Analog circuits miscellaneous


Analog circuits miscellaneous

  1. If an npn transistor has a beta cut-off frequency fB of 1MHz, and emitter short circuit low frequency current gain β0 of 200, then unity gain frequency fT and alpha cut off frequency fα respectively are









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    f T = β0 × fβ = (200) × 1 MHz = 200 MHz

    and fα =
    fβ
    =
    fβ
    1 - α1 -
    β
    1 + β

    = f β (β+ 1) = 106 (200 + 1) 106 × 201 = 201 MHz

    Correct Option: A

    f T = β0 × fβ = (200) × 1 MHz = 200 MHz

    and fα =
    fβ
    =
    fβ
    1 - α1 -
    β
    1 + β

    = f β (β+ 1) = 106 (200 + 1) 106 × 201 = 201 MHz


  1. If an npn transistor (with C = 0.3 pF) has a unity gain cut-off frequency fT of 400 MHz at a d.c. bias current Ic = 1 mA, then value of its C is approximately (VT = 26 mV)









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    Given : C = 0.3 pF

    ∴ gm =
    |IC|(mA)
    =
    1
    A/V
    26 mV26

    ⇒ C =
    gm
    =
    1 / 26
    = 15 pF
    2π fT2π × 4 × 108

    Correct Option: A

    Given : C = 0.3 pF

    ∴ gm =
    |IC|(mA)
    =
    1
    A/V
    26 mV26

    ⇒ C =
    gm
    =
    1 / 26
    = 15 pF
    2π fT2π × 4 × 108



  1. An amplifier has an open loop gain of 100, an input impedance of 1kΩ, and an out put impedance of 100 Ω. If a feedback network with a feedback factor of 0.99 is connected in a voltage series feedback mode then new input and output impedances are
    (a)
    (b)
    (c)









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    Zif = Zi (1 + βA) = 1 k (1 + 100 × 0.99) = 100 kΩ

    Zof =
    Z0
    =
    100
    = 1 Ω
    1 + βA100

    Correct Option: C

    Zif = Zi (1 + βA) = 1 k (1 + 100 × 0.99) = 100 kΩ

    Zof =
    Z0
    =
    100
    = 1 Ω
    1 + βA100


  1. RC network shown in the given figure can provide maximum theoretical phase shift of










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    Each section having phase shift of 60°, adding up to 180°

    Correct Option: B

    Each section having phase shift of 60°, adding up to 180°



  1. The resonant frequency of the Wein bridge oscillator of given in figure, is










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    f =
    1
    =
    1
    = 3120.7 Hz
    2πRC2π(51 × 103)(0.001 × 10-6)

    Correct Option: C

    f =
    1
    =
    1
    = 3120.7 Hz
    2πRC2π(51 × 103)(0.001 × 10-6)