Analog electronics circuits miscellaneous


Analog electronics circuits miscellaneous

Analog Electronic Circuits

  1. In the circuit of fig. below Zener voltage is Vz = 5 V and β = 100. The value of ICQ and VCEQ are:











  1. View Hint View Answer Discuss in Forum

    From figure applying KVL
    12 – 500 (Ic + Ib) – Vz – VBE = 0
    12 = 500 IE + 5 + 0.7
    or IE = I2.6 mA

    or IC = IE – IB = IE
    IC
    β

    or IC =
    β
    · IE =
    100
    (12.6) mA
    1 + β1 + 100

    = 12.47 mA = ICQ
    VCE = 12 – 500. (IE) = 12 – 500 × 12.6 × 10–3 = 5.7
    V = VCEQ
    Hence alternative (B) is the correct choice.


    Correct Option: B

    From figure applying KVL
    12 – 500 (Ic + Ib) – Vz – VBE = 0
    12 = 500 IE + 5 + 0.7
    or IE = I2.6 mA

    or IC = IE – IB = IE
    IC
    β

    or IC =
    β
    · IE =
    100
    (12.6) mA
    1 + β1 + 100

    = 12.47 mA = ICQ
    VCE = 12 – 500. (IE) = 12 – 500 × 12.6 × 10–3 = 5.7
    V = VCEQ
    Hence alternative (B) is the correct choice.



  1. The two transistor in fig. below are identical. If β = 25, the current IC2 is:











  1. View Hint View Answer Discuss in Forum

    Since both the transistor are identical, and both the transistor are in the forward active region.
    IB1 + IB2 + IC1 = 25 mA ....(i)
    VBE is same for both the transistor since both are identical
    i.e. IC1 = IC2 and
    IB1 = IB2
    so, equation reduces
    2IB1 + IC1 = 25 µA
    or 2IB1 + IB1 . β = 25 µA
    or IB1 (β + 2) = 25 µA

    or IB1 =
    25
    = IB2
    β + 2

    or IC2 = β. IB2 = β.
    25
    β + 2

    = 25.
    25
    = 23.2 µA
    (25 + 2)

    Hence alternative (B) is the correct choice.


    Correct Option: B

    Since both the transistor are identical, and both the transistor are in the forward active region.
    IB1 + IB2 + IC1 = 25 mA ....(i)
    VBE is same for both the transistor since both are identical
    i.e. IC1 = IC2 and
    IB1 = IB2
    so, equation reduces
    2IB1 + IC1 = 25 µA
    or 2IB1 + IB1 . β = 25 µA
    or IB1 (β + 2) = 25 µA

    or IB1 =
    25
    = IB2
    β + 2

    or IC2 = β. IB2 = β.
    25
    β + 2

    = 25.
    25
    = 23.2 µA
    (25 + 2)

    Hence alternative (B) is the correct choice.




  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