Analog electronics circuits miscellaneous


Analog electronics circuits miscellaneous

Analog Electronic Circuits

  1. If the input to the circuit given below is a sine wave the output will be a:











  1. View Hint View Answer Discuss in Forum

    Since the given circuit arrangement represent open loop system, so the output will be ± Vsat, which resembles a square wave.

    Correct Option: D

    Since the given circuit arrangement represent open loop system, so the output will be ± Vsat, which resembles a square wave.


  1. The op-amp of figure has a very poor open loop voltage gain of 45 but is otherwise ideal.
    The gain of the amplifier equals:











  1. View Hint View Answer Discuss in Forum

    A closed loop gain,

    ACL =
    V0
    =
    AOL
    Vi1 + AOL

    where, AOL = 45 (given)
    β =
    2k
    (from given figure)
    2k + 8k

    or β = 0.2
    Now, ACL =
    45
    1 + 45 × ·2

    =
    45
    = 4.5
    10


    Correct Option: D

    A closed loop gain,

    ACL =
    V0
    =
    AOL
    Vi1 + AOL

    where, AOL = 45 (given)
    β =
    2k
    (from given figure)
    2k + 8k

    or β = 0.2
    Now, ACL =
    45
    1 + 45 × ·2

    =
    45
    = 4.5
    10




  1. For the op-amp circuit shown below the voltage gain Aν = ν0i is:











  1. View Hint View Answer Discuss in Forum

    Apply KCL at node A

    VA – VC
    +
    VA
    +
    VA – VB
    = 0 ............(i)
    RRR

    KCL at node B
    VB – VA
    +
    VB – V0
    +
    VB
    = 0 ............(ii)
    RRR

    KCL at node C
    VC – Vi
    +
    VC – VA
    = 0 .............(iii)
    RR

    on putting VC = 0 (due to virtual ground) and eliminating VA and VB from equation (i), (ii) and (iii)
    V0
    = – 8
    Vi


    Correct Option: A

    Apply KCL at node A

    VA – VC
    +
    VA
    +
    VA – VB
    = 0 ............(i)
    RRR

    KCL at node B
    VB – VA
    +
    VB – V0
    +
    VB
    = 0 ............(ii)
    RRR

    KCL at node C
    VC – Vi
    +
    VC – VA
    = 0 .............(iii)
    RR

    on putting VC = 0 (due to virtual ground) and eliminating VA and VB from equation (i), (ii) and (iii)
    V0
    = – 8
    Vi



  1. For the circuit shown below gain is Aν = ν0i = – 10. The value of R is:











  1. View Hint View Answer Discuss in Forum

    Given

    V0
    = – 10
    Vi

    KCL at node A
    VA – Vi
    +
    VA – VB
    = 0 ...........(1)
    100R

    KCL at node B
    VB – VA
    +
    VB
    +
    VB – V0
    = 0 ...........(2)
    R100100

    on manipulating
    VA = 0. (due to virtual ground)
    V0 = – 10 Vi
    in equation (1) and (2) we get
    0 – Vi
    +
    0 – VB
    = 0
    100R

    or Vi = – VB.
    100
    .............(3)
    R

    VB – 0
    +
    VB
    +
    VB
    =
    V0
    R100100100

    or VB
    1
    +
    1
    +
    1
    =
    –10Vi
    R100100100

    or VB
    1
    +
    2
    =
    –16
    ·–VB
    100
    R100100R

    or
    1
    +
    2
    =
    10
    R100R

    or
    10
    1
    =
    2
    RR100

    or
    10R – R
    =
    2
    R2100

    or
    9
    =
    2
    R100

    or R = 450 kΩ


    Correct Option: B

    Given

    V0
    = – 10
    Vi

    KCL at node A
    VA – Vi
    +
    VA – VB
    = 0 ...........(1)
    100R

    KCL at node B
    VB – VA
    +
    VB
    +
    VB – V0
    = 0 ...........(2)
    R100100

    on manipulating
    VA = 0. (due to virtual ground)
    V0 = – 10 Vi
    in equation (1) and (2) we get
    0 – Vi
    +
    0 – VB
    = 0
    100R

    or Vi = – VB.
    100
    .............(3)
    R

    VB – 0
    +
    VB
    +
    VB
    =
    V0
    R100100100

    or VB
    1
    +
    1
    +
    1
    =
    –10Vi
    R100100100

    or VB
    1
    +
    2
    =
    –16
    ·–VB
    100
    R100100R

    or
    1
    +
    2
    =
    10
    R100R

    or
    10
    1
    =
    2
    RR100

    or
    10R – R
    =
    2
    R2100

    or
    9
    =
    2
    R100

    or R = 450 kΩ




  1. In circuit shown below, the input voltage νi is 0.2 V. The output voltage ν0 is:











  1. View Hint View Answer Discuss in Forum

    From figure.

    VA = –
    50
    . Vi
    10

    or VA = – 5 Vi
    and V0 = –
    150
    (VA)
    25

    V0 = – 6. (– 5 Vi)
    V0 = 30 Vi
    V0 = 30 × 0.2
    V0 = 6 V.


    Correct Option: A

    From figure.

    VA = –
    50
    . Vi
    10

    or VA = – 5 Vi
    and V0 = –
    150
    (VA)
    25

    V0 = – 6. (– 5 Vi)
    V0 = 30 Vi
    V0 = 30 × 0.2
    V0 = 6 V.