{"id":23965,"date":"2020-02-22T14:50:02","date_gmt":"2020-02-22T09:20:02","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=23965"},"modified":"2021-07-02T15:24:39","modified_gmt":"2021-07-02T09:54:39","slug":"2nd-puc-electronics-question-bank-chapter-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-electronics-question-bank-chapter-4\/","title":{"rendered":"2nd PUC Electronics Question Bank Chapter 4 Feedback in Amplifiers"},"content":{"rendered":"

Students can Download 2nd PUC Electronics Chapter 4 Feedback in Amplifiers Questions and Answers, Notes Pdf, 2nd PUC Electronics Question Bank with A nswers<\/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 4 Feedback in Amplifiers<\/h2>\n

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

Question 1.
\nWhat is feedback?
\nAnswer:
\nFeedback is the process of taking a part of output signal and feeding it back to input circuit.<\/p>\n

Question 2.
\nMention the types of feedback.
\nAnswer:<\/p>\n

    \n
  • Positive feedback.<\/li>\n
  • Negative feedback.<\/li>\n<\/ul>\n

    Question 3.
    \nDefine positive feedback.
    \nAnswer:
    \nPositive feedback is applying the feedback signal in phase with applied input.<\/p>\n

    Question 4.
    \nDefine negative feedback.
    \nAnswer:
    \nNegative feedback is applying the feedback signal 180\u00b0 out of phase with the applied input.<\/p>\n

    \"2nd<\/p>\n

    Question 5.
    \nWhat happens to input impedance of an amplifier when negative feedback is applied? Answer:
    \nInput impedance increases.<\/p>\n

    Question 6.
    \nWhat happens to voltage gain of an amplifier when negative feedback is applied?
    \nAnswer:
    \nWith negative feedback, gain decreases.<\/p>\n

    Question 7.
    \nWhat is meant by loop gain?
    \nAnswer:
    \nThe product of open loop gain and feedback ratio is called the loop gain.<\/p>\n

    Question 8.
    \nWhat is meant by closed loop gain?
    \nAnswer:
    \nClosed loop gain is the gain of an amplifier with negative feedback.<\/p>\n

    Question 9.
    \nDefine feedback factor?
    \nAnswer:
    \nFeedback factor is the ratio of feedback voltage to output voltage.<\/p>\n

    Question 10.
    \nName the type of feedback that is preferred for an amplifier?
    \nAnswer:
    \nNegative feedback.<\/p>\n

    \"2nd<\/p>\n

    Question 11.
    \nWhich type of feedback is required in an oscillator?
    \nAnswer:
    \nPositive feedback.<\/p>\n

    Question 12.
    \nWhat is the effect of negative feedback on the bandwidth of an amplifier?
    \nAnswer:
    \nBandwidth increases.<\/p>\n

    Question 13.
    \nWhat is the effect of negative feedback on stability of an amplifier?
    \nAnswer:
    \nStability increases.<\/p>\n

    Question 14.
    \nWhich type of negative feedback decreases both input and output impedances of an amplifier?
    \nAnswer:
    \nVoltage shunt negative feedback.<\/p>\n

    Question 15.
    \nWhich type of negative feedback increases both input and output impedances of an amplifier?
    \nAnswer:
    \nCurrent series negative feedback.<\/p>\n

    Question 16.
    \nName the type of feedback used in oscillator.
    \nAnswer:
    \nPositive feedback.<\/p>\n

    \"2nd<\/p>\n

    Question 17.
    \nWhat is net input to basic amplifier with positive feedback?
    \nAnswer:
    \nVi = Vs<\/sub> + VF<\/sub><\/p>\n

    Question 18.
    \nWhat is the net input to basic amplifier with negative feedback?
    \nAnswer:
    \nVi = Vs<\/sub>-VF<\/sub><\/p>\n

    Question 19.
    \nHow is the gain bandwidth product of an amplifier?
    \nAnswer:
    \nGain bandwidth of an amplifier is always a constant.<\/p>\n

    Question 20.
    \nFind output voltage if feedback fraction is 0.2 and feedback voltage is 0.02V?
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 21.
    \nMention four types of negative feedback. Ans. 1. Voltage series negative feedback.
    \nAnswer:<\/p>\n

      \n
    • Voltage shunt negative feedback.<\/li>\n
    • Current series negative feedback.<\/li>\n
    • Current shunt negative feedback.<\/li>\n<\/ul>\n

      \"2nd<\/p>\n

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

      Question 1.
      \nDistinguish between open loop and closed loop gain.
      \nAnswer:
      \nOpen loop gain of an amplifier is gain of the amplifier without feedback. Closed loop gain of an amplifier is gain of the amplifier with feedback.<\/p>\n

