Semiconductor Electronics — Practice Quiz
ICSE · Class 12 · Physics
Try a 4-question quiz on Semiconductor Electronics for ICSE Class 12 Physics: tap an answer to check it and see why. 66 questions in the full chapter test.
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Quick Quiz: Semiconductor Electronics
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An intrinsic semiconductor has electron concentration equal to hole concentration, both equal to 2.4 × 10^19 m^-3. If it is doped so that the hole concentration becomes 4.4 × 10^22 m^-3, what is the electron concentration in m^-3?
A pure Ge sample has n_i = 2.4 × 10^19 m^-3. It is doped with indium at a concentration of 4.4 × 10^22 m^-3. If μ_e = 0.39 m^2 V^-1 s^-1 and μ_h = 0.19 m^2 V^-1 s^-1, what is the conductivity of the doped Ge in Ω^-1 m^-1?
A semiconductor has intrinsic carrier concentration 2 × 10^8 m^-3. After doping, the hole concentration becomes 4 × 10^10 m^-3. What is the new electron concentration in m^-3?
A pure Si sample at 300 K has n_i = 1.5 × 10^16 m^-3. If doping by indium raises the hole concentration to 4.5 × 10^22 m^-3, what is the electron concentration?
Sample Questions
In a doped semiconductor, n_e = 2 × 10^24 m^-3 and n_i = 2 × 10^19 m^-3. What is the hole concentration?
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2 × 10^14 m^-3
Use n_e n_h = n_i^2. So n_h = (2 × 10^19)^2 / (2 × 10^24) = 4 × 10^38 / 2 × 10^24 = 2 × 10^14 m^-3.
A semiconductor has an intrinsic carrier density of 1.5 × 10^16 m^-3 at 300 K. If the hole concentration in a doped sample is 4.5 × 10^22 m^-3, which statement about the sample is correct?
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It is a p-type semiconductor, Holes are the majority carriers
A higher hole concentration shows p-type behaviour. In p-type semiconductors, holes are the majority carriers and electrons are the minority carriers. The concentrations are not equal in a doped semiconductor.
A semiconductor obeys σ = e(n_eμ_e + n_hμ_h). If n_e = 1.31 × 10^16 m^-3, n_h = 4.4 × 10^22 m^-3, μ_e = 0.39 m^2 V^-1 s^-1, μ_h = 0.19 m^2 V^-1 s^-1, and e = 1.6 × 10^-19 C, what is the contribution of holes alone to conductivity?
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1.34 × 10^3 Ω^-1 m^-1
The hole contribution is σ_h = e n_h μ_h. Substituting values gives σ_h = (1.6 × 10^-19)(4.4 × 10^22)(0.19) = 1.34 × 10^3 Ω^-1 m^-1.
In an n-type semiconductor, n_e ≈ 5 × 10^22 m^-3 and n_i = 1.5 × 10^16 m^-3. What is the hole concentration approximately?
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4.5 × 10^9 m^-3
Use n_e n_h = n_i^2. So n_h = n_i^2 / n_e = (1.5 × 10^16)^2 / (5 × 10^22) = 2.25 × 10^32 / 5 × 10^22 = 4.5 × 10^9 m^-3.
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