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. For the zener diode shown in the figure, the zener voltage at knee is 7V, the knee current is negligible and the zener dynamic resistance is 10Ω. If the input voltage (V1) range is from 10 to 16V, the output voltage (V0)—










  1. View Hint View Answer Discuss in Forum

    Given, rz = 10Ω (zener dynamic resistance) (i.e., fig)
    VZ = 7V

    I =
    Vi – VZ
    (200 + 10)

    Imax =
    Vi (max) – VZ
    		=
    16 – 7
    		= 0.043A
    210210

    Imax =
    Vi (max) – VZ
    		=
    10 – 7
    		= 0.0143A
    210210

    Output voltage is given by
    V0 = IrZ + VZ
    Now V0 (max) = Imax rZ + VZ
    or V0 (max) = 0.043 × 10 + 7 = 7.43 V
    and V0 (min) = Imin rZ + VZ = 0.0143 × 10 + 7
    = 7.143 ≈ 7.15V.
    Hence alternative (C) is the correct choice.


    Correct Option: C

    Given, rz = 10Ω (zener dynamic resistance) (i.e., fig)
    VZ = 7V

    I =
    Vi – VZ
    (200 + 10)

    Imax =
    Vi (max) – VZ
    		=
    16 – 7
    		= 0.043A
    210210

    Imax =
    Vi (max) – VZ
    		=
    10 – 7
    		= 0.0143A
    210210

    Output voltage is given by
    V0 = IrZ + VZ
    Now V0 (max) = Imax rZ + VZ
    or V0 (max) = 0.043 × 10 + 7 = 7.43 V
    and V0 (min) = Imin rZ + VZ = 0.0143 × 10 + 7
    = 7.143 ≈ 7.15V.
    Hence alternative (C) is the correct choice.


  1. In a regulated power supply using zener diode the unregulated input voltage as compared to the regulated output voltage must be—








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    In a regulated power supply using zener diode the unregulated input voltage must be larger as compared to the regulated output voltage.

    Correct Option: C

    In a regulated power supply using zener diode the unregulated input voltage must be larger as compared to the regulated output voltage.

  1. The current in the diode D1 of question no. 18 if V = 110 V is—








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    At 110V, diode D1 undergoes breakdown first. As a result diode D2 which remains reverse biased limits the current in circuit to its saturation current value i.e. 2mA.

    Correct Option: C

    At 110V, diode D1 undergoes breakdown first. As a result diode D2 which remains reverse biased limits the current in circuit to its saturation current value i.e. 2mA.

  1. The saturation currents of the two diodes are 1 and 2mA. The breakdown voltages of the diodes are the same and are equal to 100V. The current in the diode D1 if V=90 V is—










  1. View Hint View Answer Discuss in Forum

    When, V = 90V, no diode reaches breakdown. Thus I is determined by smallest saturation current i.e. I = 1m A.

    Correct Option: B

    When, V = 90V, no diode reaches breakdown. Thus I is determined by smallest saturation current i.e. I = 1m A.

  1. In the figure shown below, D1 has reverse saturation current of 20 mA and reverse breakdown voltage of 100V, corresponding values for D2 are 50 mA and 50V. The current in the circuit is the circuit is—










  1. View Hint View Answer Discuss in Forum

    At 40V input no zener diode undergoes breakdown D 1 is forward biased and D2 reverse biased. So saturation current of D2 flows into the circuit. Thus 50 mA from A to B.

    Correct Option: C

    At 40V input no zener diode undergoes breakdown D 1 is forward biased and D2 reverse biased. So saturation current of D2 flows into the circuit. Thus 50 mA from A to B.