      Question 2.
      \nWhat are advantages of negative feedback.
      \nAnswer:<\/p>\n

        \n
      • Stabilisation of the gain.<\/li>\n
      • Increase in input impedance.<\/li>\n
      • Decrease in output impedance.<\/li>\n
      • Increase in bandwidth.<\/li>\n
      • Reduction in noise and distortion.<\/li>\n<\/ul>\n

        Question 3.
        \nMention the disadvantages of positive feedback.
        \nAnswer:<\/p>\n

          \n
        • Poor gain stability<\/li>\n
        • More distortion and noise<\/li>\n
        • Poor frequency response.<\/li>\n<\/ul>\n

          Question 4.
          \nExplain the effect of positive and negative feedback on the gain of an amplifier.
          \nAnswer:
          \nWith positive feedback, gain of an amplifier increases and with negative feedback, gain an amplifier decreases.<\/p>\n

          Question 5.
          \nDraw the block diagrams of voltage shunt and current series negative feedback.
          \nAnswer:
          \n\"2nd
          \nCurrent series negative feedback.<\/p>\n

          Question 6.
          \nWhy is positive feedback seldom used in amplifier?
          \nAnswer:
          \nPositive feedback is seldom used in amplifiers because the positive feedback increases distortion and noise, decreases the gain stability and results in poor frequency response.<\/p>\n

          \"2nd<\/p>\n

          Question 7.
          \nName the type of feedback in an amplifier and in an oscillator.
          \nAnswer:<\/p>\n

            \n
          1. Negative feedback is used in an amplifier.<\/li>\n
          2. Positive feedback is used in an oscillator.<\/li>\n<\/ol>\n

            Question 8.
            \nWhich type of negative feedback increases the input impedance and decreases output impedance of an amplifier?
            \nAnswer:
            \nVoltage series negative feedback.<\/p>\n

            Question 9.
            \nAn amplifier has a gain of 900. When negative feedback is applied, the gain reduces to 250. Find the feedback fraction.
            \nAnswer:
            \n\"2nd
            \n\"2nd<\/p>\n

            Question 10.
            \nAn amplifier has a bandwidth of300 KHz and voltage gain of 100. Calculate its bandwidth if 10% negative feedback is introduced.
            \nAnswer:
            \nBW\/ = (1 + A\u03b2)BW
            \n= (1 +100 x 0.1) x 300kHz
            \n= 3,300kHz<\/p>\n

            \"2nd<\/p>\n

            Question 11.
            \nIf the bandwidth of an amplifier is 20 kHe<\/sub> and open-loop gain is 200 and \u03b2 = 0.1, find
            \nthe gain of the amplifier with negative feedback.
            \nAnswer:
            \nA= 200,
            \n\u03b2 = 0.1,
            \nBW = 20 kHz .
            \nGain with feedback is BWF
            \n<\/sub>= (1+A\u03b2) BW
            \n= (1+200 x 0.1) x 20 kHz
            \n= 21 x 20 kHz
            \n= 420 kHz\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\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 12.
            \nFind voltage of an amplifier in dB scale if its voltage gain is 10s<\/sup>.
            \nAnswer:
            \nGain in dB = 20 log Av
            \n<\/sub>= 20 log 105<\/sup>
            \n= 20 x 5 log10<\/sub> 10
            \n= 100 x 1 = 100 dB.<\/p>\n

            Question 13.
            \nFind the power gain of an amplifier whose power gain in 50dB.
            \nAnswer:
            \nGain in dB = 10 log10<\/sub> A
            \n50 = 10 log10<\/sub>Ap<\/sub>
            \n5 = log10 <\/sub>Ap<\/sub>
            \nlog10 <\/sub>Ap<\/sub> = 5
            \nAp<\/sub> = 105<\/sup>
            \n\u2234 loga<\/sub> N = x =) ax<\/sup> = N.<\/p>\n

            Question 14.
            \nWrite the expressions for the input voltage
            \n(i) with positive feedback and
            \n(ii) with negative feedback voltage.
            \nAnswer:
            \n(i) V = V + Vf
            \n<\/sub>(ii)Y = Vs <\/sub>– Vf<\/sub><\/p>\n

            \"2nd<\/p>\n

            Question 15.
            \nMention the four types of negative feedback.
            \nAnswer:<\/p>\n

              \n
            • Voltage series negative feedback.<\/li>\n
            • Voltage shunt negative feedback.<\/li>\n
            • Current series negative feedback.<\/li>\n
            • Current shunt negative feedback.<\/li>\n<\/ul>\n

