Analog electronics circuits miscellaneous


Analog electronics circuits miscellaneous

Analog Electronic Circuits

  1. The parameters for the transistor in circuit of fig. below are VTN = 2 V and Kn = 0.2 mA/V2. The power dissipated in the transistor is:











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    From fig. since gate is connected to the drain. Hence transistor will always in saturation.

    ID =
    10 – VGS
    = Kn (VGS – VTn)2
    10 kΩ

    10 – VGS = 0.2 × 10–3 × 10 × 103 (VGS – 2)2
    or 10 – VGS = 2 (VGS – 2)2
    After solving quadratic equation, we get
    VGS = – 0.27 V, 3.77 V VGS = – 0.27 is not possible since gate terminal is at 10 V. So, VGS = 3.77 V is taken
    i.e. VGS = VDS = 3.77 V
    ID =
    10 – 3·77
    = 0.623 mA
    10 kΩ

    Power = ID VDS = 0.623 × 10–3 × 3.77 = 2.35 mW


    Correct Option: B

    From fig. since gate is connected to the drain. Hence transistor will always in saturation.

    ID =
    10 – VGS
    = Kn (VGS – VTn)2
    10 kΩ

    10 – VGS = 0.2 × 10–3 × 10 × 103 (VGS – 2)2
    or 10 – VGS = 2 (VGS – 2)2
    After solving quadratic equation, we get
    VGS = – 0.27 V, 3.77 V VGS = – 0.27 is not possible since gate terminal is at 10 V. So, VGS = 3.77 V is taken
    i.e. VGS = VDS = 3.77 V
    ID =
    10 – 3·77
    = 0.623 mA
    10 kΩ

    Power = ID VDS = 0.623 × 10–3 × 3.77 = 2.35 mW



  1. The PMOS transistor shown below has parameters:
    VTP = - 1.2 V,
    W
    = 20, and KP = 30 µA/V2
    L

    If ID = 0.5 mA and VD = – 3 V, then value of RS and RD are:











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    NA

    Correct Option: D

    NA



  1. In the circuit given below the transistor parameters are:
    VTN = 1.7 V and Kn = 0.4 mA/V2.
    If ID = 0.8 mA and VD = 1 V, then value of resistor RS and RD are respectively:











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    From figure

    ID =
    5 – VD
    RD

    0.8 × 10–3 =
    5 – 1
    RD

    or RD =
    4
    = 5 kΩ
    0·8 × 10–3

    ID = Kn (VGS – VTn)2
    or 0.8 × 10–3 = 0.4 × 10–3 (VGS – 1.7)2
    or VGS = √2 + 1.7 = 3.11 V VG – VS = 3.11
    or VS = VG – 3.11 = 0 – 3.11 = – 3.11 V
    Again, ID =
    VS – (–5)
    =
    –3·11 + 5
    RSRS

    or RS =
    1·89
    = 2.3625 kΩ
    0·8 × 10–3


    Correct Option: A

    From figure

    ID =
    5 – VD
    RD

    0.8 × 10–3 =
    5 – 1
    RD

    or RD =
    4
    = 5 kΩ
    0·8 × 10–3

    ID = Kn (VGS – VTn)2
    or 0.8 × 10–3 = 0.4 × 10–3 (VGS – 1.7)2
    or VGS = √2 + 1.7 = 3.11 V VG – VS = 3.11
    or VS = VG – 3.11 = 0 – 3.11 = – 3.11 V
    Again, ID =
    VS – (–5)
    =
    –3·11 + 5
    RSRS

    or RS =
    1·89
    = 2.3625 kΩ
    0·8 × 10–3



  1. In the circuit given below, the transistor T3 is used as:











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    NA

    Correct Option: B

    NA



  1. In which one of the following transistor configuration is the input amplifier least dependent on the load resistance?









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    NA

    Correct Option: D

    NA