Conductivity Of Semiconductors
Intrinsic germanium has a carrier concentration of (2.4 \times 10^{19},\text{m}^{-3}), with electron and hole mobilities of (0.39) and (0.19,\text{m}^2\text{V}^{-1}\text{s}^{-1}) respectively. What is the electrical conductivity of this pure sample?
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Solution
\(2.23\,\text{S m}^{-1}\)
Conduction in a semiconductor is carried by both electrons and holes drifting under the applied field, so the total conductivity is (\sigma = e,(n_e \mu_e + n_h \mu_h)). For an intrinsic sample the two carrier densities are equal, (n_e = n_h = n_i), allowing the expression to be written as (\sigma = n_i,e,(\mu_e + \mu_h)). It is important to keep both terms because, although electrons and holes carry opposite charge and drift in opposite directions under the field, their currents add rather than cancel; a positive hole moving one way and a negative electron moving the other both transport positive charge in the same sense. Inserting (n_i = 2.4 \times 10^{19},\text{m}^{-3}), (e = 1.6 \times 10^{-19},\text{C}) and ((\mu_e + \mu_h) = 0.58,\text{m}^2\text{V}^{-1}\text{s}^{-1}) gives (\sigma = (2.4 \times 10^{19})(1.6 \times 10^{-19})(0.58) \approx 2.23,\text{S m}^{-1}). The value 1.50 wrongly uses only the electron mobility, dropping the hole contribution. The value 0.89 mistakenly subtracts the mobilities rather than adding them. The value 4.46 double counts by using (2n_i) for each carrier. As a sanity check, the corresponding resistivity is about (0.45,\Omega\text{m}), which lies sensibly between metals and insulators, confirming a semiconductor-scale conductivity.
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About This Question
- Subject
- physics
- Chapter
- semiconductor electronics
- Topic
- conductivity of semiconductors
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
\(2.23\,\text{S m}^{-1}\)
Conduction in a semiconductor is carried by both electrons and holes drifting under the applied field, so the total conductivity is (\sigma = e,(n_e \mu_e + n_h \mu_h)). For an intrinsic sample the two carrier densities are equal, (n_e = n_h = n_i), allowing the expression to be written as (\sigma = n_i,e,(\mu_e + \mu_h)). It is important to keep both terms because, although electrons and holes carry opposite charge and drift in opposite directions under the field, their currents add rather than cancel; a positive hole moving one way and a negative electron moving the other both transport positive charge in the same sense. Inserting (n_i = 2.4 \times 10^{19},\text{m}^{-3}), (e = 1.6 \times 10^{-19},\text{C}) and ((\mu_e + \mu_h) = 0.58,\text{m}^2\text{V}^{-1}\text{s}^{-1}) gives (\sigma = (2.4 \times 10^{19})(1.6 \times 10^{-19})(0.58) \approx 2.23,\text{S m}^{-1}). The value 1.50 wrongly uses only the electron mobility, dropping the hole contribution. The value 0.89 mistakenly subtracts the mobilities rather than adding them. The value 4.46 double counts by using (2n_i) for each carrier. As a sanity check, the corresponding resistivity is about (0.45,\Omega\text{m}), which lies sensibly between metals and insulators, confirming a semiconductor-scale conductivity.
This hard difficulty physics question is from the chapter semiconductor electronics, covering the topic of conductivity of semiconductors. It appeared in the 2025 exam.
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