Q6Electrical Materials
Question
Q.6. Calculate the conductivity of a semi-conductor material which has:
- Hole density = 10¹³/cm³
- Electron density = 10¹⁵/cm³
- Electron mobility = 1300 cm²/V-sec
- Hole mobility = 600 cm²/V-sec
Answer
With n = 10¹⁵/cm³ (μe = 1300 cm²/V·s) and p = 10¹³/cm³ (μh = 600 cm²/V·s): σ = e(nμe + pμh) = 1.6×10⁻¹⁹(1.3×10¹⁸ + 6×10¹⁵) ≈ 0.209 S/cm.
Both carrier types contribute to the conductivity of a semiconductor, each weighted by its concentration and mobility:
Substituting the given data (n = 10¹⁵ cm⁻³, μe = 1300 cm²V⁻¹s⁻¹, p = 10¹³ cm⁻³, μh = 600 cm²V⁻¹s⁻¹):
The conductivity of the sample is approximately 0.209 S/cm (equivalently, resistivity ρ = 1/σ ≈ 4.8 Ω·cm). Note that the electron term dominates by more than two orders of magnitude — the material is effectively N-type (electron majority 100:1, and electron mobility roughly double the hole mobility), so the hole contribution (6×10¹⁵ against 1.3×10¹⁸) alters the result by well under 1%, again illustrating that in extrinsic material the majority-carrier product nμ alone gives an excellent approximation to the total conductivity.
Physical interpretation of the mobility values: the given electron mobility (1300 cm²/V·s) and hole mobility (600 cm²/V·s) correspond closely to the accepted room-temperature values for lightly doped silicon, confirming the sample is silicon-based; the electron/hole mobility ratio of roughly 2.2:1 in silicon arises from the differing effective masses and scattering behaviour of electrons in the conduction band versus holes in the valence band — electrons in silicon's conduction-band valleys have a smaller conductivity effective mass, allowing them to accelerate more readily between scattering events (by phonons and ionized impurities) for the same applied field, and hence achieve higher drift velocity per unit field. This mobility asymmetry is precisely why, all else equal, N-type silicon is intrinsically more conductive than P-type silicon of the same doping density — a fact that has practical consequences in device design, such as the tendency to favour NMOS-type conduction channels for higher current drive in a given device area.
Comparison with the sister calculation: this problem is structurally identical to the resistivity calculation in Part C, Q.2(b) of the 2024 paper, except with electron and hole densities roughly two orders of magnitude lower here (n = 10¹⁵ cm⁻³ versus 8×10¹³ cm⁻³, but with an inverted electron-to-hole ratio of only 100:1 rather than 16:1) and a markedly lower assumed electron mobility (1300 cm²/V·s here versus 2.3×10⁴ cm²/V·s there, the latter being unrealistically high for silicon and more typical of a high-mobility compound semiconductor such as InSb or a very lightly doped, low-temperature sample). The resulting conductivities nonetheless both fall in the same broad 0.1-0.3 S/cm range typical of moderately doped semiconductor material used in practical devices, illustrating that conductivity is set by the product of density and mobility rather than by either factor alone — a sample with an order-of-magnitude lower doping density can still reach comparable conductivity if its mobility is correspondingly higher, which is exactly the trade-off exploited when selecting compound semiconductors over silicon for high-frequency and high-speed device applications where mobility, not just doping, is the limiting design parameter.
Temperature sensitivity of the result: both n and the mobilities used here are implicitly room-temperature (300 K) values; raising the temperature would leave the extrinsic donor/acceptor density n, p essentially unchanged (full ionization already holds at 300 K) but would reduce both μe and μh through increased phonon scattering, so σ computed by this same formula would decrease somewhat at higher temperature — the extrinsic-conduction regime of a lightly to moderately doped semiconductor, distinct from the strongly rising conductivity of the freeze-out or intrinsic regimes discussed elsewhere in this paper.