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 Incoming links: Conductivity ratio of doped and intrinsic semiconductors

Past exam of the physics A course of the University of Cambridge / 2017 / Paper 2 / Section C / C11 / Carrier density and the conductivity-ratio assumption Created 2026-10-18 Updated 2026-10-20

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The atomic number density is

n_{at} = \frac{5300 kg m ^{- 3}}{0.073 kg mo l ^{- 1}} N_{A} = 4.372 \times 1 0^{28} m^{- 3} .

(1)

One fully ionised single-electron donor per million atoms gives

N_{D} = 4.372 \times 1 0^{22} m^{- 3}, N_{D} / n_{i} = 1.46 \times 1 0^{3}, n_{i} = 3 \times 1 0^{19} m^{- 3} .

(2)

The usual simplified answer treats conductivity as proportional to electron concentration and assumes unchanged electron mobility, giving about $1.46 \times 1 0^{3}$ times the undoped conductivity. There is a qualification: the conductivity ratio of doped and intrinsic semiconductors is not fixed by electron counts alone. Intrinsic material also has

p = n_{i}

, whereas the heavily n-type sample has

n ≃ N_{D}

and

p ≃ n_{i}^{2} / N_{D}

. Therefore

\frac{σ _{d}}{σ _{i}} ≃ \frac{N _{D} μ _{e, d}}{n _{i} ( μ _{e, i} + μ _{h, i} )} .

(3)

With unchanged electron mobility and negligible intrinsic hole conductivity this reduces to

1.46 \times 1 0^{3}

; with equal, unchanged electron and hole mobilities it is about

7.29 \times 1 0^{2}

. The problem supplies no mobilities, so the density-ratio estimate requires its stated approximation. Ionised impurities may additionally reduce mobility.

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