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 silicon crystal having a cross-sectional area of 0.001 cm2 and a length of 20 µm is connected to its ends to a 20 V battery. At T = 300 K, we want a current of 100 mA in crystal. The concentration of donor atoms to be added is—








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    R =
    V
    		=
    20V
    		 =
    20
    		= 200 Ω and from relation
    I100mA100 × 10–3

    R =
    ρl
    		=
    1
    		 ·
    1

    AσA

    or σ =
    I
    		=
    2 × 10–3

    RA200 × 0.001

    = 0.01 (Ω-cm)–1
    and σ = n0 q. µn for Si, µn = 1350
    0.01 = n0. (1.6 × 10–19). 1350
    n0 =
    0.01
    1.6 × 10–19 × 1350

    = 4.6 × 1013 cm–3
    since n0 >> ni (intrinsic concentration)
    it means n0 = Nd
    Hence alternative (B) is the correct choice.

    Correct Option: B

    R =
    V
    		=
    20V
    		 =
    20
    		= 200 Ω and from relation
    I100mA100 × 10–3

    R =
    ρl
    		=
    1
    		 ·
    1

    AσA

    or σ =
    I
    		=
    2 × 10–3

    RA200 × 0.001

    = 0.01 (Ω-cm)–1
    and σ = n0 q. µn for Si, µn = 1350
    0.01 = n0. (1.6 × 10–19). 1350
    n0 =
    0.01
    1.6 × 10–19 × 1350

    = 4.6 × 1013 cm–3
    since n0 >> ni (intrinsic concentration)
    it means n0 = Nd
    Hence alternative (B) is the correct choice.

  1. A silicon semiconductor sample at T = 300 K is doped with phosphorus atoms at a concentrations of 1015 cm– 3. The position of the Fermi level with respect to the intrinsic Fermi level is—








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    We know that,

    EF – EFi = kT InNd
    Ni

    = 26 meV In
    1015
    1.5 × 1010

    = 26 × 10–3
    1015
    		 eV
    1.5 × 1010

    = 0.287 eV
    ≅ 0.3 eV
    Hence alternative (A) is the correct choice.

    Correct Option: A

    We know that,

    EF – EFi = kT InNd
    Ni

    = 26 meV In
    1015
    1.5 × 1010

    = 26 × 10–3
    1015
    		 eV
    1.5 × 1010

    = 0.287 eV
    ≅ 0.3 eV
    Hence alternative (A) is the correct choice.

  1. A silicon sample contains acceptor atoms at a concentration of Na = 5 × 1015 cm–3. Donor atoms are added forming and n-type compensated semiconductor such that the Fermi level is 0.215 eV below the conduction band edge. The concentration of donors atoms added are—








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    We know that, n0 = Nd – Na = Nc e – {( EC – EF) / kT}
    for Si, Nc = 2.8 × 1019 cm–3
    given Na = 5 × 1015 cm–3
    Now, Nd = Na + Nc e – {( EC – EF) / kT}
    or Nd = 5 × 1015 + 2.8 × 1019 e – (215eV / 26 x 10-3eV)
    {˙.˙ given that Fermi level is 0.215 eV below the or Nd = 1.19 × 1016 cm–3 conduction band i.e. Ec – EF = 0.215 eV}
    Hence alternative (A) is the correct choice.

    Correct Option: A

    We know that, n0 = Nd – Na = Nc e – {( EC – EF) / kT}
    for Si, Nc = 2.8 × 1019 cm–3
    given Na = 5 × 1015 cm–3
    Now, Nd = Na + Nc e – {( EC – EF) / kT}
    or Nd = 5 × 1015 + 2.8 × 1019 e – (215eV / 26 x 10-3eV)
    {˙.˙ given that Fermi level is 0.215 eV below the or Nd = 1.19 × 1016 cm–3 conduction band i.e. Ec – EF = 0.215 eV}
    Hence alternative (A) is the correct choice.

  1. A sample of silicon at T = 300 K is doped with boron at a concentration of 2.5 × 103 cm–3 and with arsenic at a concentration of 1 × 103 cm–3. The material is—








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    Since Na > Nd → It means material is p-type
    Po = Na – Nd
    or Po = (2.5 × 103 – 1.0 × 103) cm–3
    or Po = 1.5 × 103 cm–3

    Correct Option: A

    Since Na > Nd → It means material is p-type
    Po = Na – Nd
    or Po = (2.5 × 103 – 1.0 × 103) cm–3
    or Po = 1.5 × 103 cm–3

  1. In a bipolar transistor at room temperature, if the emitter current is doubled the voltage across its base-emitter junction—








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    NA

    Correct Option: C

    NA