Physical electronics devices and ics miscellaneous

Physical electronics devices and ics miscellaneous

Easy Questions

Physical Electronics Devices and ICs

Physical electronics devices and ics miscellaneous
  1. The operation of a JEFT involves—








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    The operation of a junction field effect transistor involves a flow of majority carriers.

    Correct Option: B

    The operation of a junction field effect transistor involves a flow of majority carriers.

  1. A Field-Effect Transistor (FET)—








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    The field-effect transistor (FET) operation depends on the variation of the depletion-layer width with reverse voltage, for its operation.

    Correct Option: D

    The field-effect transistor (FET) operation depends on the variation of the depletion-layer width with reverse voltage, for its operation.

  1. When a positive voltage signal is applied to the base of a normally biased NPN common-emitter transistor amplifier—








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    NA

    Correct Option: B

    NA

  1. A transistor is said to be in a quiescent state when—








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    A transistor is said to be quiescent state when no signal is applied to the input.

    Correct Option: A

    A transistor is said to be quiescent state when no signal is applied to the input.

  1. If the transistors in figure have high values of β and a VBE of 0.65V, the current I flowing through the 2 kΩ resistance will be—










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    For DC current
    IE2 ≈ IE1
    or IC2 ≈ IC1
    Neglecting IB1 which will be very small, β being very high, the base voltage of Q1 is

    VB1 =
    VCC
    		(by using potential divider method)
    (1.65 kΩ) 6.5 kΩ + 1.85 kΩ + 1.65 kΩ

    = 10 ×
    1.65
    		= 1.65 V IE1
    10

    =
    VB1 – VBE
    		=
    = (1.65 – .65) V
    		mA.
    1 kΩ1 kΩ


    The current flowing through the 2 kΩ resistor is 1 mA.
    Since IC2 ≈ IC1.
    Hecne alternative (A) is the correct answer.

    Correct Option: A

    For DC current
    IE2 ≈ IE1
    or IC2 ≈ IC1
    Neglecting IB1 which will be very small, β being very high, the base voltage of Q1 is

    VB1 =
    VCC
    		(by using potential divider method)
    (1.65 kΩ) 6.5 kΩ + 1.85 kΩ + 1.65 kΩ

    = 10 ×
    1.65
    		= 1.65 V IE1
    10

    =
    VB1 – VBE
    		=
    = (1.65 – .65) V
    		mA.
    1 kΩ1 kΩ


    The current flowing through the 2 kΩ resistor is 1 mA.
    Since IC2 ≈ IC1.
    Hecne alternative (A) is the correct answer.