Semiconductors — Question 10

Section 41-5 · Physics pages 6890–6891

Question 10 Surface recombination shortens the measured carrier lifetime A thin semiconductor slab of thickness d has nearly uniform excess density ∆n. Bulk recombination rate is ∆n/τb per volume. Each of its two broad faces removes carriers at flux S∆n, where S is the surface recombination velocity; edge losses are negligible. S∆n S∆n d Schematic; labels, not drawing lengths, specify the data. (a) Derive the effective lifetime using a carrier balance. (b) For d = 100 µm, τb = 10.0 µs, and S = 10.0 m/s, evaluate the lifetime and the fraction of losses occurring at the surfaces.
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Question 10 - Solution Surface recombination shortens the measured carrier lifetime S∆n S∆n d (a) Method and calculation Add∆n dt = −Ad∆n τb −2AS∆n. τ −1 eff= τ −1 b + 2S/d . (b) Interpretation and further calculation τeff= 1/(105 + 2 × 105) = 3.333 µs . fs = 2S/d 1/τb + 2S/d = 2/3 . Physical check The uniform-density approximation requires diffusion across the slab to be fast enough to prevent a strong depth gradient. Passivation increases lifetime by reducing S.

Question 10

Surface recombination shortens the measured carrier lifetime

A thin semiconductor slab of thickness d has nearly uniform excess density Δn\Delta n. Bulk recombination rate is Δn/τb\Delta n/\tau_b per volume. Each of its two broad faces removes carriers at flux SΔnS\Delta n, where S is the surface recombination velocity; edge losses are negligible.

Diagram shown in the original SVG worksheet.

  1. Derive the effective lifetime using a carrier balance.

  2. For d=100μmd=100\,\mu\mathrm m, τb=10.0μs\tau_b=10.0\,\mu\mathrm s, and S=10.0m/sS=10.0\,\mathrm{m/s}, evaluate the lifetime and the fraction of losses occurring at the surfaces.

Question 10 - Solution

Surface recombination shortens the measured carrier lifetime

Diagram shown in the original SVG worksheet.

(a) Method and calculation

AddΔndt=−AdΔnτb−2ASΔn.Ad\frac{d\Delta n}{dt}=-Ad\frac{\Delta n}{\tau_b}-2AS\Delta n.τeff−1=τb−1+2S/d.\boxed{\tau_{\mathrm{eff}}^{-1}=\tau_b^{-1}+2S/d}.

(b) Interpretation and further calculation

τeff=1/(105+2×105)=3.333μs.\boxed{\tau_{\mathrm{eff}}=1/(10^5+2\times10^5)=3.333\,\mu\mathrm s}.fs=2S/d1/τb+2S/d=2/3.\boxed{f_s=\frac{2S/d}{1/\tau_b+2S/d}=2/3}.

Physical check

The uniform-density approximation requires diffusion across the slab to be fast enough to prevent a strong depth gradient. Passivation increases lifetime by reducing S.