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 built-in potential (Diffusion Potential) in a p-n junction—










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    (A), (B), (C) Since the built in potential (Diffusion potential) in a p-n junction
    * is equal to the difference in Fermi level of two sides
    * increases with increase in the doping level of two sides since

    VT =
    kT
    		In
    NAND

    qm2

    VT = kT q In NAND m2
    * increases with increase in temperature, since VT ∝ T.
    Hence alternative (A), (B) and (C) are the correct answer.

    Correct Option: E

    (A), (B), (C) Since the built in potential (Diffusion potential) in a p-n junction
    * is equal to the difference in Fermi level of two sides
    * increases with increase in the doping level of two sides since

    VT =
    kT
    		In
    NAND

    qm2

    VT = kT q In NAND m2
    * increases with increase in temperature, since VT ∝ T.
    Hence alternative (A), (B) and (C) are the correct answer.

  1. α cut-off frequency of a bipolar junction transistor increases with the—








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    We know that,

    fα = fT = hfe.
    1
    		…(i)
    2πRC

    where, fα = cut-off frequency of a BJT.
    hfe = Gain factor
    Capacitance C is given as
    C =
    ∈0A
    		…(ii)
    d

    Where
    ∈0 = free space permittivity
    C = capacitance.
    A = Area
    d = deplection width
    from the relation (ii) it is clear that with increase in base width C decreases which increases the fα i.e. α-cut off frequency.
    Hence alternative (A) is the correct answer.

    Correct Option: A

    We know that,

    fα = fT = hfe.
    1
    		…(i)
    2πRC

    where, fα = cut-off frequency of a BJT.
    hfe = Gain factor
    Capacitance C is given as
    C =
    ∈0A
    		…(ii)
    d

    Where
    ∈0 = free space permittivity
    C = capacitance.
    A = Area
    d = deplection width
    from the relation (ii) it is clear that with increase in base width C decreases which increases the fα i.e. α-cut off frequency.
    Hence alternative (A) is the correct answer.

  1. In the figure Vi is the input and Vo the output—










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    From above circuit it is clear that, the circuit clipps off the input Vi for Vi > VR.

    Correct Option: A


    From above circuit it is clear that, the circuit clipps off the input Vi for Vi > VR.


  1. In the circuit the input Vi has positive and negative swings. Vo is the output—








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    From the given circuit it is clear that the output will appear only when the diode will be in the conducting situation, and it will be possible only when Vi > VR, than V0 = VR.

    Hence alternative (C) is the correct answer.

    Correct Option: C

    From the given circuit it is clear that the output will appear only when the diode will be in the conducting situation, and it will be possible only when Vi > VR, than V0 = VR.

    Hence alternative (C) is the correct answer.


  1. Given gm = 2 mA/V, rd → ∞, |AV| =Vois given by—
    VS








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    Given gm = 2 mA/V, rd → ∞, |AV| =Vo=?
    VS

    AV =
    Vo
    		= gm R′L where R′L = (5 || 5) k = 2.5 k.
    VS

    or AV = 2 × 10– 3 × 2.5 × 103 = 5
    Hence alternative (B) is the correct answer.

    Correct Option: B

    Given gm = 2 mA/V, rd → ∞, |AV| =Vo=?
    VS

    AV =
    Vo
    		= gm R′L where R′L = (5 || 5) k = 2.5 k.
    VS

    or AV = 2 × 10– 3 × 2.5 × 103 = 5
    Hence alternative (B) is the correct answer.