              Question 16.
              \nWrite the expressions for
              \n(i) open loop gain
              \n(ii) Closed loop gain
              \nAnswer:
              \ni. Open loop gain, \\(A=\\frac{V_{o}}{V_{i}}\\)
              \nii. Closed loop gain, \\(\\mathrm{A}_{f}=\\frac{\\mathrm{V}_{o}}{\\mathrm{V}_{s}}\\)<\/p>\n

              Question 17.
              \nDraw block diagrams to show
              \n(i) current shunt negative feedback and
              \n(ii) voltage series negative feedback.
              \nAnswer:
              \n(i) Current shunt negative feedback.
              \n\"2nd<\/p>\n

              (ii) Voltage series negative feedback.
              \n\"2nd<\/p>\n

              Question 18.
              \nDistinguish between positive feedback and negative feedback.
              \nAnswer:<\/p>\n

                \n
              • Positive feedback is the process of applying a portion of output in phase with the input.<\/li>\n
              • Negative feedback is the process of applying a portion of output out of phase with the input.<\/li>\n<\/ul>\n

                Question 19.
                \nExplain how stability is achieved in negative feedback amplifier.
                \nAnswer:
                \nFluctuation in gain with negative feedback is:
                \n\"2nd
                \nFor negative feedback, 1+A \u03b2> 1 and hence \\(A \\oplus B=A \\bar{B}+\\bar{A} B\\) . With negative feedback, fluctuation in the gain decreases and the stability is achieved<\/p>\n

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

                Question 1.
                \nExplain the terms feedback factor, loop gain and closed loop gain.
                \nAnswer:<\/p>\n

                  \n
                • Feedack factor is the ratio of feedback voltage to the output voltage.<\/li>\n
                • Loop gain of an amplifier is the product of gain and feedback fraction.<\/li>\n
                • Closed loop gain is the gain of an amplifier with negative feedback.<\/li>\n<\/ul>\n

                  \"2nd<\/p>\n

                  Question 2.
                  \nDraw the frequency response curves of an amplifier with and without negative feedback.
                  \nAnswer:
                  \n\"2nd<\/p>\n

                  Question 3.
                  \nRelate the bandwidth of an amplifier with and without feedback and comment on the gain-bandwidth product of an amplifier.
                  \nAnswer:
                  \nBWf <\/sub>= BW(1 + A\u03b2)
                  \nThe gain-bandwidth product of an amplifier is always a constant ie B W x A = BWf<\/sub> x Af<\/sub><\/p>\n

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

                  Question 1.
                  \nDerive the expression for the gain of an amplifier with negative feedback.
                  \nAnswer:
                  \nThe open loop gain of an amplifier is
                  \n\"2nd
                  \n\"2nd<\/p>\n

                  Question 2.
                  \nExplain the effect of negative feedback on the stability of the gain.
                  \nAnswer:
                  \nThe voltage gain of an amplifier with negative feedback is
                  \n\"2nd
                  \nAs 1+A\u03b2 >1, the percentage change in Af<\/sub> is less than percentage change in A.<\/p>\n

                  Question 3.
                  \nDraw the block diagrams for the four types of negative feedback.
                  \nAnswer:
                  \n1.Voltage series negative feedback.
                  \n\"2nd<\/p>\n

                  2. Voltage shunt negative feedback:
                  \n\"2nd<\/p>\n

                  3.Current series negative feedback:
                  \n\"2nd<\/p>\n

                  4. Current shunt negative feedback:
                  \n\"2nd<\/p>\n

                  \"2nd<\/p>\n

                  Question 4.
                  \nDerive an expression for the input impedance of an amplifier with negative feedback.
                  \nAnswer:
                  \nAn amplifier should have high input impedance so that it will not load the preceding stage or input voltage source. High input impedance for an amplifier can be achieved with Voltage series negative feedback.<\/p>\n

                  Let Ii and Vi be input current and input voltage to the basic amplifier. Vs is input to feedback amplifier, Vf is feedback voltage and vo is the output voltage.<\/p>\n

                  Input impedance of basic amplifier is
                  \n\\(z_{i}=\\frac{V_{i}}{I_{i}}-(1)\\)<\/p>\n

                  Input impedance of feedback amplifier is
                  \n\\(z_{v}=\\frac{V_{s}}{I_{i}} \\rightarrow(2)\\)<\/p>\n

                  Input voltage to basic amplifier with negative feedback is
                  \n\"2nd
                  \n\\(=\\left(\\frac{V_{i}}{I_{i}}\\right)(1+A \\beta)\\)
                  \nZif<\/sub> = Zi <\/sub>(1+A\u03b2)from eq ………… (1)
                  \nAs (1 + A\u03b2) > 1 for negative feedback, Zif<\/sub> > Zi
                  \n<\/sub>Hence with negative feedback, input impedance of an amplifier increases.<\/p>\n

