Analog electronics circuits miscellaneous


Analog electronics circuits miscellaneous

Analog Electronic Circuits

  1. The dissipation at the collector is zero in the quiescent state and increases with excitation in the case of:









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    NA

    Correct Option: A

    NA


  1. The change in the value of the emitter resistance RE in a difference amplifier:









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    ACM =
    –hfe · Rc
    hie + Rs + (1 + hfe)RE

    and ADM =
    1
    ×
    –hfe·Rc
    2(Rs + hie)

    Hence alternative (B) is the correct choice.

    Correct Option: B

    ACM =
    –hfe · Rc
    hie + Rs + (1 + hfe)RE

    and ADM =
    1
    ×
    –hfe·Rc
    2(Rs + hie)

    Hence alternative (B) is the correct choice.



  1. A op-amp has an offset voltage of 1 mv and is ideal in all respects. If this op-amp is used in the circuit shown below the output voltage will be (select the nearest value)











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    Output of the op-amp due to offset voltage of 1 mV

    Vo = –
    1 MΩ
    Vi
    1 kΩ

    or Vo = – (103) (± 1 × 10–3)
    or Vo = ± 1 V

    Correct Option: C

    Output of the op-amp due to offset voltage of 1 mV

    Vo = –
    1 MΩ
    Vi
    1 kΩ

    or Vo = – (103) (± 1 × 10–3)
    or Vo = ± 1 V


  1. Assume that the op-amp shown below is ideal. The current I through the 1 kΩ resistor is:











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    Due to virtual ground voltage at terminal A is 0 V. KCL at node A

    2 =
    VA – VB
    2

    2 =
    0 – VB
    2

    or VB = – 4 V
    Again KCL at B
    I =
    VB – VA
    +
    VB – 0
    22

    or I =
    (–4 – 0)
    +
    –4
    22

    or I = – 2 – 2 I = – 4 mA
    Hence alternative (A) is the correct choice.


    Correct Option: A

    Due to virtual ground voltage at terminal A is 0 V. KCL at node A

    2 =
    VA – VB
    2

    2 =
    0 – VB
    2

    or VB = – 4 V
    Again KCL at B
    I =
    VB – VA
    +
    VB – 0
    22

    or I =
    (–4 – 0)
    +
    –4
    22

    or I = – 2 – 2 I = – 4 mA
    Hence alternative (A) is the correct choice.




  1. In the op-amp circuit V0 is given by:











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    Due to virtual ground voltage at terminal A is Vs KCL at node A.

    Vs
    +
    Vs – Vo
    = 0
    R2R

    Vs
    +
    Vs
    =
    Vo
    R2R2R

    Vs
    3R
    =
    Vo
    2R22R

    or V0 = 3 Vs
    Alternative method
    V0 = Vs 1 +
    2R
    = 3 Vs
    R


    Correct Option: C

    Due to virtual ground voltage at terminal A is Vs KCL at node A.

    Vs
    +
    Vs – Vo
    = 0
    R2R

    Vs
    +
    Vs
    =
    Vo
    R2R2R

    Vs
    3R
    =
    Vo
    2R22R

    or V0 = 3 Vs
    Alternative method
    V0 = Vs 1 +
    2R
    = 3 Vs
    R