Analog electronics circuits miscellaneous

Analog electronics circuits miscellaneous

Easy Questions

Analog Electronic Circuits

Analog electronics circuits miscellaneous
  1. An amplifier has a gain of 1000 ± 10. Negative feedback is provided such that the gain variation is within 0.1%. Then the amount of feedback βf is:








  1. View Hint View Answer Discuss in Forum

    Given that A = 1000 ± 10

    dA
    		=
    10
    		=
    1
    		= 0.01
    A1000100

    dAf
    		= 0.1% =
    .1
    		= 0.001
    Af100

    Now, from relation
    dAf
    		=
    1
    		x
    dA
    dA(1 + Aβ)A

    0.001 =
    1
    		× 0.01
    (1 + 1000 β)

    or 1 + 1000 β = 10
    or 1000 β = 9
    or β =
    9
    1000

    Hence alternative (D) is the correct choice.

    Correct Option: D

    Given that A = 1000 ± 10

    dA
    		=
    10
    		=
    1
    		= 0.01
    A1000100

    dAf
    		= 0.1% =
    .1
    		= 0.001
    Af100

    Now, from relation
    dAf
    		=
    1
    		x
    dA
    dA(1 + Aβ)A

    0.001 =
    1
    		× 0.01
    (1 + 1000 β)

    or 1 + 1000 β = 10
    or 1000 β = 9
    or β =
    9
    1000

    Hence alternative (D) is the correct choice.

  1. The high frequency response of an amplifier is given by
    A =
    A0
    1 + j
    ω
    ω2

    where A0 = 1000 and ω2 = 104. Negative feedback with β = 9/1000 is now given. The value of ω2 now is:








  1. View Hint View Answer Discuss in Forum

    From given high frequency response

    A =
    A0
    1 + j
    ω
    ω2

    where, ω2 is the higher cut-off frequency.
    and due to negative feedback ω2 increases, so the new,
    ω2* = ω2 (1 + Aβ)
    = 1041 + 1000 ×
    9
    		= 104. 10 = 105
    1000

    Hence alternative (C) is the correct choice.

    Correct Option: C

    From given high frequency response

    A =
    A0
    1 + j
    ω
    ω2

    where, ω2 is the higher cut-off frequency.
    and due to negative feedback ω2 increases, so the new,
    ω2* = ω2 (1 + Aβ)
    = 1041 + 1000 ×
    9
    		= 104. 10 = 105
    1000

    Hence alternative (C) is the correct choice.

  1. An amplifier with mid band gain A = 500 has negative feedback applied of value β = 1/100. Given the upper cut off frequency without feedback was 60 kHz. With feedback it becomes:








  1. View Hint View Answer Discuss in Forum

    Given that

    β =
    1
    100

    A = 500
    fH = 60 kHz
    After feedback. New upper cut-off frequency increases as bandwidth increases after feedback.
    fH* = fH (1 + Aβ)
    = 601 +
    500
    		= 60 × 6 = 360 kHz
    100

    Correct Option: D

    Given that

    β =
    1
    100

    A = 500
    fH = 60 kHz
    After feedback. New upper cut-off frequency increases as bandwidth increases after feedback.
    fH* = fH (1 + Aβ)
    = 601 +
    500
    		= 60 × 6 = 360 kHz
    100

  1. The amount of feedback applied to an amplifier reduces the gain by a factor of 10. The bandwidth:








  1. View Hint View Answer Discuss in Forum

    Since the product of gain and bandwidth remains constant. Therefore if gain reduces by factor of 10, then surely bandwidth will increase by a factor of 10.

    Correct Option: B

    Since the product of gain and bandwidth remains constant. Therefore if gain reduces by factor of 10, then surely bandwidth will increase by a factor of 10.

  1. The feedback amplifier shown in figure has:










  1. View Hint View Answer Discuss in Forum

    Here the output current is feedback in the node. Hence the type of feedback is current-shunt.

    Correct Option: D

    Here the output current is feedback in the node. Hence the type of feedback is current-shunt.