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Diffusion coefficient of electrons at 30°C

Short answer

Using the Einstein relation D = μ kT/e. At 30°C kT/e = 0.02612 V, so D_e = 1300 × 0.02612 = 34.0 cm²/s.

Calculate the diffusion coefficient of electrons at 30°C in silicon doped with 10¹⁵ Arsenic atoms cm⁻³. Given that the drift mobility of electron with 10¹⁵ cm⁻³ dopants is 1300 cm²V⁻¹s⁻¹.

The diffusion coefficient is found from the Einstein relation, which links diffusion and drift of the same carrier:

Deμe=kTe\frac{D_e}{\mu_e} = \frac{kT}{e}

Given: μe=1300 cm2V−1s−1\mu_e = 1300\ \text{cm}^2\text{V}^{-1}\text{s}^{-1} (for 101510^{15} cm⁻³ As), T=30∘C=303.15T = 30^\circ\text{C} = 303.15 K, k=8.617×10−5k = 8.617\times10^{-5} eV K⁻¹.

Thermal voltage

kTe=8.617×10−5×303.15=0.02612 V\frac{kT}{e} = 8.617\times10^{-5}\times303.15 = 0.02612\ \text{V}

Diffusion coefficient

De=μekTe=1300×0.02612=33.96 cm2s−1\begin{aligned} D_e &= \mu_e\frac{kT}{e} = 1300\times0.02612\\ &= 33.96\ \text{cm}^2\text{s}^{-1} \end{aligned}

Answer: De≈34.0 cm2s−1=3.40×10−3 m2s−1D_e \approx 34.0\ \text{cm}^2\text{s}^{-1} = 3.40\times10^{-3}\ \text{m}^2\text{s}^{-1}.

Arsenic doping of 101510^{15} cm⁻³ is light, so the material is non-degenerate and the Einstein relation in this simple form is valid. The doping matters only through the mobility value given.

✓ Answer checked by Claude Opus 5.5 (high effort)

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