Physical electronics devices and ics miscellaneous
Direction: A semiconductor has following parameters. µa = 7500 cm2/ V - s,
µp = 300 cm2/ V - s,
ni = 3.6 × 1012 cm–3
- When conductivity is minimum, the hole concentration is—
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Given data
µn = 7500 cm2/ V-s
µP = 300 cm2/ V-s
ni = 3.6 × 1012 cm–3
we know that overall conductivity is given by relationσ = q (µn n + µP P) where, n = n2i P or σ = 9 µn. n2i + 9. µP. P …(A) P
for max. or min. value of conductivity, differentiate equation (A) w.r.t. hole or electron concentration and put equals to zero.dσ = - 9µnni 2
+ 9µPdP P2 0 = – 9µnni 2 + 9µP P2 or P = ni 
µn 
1/2 µP or P = 3.6 × 1012 7500 1/2 300
or P = 7.2 × 1012 (25)1/2
or P = 7.2 × 5 × 1012
or P = 36.0 × 1012 cm–3
Hence alternative (A) is the correct choiceCorrect Option: A
Given data
µn = 7500 cm2/ V-s
µP = 300 cm2/ V-s
ni = 3.6 × 1012 cm–3
we know that overall conductivity is given by relationσ = q (µn n + µP P) where, n = n2i P or σ = 9 µn. n2i + 9. µP. P …(A) P
for max. or min. value of conductivity, differentiate equation (A) w.r.t. hole or electron concentration and put equals to zero.dσ = - 9µnni 2
+ 9µPdP P2 0 = – 9µnni 2 + 9µP P2 or P = ni µn 1/2 µP or P = 3.6 × 1012 7500 1/2 300
or P = 7.2 × 1012 (25)1/2
or P = 7.2 × 5 × 1012
or P = 36.0 × 1012 cm–3
Hence alternative (A) is the correct choice
- A gallium arsenide semiconductor at T = 300 K is doped with impurity concentration Nd = 1016 cm–3. The mobility µa is 7500 cm2/V - s. For an applied field of 10 V/cm the drift current density is—
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Given data
Nd = 1016 cm–3
µn = 7500 cm2/ V-s
E = 10 v/cm
Jdrift =?
Jdrift = 9. µn. n.
E = 1.6 × 10–19 × 7500 × 1016 × 10
∴ Nd >> ni = 120 A/ cm2Correct Option: A
Given data
Nd = 1016 cm–3
µn = 7500 cm2/ V-s
E = 10 v/cm
Jdrift =?
Jdrift = 9. µn. n.
E = 1.6 × 10–19 × 7500 × 1016 × 10
∴ Nd >> ni = 120 A/ cm2
- In a GaAs sample the electrons are moving under an electric field of 5 kV/cm and the carrier concentration is uniform at 1016 cm– 3. The electron velocity is the saturated velocity of 107 cm/s. The drift current density is—
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Given data, E = 5 kV/ cm = 5000 V/ cm
n = 1016 cm–3
ν d = 107 cm/s
Jdrift =?
Jdrift = 9 µn. n. E = 9. n. νd (˙.˙ νd = µn. E)
= 1.6 × 10–19 × 1016 × 107
= 1.6 × 104 A/ cm2Correct Option: A
Given data, E = 5 kV/ cm = 5000 V/ cm
n = 1016 cm–3
ν d = 107 cm/s
Jdrift =?
Jdrift = 9 µn. n. E = 9. n. νd (˙.˙ νd = µn. E)
= 1.6 × 10–19 × 1016 × 107
= 1.6 × 104 A/ cm2
- In a silicon sample the electron concentration drops linearly from 1018 cm– 3 to 1016 cm– 3 over a length of 2.0 µm. The current density due to the electron diffusion current is (Dn = 35 cm2/s)—
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We know that electron current density
Jn = q Dn dn dx or Jn = 1.6 × 10–19 × 35 1018 – 1016 2 × 10–4
= 2.8 × 104 A/ cm2
Hence alternative (B) is the correct choice.Correct Option: B
We know that electron current density
Jn = q Dn dn dx or Jn = 1.6 × 10–19 × 35 1018 – 1016 2 × 10–4
= 2.8 × 104 A/ cm2
Hence alternative (B) is the correct choice.
- An n-type silicon sample has a resistivity of 5 Ω - cm at T = 300 K. The mobility is µa = 1350 cm2/ V - s. The donor impurity concentration is—
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Given data, ρ = 5 Ω-cm
µn = 1350 cm2/ V-s
Nd =?ρ = 1 = 1 σ 9µnNd 5 = 1 1.6 × 10–19 × 1350 × Nd
or Nd = 9.25 × 1014 cm–3Correct Option: B
Given data, ρ = 5 Ω-cm
µn = 1350 cm2/ V-s
Nd =?ρ = 1 = 1 σ 9µnNd 5 = 1 1.6 × 10–19 × 1350 × Nd
or Nd = 9.25 × 1014 cm–3
