Analog electronics circuits miscellaneous

Analog electronics circuits miscellaneous

Easy Questions

Analog Electronic Circuits

Analog electronics circuits miscellaneous
  1. The parameter of the transistor in fig. below are VTN = 0.6 V and Kn = 0.2 mA/V2. The voltage Vs is:










  1. View Hint View Answer Discuss in Forum

    Given that VTN = 0.6 V
    Kn = 0.2 mA/V2
    ID = 0.25 mA
    ID = Kn (VGS – VTn)2
    0.25 × 10– 3 = 0.2 × 10– 3 (VGS – 0.6)2

    25
    		= (VGS – 0.6)2
    20

    VGS =
    5
    		+ 0.6 = 1.718 ≈ 1.72
    √20

    VG = 0 V
    VGS = VG – VS
    or VS = VG – VGS = 0 – 1.72 = – 1.72 V
    Hence (B) is the correct choice.


    Correct Option: B

    Given that VTN = 0.6 V
    Kn = 0.2 mA/V2
    ID = 0.25 mA
    ID = Kn (VGS – VTn)2
    0.25 × 10– 3 = 0.2 × 10– 3 (VGS – 0.6)2

    25
    		= (VGS – 0.6)2
    20

    VGS =
    5
    		+ 0.6 = 1.718 ≈ 1.72
    √20

    VG = 0 V
    VGS = VG – VS
    or VS = VG – VGS = 0 – 1.72 = – 1.72 V
    Hence (B) is the correct choice.


  1. In the regulator circuit of fig. below Vz = 12 V, β = 50, VBE = 0.7 V. The Zener current is:










  1. View Hint View Answer Discuss in Forum

    V0 = VZ – VBE
    = 12 – 0.7 = 11.3 V
    VCE = 20 – 11.3 = 8.7 V
    Current in 220 Ω resistor is

    I220 Ω =
    20 – 12
    		= 36.4 mA
    220

    Current in 1 kΩ resistor is
    11kΩ =
    V0
    		=
    11·3
    		= 11.3 mA = IC
    1 kΩ1 kΩ

    IB =
    IC
    		=
    11·3
    		= 0.226 mA
    β50

    KCL at node A, we get
    IZ = I220 Ω – IB = 36.4 – 0.226 = 36.17 mA
    Hence alternative (B) is the correct choice.


    Correct Option: B

    V0 = VZ – VBE
    = 12 – 0.7 = 11.3 V
    VCE = 20 – 11.3 = 8.7 V
    Current in 220 Ω resistor is

    I220 Ω =
    20 – 12
    		= 36.4 mA
    220

    Current in 1 kΩ resistor is
    11kΩ =
    V0
    		=
    11·3
    		= 11.3 mA = IC
    1 kΩ1 kΩ

    IB =
    IC
    		=
    11·3
    		= 0.226 mA
    β50

    KCL at node A, we get
    IZ = I220 Ω – IB = 36.4 – 0.226 = 36.17 mA
    Hence alternative (B) is the correct choice.


  1. The parameter of the transistor in fig given below VTN = 1.2 V, Kn = 0.5 mA/V2. The voltage VDS is:










  1. View Hint View Answer Discuss in Forum

    Given that VTN = 1.2 V
    Kn = 0.5 mA/V2
    ID = Kn (VS – VTN)2
    50 × 10– 6 = 0.5 × 10– 3 (VGS – 1.2)2
    VGS = √.1 + 1.2 = 1.516 V
    VG = 0
    so VGS = VG – VS
    or VS = VG – VGS = 0 – 1.516
    = – 1.516 V
    VDS = VD – VS = 5 – (– 1.516)
    = 6.516 V


    Correct Option: B

    Given that VTN = 1.2 V
    Kn = 0.5 mA/V2
    ID = Kn (VS – VTN)2
    50 × 10– 6 = 0.5 × 10– 3 (VGS – 1.2)2
    VGS = √.1 + 1.2 = 1.516 V
    VG = 0
    so VGS = VG – VS
    or VS = VG – VGS = 0 – 1.516
    = – 1.516 V
    VDS = VD – VS = 5 – (– 1.516)
    = 6.516 V


Direction: Statement for :
In the circuit of fig. below the transistor parameters are VTP = – 0.8 V and KP = 200 µA/V2

  1. VSD =?








  1. View Hint View Answer Discuss in Forum

    VD = ID RD – 5 = 0.4 × 10– 3 × 5 × 103 – 5 = – 3 V
    VSD = VS – VD = 2.21 – (– 3) = 5.21 V
    Hence alternative (A) is the correct choice.

    Correct Option: A

    VD = ID RD – 5 = 0.4 × 10– 3 × 5 × 103 – 5 = – 3 V
    VSD = VS – VD = 2.21 – (– 3) = 5.21 V
    Hence alternative (A) is the correct choice.

  1. VS =?








  1. View Hint View Answer Discuss in Forum

    Assume transistor in saturation
    given, ID = 0.4 mA
    VTP = – 0.8 V
    kP = 200 µA/V2
    ID = kP [VSG + (– VTP)]2
    0.4 × 10–3 = 200 × 10–6 [VSG + 0.8]2
    or [VSG + 0.8]2 = 2
    or VSG = √2 + 0.8 = 2.21 V
    or VSG = VS – VG = VS – 0 = VS
    = 2.21 V


    Correct Option: C

    Assume transistor in saturation
    given, ID = 0.4 mA
    VTP = – 0.8 V
    kP = 200 µA/V2
    ID = kP [VSG + (– VTP)]2
    0.4 × 10–3 = 200 × 10–6 [VSG + 0.8]2
    or [VSG + 0.8]2 = 2
    or VSG = √2 + 0.8 = 2.21 V
    or VSG = VS – VG = VS – 0 = VS
    = 2.21 V