}}=0.2 \\mathrm{mA}\\)
                          \nV0<\/sub> = I0<\/sub> R0<\/sub>
                          \n= I0<\/sub> R0 <\/sub>(\u2235 In CB amplifier, Ii<\/sub> = Ii<\/sub>)
                          \n= 0.2 x 10-3<\/sup> x 5 x 103
                          \n<\/sup>= 1V.<\/p>\n

                          \"2nd<\/p>\n

                          Question 2.
                          \nCalculate the input impedance of CE amplifier given IE<\/sub>=4.5mA, \u03b2 = 180 for a Si transistor.
                          \nAnswer:
                          \nFor Si transistor, re1
                          \n\"2nd<\/sup>
                          \nInput impedance,\u03b2 re1<\/sup> = 180 x 5.77
                          \n= 1038.6Q
                          \n= 1.038k\u00a3L<\/p>\n

                          Question 3.
                          \nFind the bandwidth and gain at 3dB line for an amplifier having mid-band gain of 600 haveing 3 dB frequencies 100Hz and 300kHz.
                          \nAnswer:
                          \nBandwidth = f2<\/sub> = f1
                          \n<\/sub>= 300kHz – 100
                          \n= 3,00,000 – 100
                          \n= 299900 Hz
                          \n=299.9kHz
                          \nGain at 3dB \\(=\\frac{A_{m}}{\\sqrt{2}}=\\frac{600}{\\sqrt{2}}\\)
                          \n= 424.32<\/p>\n

                          Question 4.
                          \nIn an amplifier, maximum voltage gain is 2000 for signal frequency of 2 It falls to 1414 at 10KHz and 50Hz. Find bandwidth, lower cut off frequency and upper cut off\u00a0 frequency.
                          \nAnswer:
                          \nBandwidth = f2<\/sub> = f1
                          \n<\/sub>= 10kHz – 50
                          \n= 10,000-50
                          \n= 9950Hz
                          \n= 9.95kHz.
                          \nf2<\/sub> = 50Hz,
                          \nf1 <\/sub>= 10kHz.<\/p>\n

                          Question 5.
                          \nThe voltage gain of two stage amplifier is 70 If the input voltage is 5|iV, find the output voltage.
                          \nAnswer:
                          \n\"2nd<\/p>\n

                          \"2nd<\/p>\n

                          Question 6.
                          \nThe DC emitter current of transistor is 3 Calculate the AC resistance of emitter diode of silicon transistor.
                          \nAnswer:
                          \n\"2nd<\/p>\n

                          Question 7.
                          \nA three stage amplifier has a first stage voltage gain of 100, second stage voltage gain of
                          \nAnswer:
                          \nA1 = 100, A2<\/sub> = 200, A3<\/sub> = 400.
                          \nATotal<\/sub> A1<\/sub> x A2<\/sub> x A3
                          \n<\/sub>= 100 X 200 X 400 = 8 x 106
                          \n<\/sup>Gain in dB = 20 log Atm
                          \n= 20 log 8 x 106<\/sup>
                          \n= 20 x (6.9031)
                          \n= 138.02dB.<\/p>\n

                          Question 8.
                          \n200 and third stage voltage gain of 400. Find the total voltage gain in dB.\u00a0 CE amplifier circuit using germaniuntransistor is as shown in the figure. Calculate
                          \ni. The voltage across 5k\u03a9
                          \nii. IE<\/sub>
                          \niii. Zin (base) <\/sub>
                          \niv. Zin<\/sub>
                          \nv. Z0<\/sub> Given re<\/span><\/sub>1<\/sup> = \\(\\frac{52 m V}{I_{E}}\\), \u03b2 = 100.
                          \nAnswer:
                          \n\"2nd<\/p>\n

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

                          \"2nd<\/p>\n

                          Question 9.
                          \nFor CE amplifier circuit using silicon transistor, find
                          \ni. r’e<\/sub> ii. Av<\/sub> iii. V0<\/sub> IV. A. V. Ap. VI. Gp<\/sub> given IE<\/sub> = 1.043mA and \u03b2 = 100.
                          \nAnswer:
                          \n\"2nd
                          \n\"2nd
                          \nV0<\/sub> = Av<\/sub> Vin
                          \n<\/sub>= 83.28 x 10 x 10-3
                          \n<\/sup>= o.8328Volt.
                          \nAi<\/sub>= 0 = 100.
                          \nAp<\/sub> = Ai Av
                          \n= 100 x 83.28 = 8328.
                          \nGp = 10 log Ap
                          \n= 10 log 8328
                          \n= 10 x 3.9205
                          \n= 39.05dB.<\/p>\n

                          Question 10.
                          \nFor CE amplifier circuit using Germanium transistor, given below, find 1<\/sup>
                          \ni. Voltage across 10k\u03a9
                          \nii. IE<\/sub>
                          \niii. r3\u00a0<\/sup>e
                          \n<\/sub> IV. Av
                          \n<\/sub>Answer:
                          \n\"2nd
                          \n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

                          Students can Download 2nd PUC Electronics Chapter 3 Transistor Amplifiers 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 3 Transistor Amplifiers 2nd PUC Electronics Transistor Amplifiers 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\/23845"}],"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=23845"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/23845\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=23845"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=23845"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=23845"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}