Analog electronics circuits miscellaneous


Analog electronics circuits miscellaneous

Analog Electronic Circuits

  1. The ideal characteristics of op-amp are:
    (i) input impedance, Ri = ∞
    (ii) output impedance, R0 = ∞
    (iii) bandwidth impedance, BW = 0
    (iv) gain impedance, A = ∞ The correct statement are:









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    NA

    Correct Option: C

    NA


  1. In the circuit shown below the input offset voltage and input offset current are Vio = 4 mV and Iio = 150 nA. The total output offset voltage is:











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    Total output offset voltage is given by relation

    VOT =1 +
    Rf
    Vos + Rf. IB
    R1

    where, VOT = total output offset voltage
    Rf = feedback resistance
    Vos = input offset voltage
    IB = input offset current Now,
    VOT =1 +
    500
    4 × 10–3 + 5000 × 103 × 150 × 10–9
    5

    VOT = 404 × 10–3 + 75 × 10–3
    or VOT = 479 × 10–3 = 479 mV.


    Correct Option: A

    Total output offset voltage is given by relation

    VOT =1 +
    Rf
    Vos + Rf. IB
    R1

    where, VOT = total output offset voltage
    Rf = feedback resistance
    Vos = input offset voltage
    IB = input offset current Now,
    VOT =1 +
    500
    4 × 10–3 + 5000 × 103 × 150 × 10–9
    5

    VOT = 404 × 10–3 + 75 × 10–3
    or VOT = 479 × 10–3 = 479 mV.




  1. In the circuit of fig. shown the op-amp slew rate is SR = 0.5 V/µs. If the amplitude of input signal is 0.02 V, then the maximum frequency that may be used is:











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    Max. output voltage

    Vo = Vi
    240
    = 24 × 0.02
    10

    = 0.48 V
    S.R. 2 π fmax. Vmax
    or S.R.. Vo
    or
    S.R.
    Vo

    or =
    0·5 × 106
    = 1.1 × 106 rad/s.
    0·48

    Correct Option: C

    Max. output voltage

    Vo = Vi
    240
    = 24 × 0.02
    10

    = 0.48 V
    S.R. 2 π fmax. Vmax
    or S.R.. Vo
    or
    S.R.
    Vo

    or =
    0·5 × 106
    = 1.1 × 106 rad/s.
    0·48


  1. An amplifier has an input power of 2 mW. The power gain of amplifier has 60 dB. The output power will be:









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    Power gain = 20 log10
    Po
    Pi

    or 60 = 20 log10
    Po
    Pi

    or 3 = log10
    Po
    Pi

    or
    Po
    = 103
    Pi

    or Po = Pi × 103 = 2 × 10–3 × 103 = 2 W

    Correct Option: D

    Power gain = 20 log10
    Po
    Pi

    or 60 = 20 log10
    Po
    Pi

    or 3 = log10
    Po
    Pi

    or
    Po
    = 103
    Pi

    or Po = Pi × 103 = 2 × 10–3 × 103 = 2 W



  1. In figure below assume that the diode and op-amps are ideal. For an input Vin = sin ωt, the output voltage V0 is:











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    The op-amp (1) circuit represent precision half wave rectifier conducting for negative cycle at the inverting input, and op-amp (2) is a unity follower.
    Hence alternative (C) is the correct choice.

    Correct Option: C

    The op-amp (1) circuit represent precision half wave rectifier conducting for negative cycle at the inverting input, and op-amp (2) is a unity follower.
    Hence alternative (C) is the correct choice.