                  Question 5.
                  \nDerive an expression for the output impedance of an amplifier with negative feedback
                  \nAnswer:
                  \n\"2nd
                  \nAn amplifier should have a low output impedance so that it can deliver power to load without much loss. With negative feedback, the output impedance of the amplifier decreases which is desirable.<\/p>\n

                  Let Zo<\/sub> be output impedance of basic amplifier and Zof<\/sub>\u00a0be the output impedance with negative feedback. A hypothetical source of voltage Vo<\/sub> is applied at the output. By applying KVL to the output loop,<\/p>\n

                  Vo<\/sub> = Io<\/sub>Zo<\/sub> +AVi<\/sub>.
                  \nWith negative feedback, input to basic amplifier with negative feedback is
                  \nVi <\/sub>= Vs<\/sub> – Vf
                  \n<\/sub>Vi\u00a0 <\/sub>= o – Vf<\/sub>
                  \nVi\u00a0 <\/sub>– Vf<\/sub>
                  \n<\/strong>{As input is short circuited, V = o}
                  \n\"2nd
                  \nAs 1+A\u03b2>1,output impedance of amplifier decreases by a factor(1+A\u03b2)<\/p>\n

                  \"2nd<\/p>\n

                  Question 6.
                  \nThe input and output voltages of an amplifier are 20 mV and 5V respectively. If the gain with nagative feedback is 100 and input impedance without feedback is 15k\u03a9, find feedback fraction and input impedance with feedback.
                  \nAnswer:
                  \nVin<\/sub> = 20mv = 20 x 10-3<\/sup>V
                  \nV0<\/sub> =5V
                  \nAf<\/sub>= 200
                  \nZif <\/sub>= 15k\u03a9
                  \n\"2nd<\/p>\n

                  \"2nd<\/p>\n

                  Question 7.
                  \nAn amplifier with loop gain of 25 is, subjected to a negative feedback of 12%. If open loop gain changes by 5% due to temperature rise, find percentage change for gain with feedback. If the gain hand with product is 1,25,000 Hz, Calculate the bandwidth with and without feedback.
                  \nAnswer:
                  \nA\u03b2 = 25
                  \n\u03b2 = 12%
                  \n\\(=\\frac{12}{100}=0.12\\)
                  \n\"2nd
                  \n\"2nd<\/p>\n

                  \"2nd<\/p>\n

                  Question 8.
                  \nThe open loop voltage gain of an amplifier is 60 dB. A negative feedback factor 0.03 is applied to it. If the open loop gain change is 10%, find
                  \n(a) closed loop gain in decibels
                  \n(b) the % change in the over all gain.
                  \nAnswer:
                  \nG\u00a0 = 6odB = 20 log A
                  \n20 log10 <\/sub>A = 60
                  \nlog10 <\/sub>A = 3
                  \nA = 103<\/sup> = 1000 {loga <\/sub>N = x = ax<\/sup> = N}
                  \n\"2nd<\/p>\n

                  Question 9.
                  \nThe gain of an amplifer falls from 100 to 40 when negative feedback is applied. If the input voltage is IV, calculate (i) feedback factor (ii). Feedback voltage.
                  \nAnswer:
                  \n\"2nd<\/p>\n

                  \"2nd<\/p>\n

                  Question 10.
                  \nAn amplifier having a gain of 700 without feedback has overall negative feedback applied which reduces the gain to 100. Calculate feedback factor. If due to ageing of components, the gain without feedback falls by 20%, calculate the percentage fall in gain with feedback.
                  \nAnswer:
                  \n\"2nd<\/p>\n

                  Question 11.
                  \nThe input impedance of an amplifier is 2k\u03a9 while its output impedance with and without feedback is 150k\u03a9 and 5k\u03a9. If the gain of the amplifier is 500, calculate
                  \n(i) Input impedance with feedback
                  \n(ii) Gain with feedback.
                  \nAnswer:
                  \n\"2nd<\/p>\n

                  Question 12.
                  \nThe gain of an amplifier is 50 and its output resistance is 2.5k\u03a9. A negative feedback is applied so that output impedance reduces to 500\u03a9. What is the value of \u03b2? It bandwidth before feedback is 200 kHz, what is the new bandwidth.
                  \nAnswer:
                  \n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

                  Students can Download 2nd PUC Electronics Chapter 4 Feedback in Amplifiers Questions and Answers, Notes Pdf, 2nd PUC Electronics Question Bank with A nswers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Electronics Question Bank Chapter 4 Feedback in Amplifiers 2nd PUC Electronics Feedback …<\/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\/23965"}],"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=23965"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/23965\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=23965"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=23965"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=23965"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}