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 a particular semiconductor material following parameters are observed—
    µa = 1000 cm2/ V-s,
    µp = 600 cm2/ V-s,
    Nc = Nv = 1019 cm–3
    These parameters are independent of temperature. The measured conductivity of the intrinsic material is σ = 10–6 (Ω - cm)– 1 at T = 300 K. The conductivity at T = 500 K is—








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    Given data, µn = 1000 cm2/ V-s
    µP = 600 cm2/ V-s
    Nc = NV = 1019 cm–3
    Conductivity, σ = 10–6 (Ω-cm)–1 at temp, T = 300 K,
    Let the conductivity at T = 500 K is σ′
    Now, σ′ = 9 n′1 (µn + µp), at T = 300 K
    10–6 = 1.6 × 10–19. ni (1000 + 600)
    or ni = 3.9 × 109 cm–3
    Now calculate new n′1
    ni 2 = Ni. NV. e– (Eg/ kT)

    where Eg = kT In
    NCNV
    ni 2

    = 26 × 10–3 In
    1019 · 1019
    (3.91 × 109) 2

    = 1.122 eV
    at T = 500 K ⇒ kT = 26 × 10–3
    500
    		 = 0.432 eV
    300

    Now, n′ 2 i = 1019.1019. e (1.122 / 0.432) = 2.29 × 1013 cm–3
    σ′ = q n′i (µn + µp)
    = 1.6 × 10–19 × 2.29 × 1013 (1000 + 600)
    = 5.86 × 10–3 (Ω-cm)–1
    Hence alternative (D) is the correct choice.

    Correct Option: D

    Given data, µn = 1000 cm2/ V-s
    µP = 600 cm2/ V-s
    Nc = NV = 1019 cm–3
    Conductivity, σ = 10–6 (Ω-cm)–1 at temp, T = 300 K,
    Let the conductivity at T = 500 K is σ′
    Now, σ′ = 9 n′1 (µn + µp), at T = 300 K
    10–6 = 1.6 × 10–19. ni (1000 + 600)
    or ni = 3.9 × 109 cm–3
    Now calculate new n′1
    ni 2 = Ni. NV. e– (Eg/ kT)

    where Eg = kT In
    NCNV
    ni 2

    = 26 × 10–3 In
    1019 · 1019
    (3.91 × 109) 2

    = 1.122 eV
    at T = 500 K ⇒ kT = 26 × 10–3
    500
    		 = 0.432 eV
    300

    Now, n′ 2 i = 1019.1019. e (1.122 / 0.432) = 2.29 × 1013 cm–3
    σ′ = q n′i (µn + µp)
    = 1.6 × 10–19 × 2.29 × 1013 (1000 + 600)
    = 5.86 × 10–3 (Ω-cm)–1
    Hence alternative (D) is the correct choice.

  1. Six volt is applied across a 2 cm long semiconductor bar. The average drift velocity is 104 cm/s. The electron mobility is—








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    Given L = 2 cm, vd = 104 cm/s, V = 6 volt. drift velocity = µn. E

    vd = µn.
    V
    L

    104 = µn.
    6
    2

    or µn =
    104
    		= 3333 cm2/ V-s
    3

    Correct Option: D

    Given L = 2 cm, vd = 104 cm/s, V = 6 volt. drift velocity = µn. E

    vd = µn.
    V
    L

    104 = µn.
    6
    2

    or µn =
    104
    		= 3333 cm2/ V-s
    3

  1. A silicon sample doped n-type at 1018 cm–3 have a resistance of 10 Ω. The sample has an area of 10–6 cm2 and a length of 10 µm. The doping efficiency of the sample is (µa = 800 cm2/ V – s)—








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    Given data,
    Nd = 1018
    R = 10 Ω
    A = 10–6 cm2
    L = 10 mm = 10 × 10–4 cm,
    µn = 800 cm2/ V–s
    % doping efficiency = n0 Nd × 100 =?

    Now, n0 =
    L
    q·µn·A·R

    =
    10 × 10–4
    1.6 × 10–19 × 800 × 10–6 × 10

    = 7.81 × 1017 cm–3
    % doping efficiency =
    7.81 × 1017
    		× 100 = 78.1%
    1 × 1018

    Hence alternative (B) is the correct choice.

    Correct Option: B

    Given data,
    Nd = 1018
    R = 10 Ω
    A = 10–6 cm2
    L = 10 mm = 10 × 10–4 cm,
    µn = 800 cm2/ V–s
    % doping efficiency = n0 Nd × 100 =?

    Now, n0 =
    L
    q·µn·A·R

    =
    10 × 10–4
    1.6 × 10–19 × 800 × 10–6 × 10

    = 7.81 × 1017 cm–3
    % doping efficiency =
    7.81 × 1017
    		× 100 = 78.1%
    1 × 1018

    Hence alternative (B) is the correct choice.

  1. A thin film resistor is to be made from a GaAs film doped n-type. The resistor is to have a value of 2 kΩ. The resistor length is to be 200 µm and area is to be 10–6 cm2. The doping efficiency is known to be 90%. The mobility of electrons is 8000 cm2/ V–s. The doping needed is—








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    Given data, R = 2 kΩ = 2000 Ω
    L = 200 µm = 200 × 10–4 cm
    A = 10–6 cm2
    n0 = .9 Nd
    Now from relation

    R =
    L
    		⇒ n0
    L

    n0q·µnARqµnA

    .9 Nd =
    20 × 10–4
    2000 × 1.6 × 10–19 × 8000 × 10–6

    or Nd = 8.7 × 1015 cm–3
    Hence alternative (A) is the correct choice.

    Correct Option: A

    Given data, R = 2 kΩ = 2000 Ω
    L = 200 µm = 200 × 10–4 cm
    A = 10–6 cm2
    n0 = .9 Nd
    Now from relation

    R =
    L
    		⇒ n0
    L

    n0q·µnARqµnA

    .9 Nd =
    20 × 10–4
    2000 × 1.6 × 10–19 × 8000 × 10–6

    or Nd = 8.7 × 1015 cm–3
    Hence alternative (A) is the correct choice.

  1. The cross sectional area of silicon bar is 100 µm2. The length of bar is 1 mm. The bar is doped with arsenic atoms. The resistance of bar is—








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    Given data,
    A = 100 × 10–8 cm2
    L = 1 mm = 10–1 cm
    µn (for arsenic atom) = 1100
    Nd >> ni ÷ n0 = Nd = 5 × 1016 cm–3
    Now,

    R =
    L
    		=
    L

    QσAn0q·µn·A

    =
    10–1
    5 × 1010 × 1.6 × 10–19 × 1100 × 100 × 10–8

    ≅ 11.36 kΩ

    Correct Option: B

    Given data,
    A = 100 × 10–8 cm2
    L = 1 mm = 10–1 cm
    µn (for arsenic atom) = 1100
    Nd >> ni ÷ n0 = Nd = 5 × 1016 cm–3
    Now,

    R =
    L
    		=
    L

    QσAn0q·µn·A

    =
    10–1
    5 × 1010 × 1.6 × 10–19 × 1100 × 100 × 10–8

    ≅ 11.36 kΩ