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. In zener and avalanche breakdown diodes, current flow is due to—








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    In zener and avalanche breakdown diodes, current flow is due to majority carriers.

    Correct Option: B

    In zener and avalanche breakdown diodes, current flow is due to majority carriers.

  1. Select the correct output (V0) wave shape for a given input (Vi) in clamping network shown below—










  1. View Hint View Answer Discuss in Forum

    The given circuit (i.e., fig -1)
    Equivalent circuit for the first half cycle is given below (i.e., fig -2)
    V0 = 2V and VC = 8V with mentioned polarity Now, the equivalent circuit for second half cycle is shown below (i.e., fig -3)
    V0 + 8 + 10 = 0
    or V0 = – 18V
    Hence alternative (D) is the correct choice.


    Correct Option: D

    The given circuit (i.e., fig -1)
    Equivalent circuit for the first half cycle is given below (i.e., fig -2)
    V0 = 2V and VC = 8V with mentioned polarity Now, the equivalent circuit for second half cycle is shown below (i.e., fig -3)
    V0 + 8 + 10 = 0
    or V0 = – 18V
    Hence alternative (D) is the correct choice.


  1. A voltage signal 10 sin t is applied to the circuit with ideal diodes, as shown in figure. The maximum and minimum values of the output waveform Vout of the circuit are respectively—










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    The given circuit (i.e., fig-1)
    For first half cycle the diode D2 will conduct while diode D1 will not conduct. Equivalent circuit for this case is shown below: (i.e., fig-2)
    For the second half cycle diode D1 will conduct while diode D1 will not conduct. Equivalent circuit for this case is shown below: (i.e., fig-3)
    From above figure
    – 10 KI – 4 – 10 KI + 10 = 0

    or I =
    6
    		mA
    20

    V0 = – 4 – 10 ×
    6
    		= – 4 – 3
    20

    = – 7V = V0 (min)
    Hence alternative (D) is the correct choice.


    Correct Option: D

    The given circuit (i.e., fig-1)
    For first half cycle the diode D2 will conduct while diode D1 will not conduct. Equivalent circuit for this case is shown below: (i.e., fig-2)
    For the second half cycle diode D1 will conduct while diode D1 will not conduct. Equivalent circuit for this case is shown below: (i.e., fig-3)
    From above figure
    – 10 KI – 4 – 10 KI + 10 = 0

    or I =
    6
    		mA
    20

    V0 = – 4 – 10 ×
    6
    		= – 4 – 3
    20

    = – 7V = V0 (min)
    Hence alternative (D) is the correct choice.


  1. At high frequencies, ordinary diodes does not work properly because of—








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    At high frequencies, ordinary diodes does not work properly because of charge storage.

    Correct Option: D

    At high frequencies, ordinary diodes does not work properly because of charge storage.

  1. In figure, a silicon diode is carrying a constant current of 1 mA. When the temperature of the diode is 20°C, VD is found to be 700 mV. If the temperature raises to 40°C, VD becomes approximately equal to—










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    We know that on increasing 1°C of temperature the cut-in-voltage Vr of the diode is approximately decreases by – 2·5 mV/°C.
    So, net increase in temperature = (40 – 20) = 20°C
    decrease in cut-in-voltage by = 700 mV – 20 × 2·5 mV = 650 mV (approximately)

    Correct Option: B

    We know that on increasing 1°C of temperature the cut-in-voltage Vr of the diode is approximately decreases by – 2·5 mV/°C.
    So, net increase in temperature = (40 – 20) = 20°C
    decrease in cut-in-voltage by = 700 mV – 20 × 2·5 mV = 650 mV (approximately)