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. A material has conductivity of a 10–2 mho/m and a relative permittivity of 4, the frequency at which the conduction current in the medium is equal to the displacement current is—








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    NA

    Correct Option: A

    NA

  1. The diffusion capacitance of a forward biased p+ – n junction diode with a steady current I depends on—








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    The diffusion capacitance of a forward biased p+–n junction is given by

    CD =
    I
    VT

    Where, = Mean life time of charge carriers
    I = Forward current
    CD = Diffusion capacitance
    VT = Volt equivalent of temperature p+ – n → means p-layer is heavily doped than n-layer. It means the current at the junction is entirely due to holes. Thus diffusion capacitance of a forward biased p+ – n junction diode with a steady current I depends on mean life time of the holes.

    Correct Option: B

    The diffusion capacitance of a forward biased p+–n junction is given by

    CD =
    I
    VT

    Where, = Mean life time of charge carriers
    I = Forward current
    CD = Diffusion capacitance
    VT = Volt equivalent of temperature p+ – n → means p-layer is heavily doped than n-layer. It means the current at the junction is entirely due to holes. Thus diffusion capacitance of a forward biased p+ – n junction diode with a steady current I depends on mean life time of the holes.

  1. The figure given below shows the transfer characteristics of which one of the following—










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    The given circuit (i.e., fig)
    Figure shows the transfer characteristics of Peak clipper.


    Correct Option: A

    The given circuit (i.e., fig)
    Figure shows the transfer characteristics of Peak clipper.


  1. In the given circuit, D1 is an ideal germanium diode and D2 is a silicon diode having its cut-in voltage as 0·7 V, forward resistance as 20 Ω and reverse saturation current (Is) as 10 nA. What are the values of I and V for this circuit, respectively?










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    The given circuit (i.e., fig-1)
    Given D1 → Ideal Germanium diode D2 → Silicon diode with
    Vr = 0·7V and
    Vf = 20Ω and I0 = 10 nA
    The equivalent circuit of the given circuit using given information can be redrawn as(i.e., fig-2)

    From above circuit, V = 0V and I =
    V
    		loge e
    2VT

    Hence alternative (A) is the correct choice.


    Correct Option: A

    The given circuit (i.e., fig-1)
    Given D1 → Ideal Germanium diode D2 → Silicon diode with
    Vr = 0·7V and
    Vf = 20Ω and I0 = 10 nA
    The equivalent circuit of the given circuit using given information can be redrawn as(i.e., fig-2)

    From above circuit, V = 0V and I =
    V
    		loge e
    2VT

    Hence alternative (A) is the correct choice.


  1. Assuming an operating temperature T = 300 K and corresponding VT = 26 mV, what is the change in semiconductor silicon diode forward voltage VD to produce a 10: 1 change in diode current ID, while operating in the forward bias region (<25mA)?








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    We know that diode current equation is given by

    I = I0 e V- 1
    VT

    I = I0 e V
    VT
    [For forward bias]
    Since, = 2 for Si diode
    I = I0 V
    2VT
    Where, I = Forward current
    I0 = Leakage current
    V = Applied forward voltage
    VT = Volt equivalent of temperature (Given VT = 26 mV)
    From equation
    I1
    		e
    V1 – V2
    I22VT

    Given
    I1
    		=
    10
    		then
    I21

    10 e
    V
    		(V = change in forward voltage)
    2V2

    or 10 loge 10 =
    V
    		loge e
    2VT

    or V 2VT loge10
    2 × 26 mV × loge10 119·7 mV
    Hence alternative (B) is the correct choice.

    Correct Option: B

    We know that diode current equation is given by

    I = I0 e V- 1
    VT

    I = I0 e V
    VT
    [For forward bias]
    Since, = 2 for Si diode
    I = I0 V
    2VT
    Where, I = Forward current
    I0 = Leakage current
    V = Applied forward voltage
    VT = Volt equivalent of temperature (Given VT = 26 mV)
    From equation
    I1
    		e
    V1 – V2
    I22VT

    Given
    I1
    		=
    10
    		then
    I21

    10 e
    V
    		(V = change in forward voltage)
    2V2

    or 10 loge 10 =
    V
    		loge e
    2VT

    or V 2VT loge10
    2 × 26 mV × loge10 119·7 mV
    Hence alternative (B) is the correct choice.