Analog electronics circuits miscellaneous

Analog electronics circuits miscellaneous

Easy Questions

Analog Electronic Circuits

Analog electronics circuits miscellaneous
  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

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










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    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:










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    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.