Doped Semiconductors — Question 10

Section 41-6 · Physics pages 6910–6911

Question 10 Quasi-Fermi separation quantifies injected carriers in a doped sample An n-type sample has dark n0 = 1021 m−3, p0 = n2 i /n0, and ni = 1016 m−3. Illumination adds equal excess densities ∆n = ∆p = 1019 m−3. Use np = n2 i e(Fn−Fp)/(kBT) and kBT = .0250 eV. Ev Ec Eg Fn Fp Schematic; labels, not drawing lengths, specify the data. (a) Find n and p and test whether low-level injection applies. (b) Calculate the electron-hole quasi-Fermi-level separation.
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Question 10 - Solution Quasi-Fermi separation quantifies injected carriers in a doped sample Ev Ec Eg Fn Fp (a) Method and calculation n = 1.01 × 1021, p = 1019 + 1011 ≃1019 m−3 . ∆n/n0 = .0100 ≪1 low-level injection. (b) Interpretation and further calculation Fn −Fp = kBT ln(np/n2 i ). Fn −Fp = .0250 ln(1.01000001 × 108) = .4608 eV . Physical check The minority population can rise enormously while the majority population changes little. A single equilibrium Fermi level cannot describe this illuminated steady state.

Question 10

Quasi-Fermi separation quantifies injected carriers in a doped sample

An n-type sample has dark n0=1021m−3n_0=10^{21}\,\mathrm{m^{-3}}, p0=ni2/n0p_0=n_i^2/n_0, and ni=1016m−3n_i=10^{16}\,\mathrm{m^{-3}}. Illumination adds equal excess densities Δn=Δp=1019m−3\Delta n=\Delta p=10^{19}\,\mathrm{m^{-3}}. Use np=ni2e(Fn−Fp)/(kBT)np=n_i^2e^{(F_n-F_p)/(k_BT)} and kBT=.0250eVk_BT=.0250\,\mathrm{eV}.

Diagram shown in the original SVG worksheet.

  1. Find n and p and test whether low-level injection applies.

  2. Calculate the electron-hole quasi-Fermi-level separation.

Question 10 - Solution

Quasi-Fermi separation quantifies injected carriers in a doped sample

Diagram shown in the original SVG worksheet.

(a) Method and calculation

n=1.01×1021,p=1019+1011≃1019m−3.\boxed{n=1.01\times10^{21},\quad p=10^{19}+10^{11}\simeq10^{19}\,\mathrm{m^{-3}}}.Δn/n0=.0100≪1low-level injection.\boxed{\Delta n/n_0=.0100\ll1}\quad\text{low-level injection}.

(b) Interpretation and further calculation

Fn−Fp=kBTln⁡(np/ni2).F_n-F_p=k_BT\ln(np/n_i^2).Fn−Fp=.0250ln⁡(1.01000001×108)=.4608eV.\boxed{F_n-F_p=.0250\ln(1.01000001\times10^8)=.4608\,\mathrm{eV}}.

Physical check

The minority population can rise enormously while the majority population changes little. A single equilibrium Fermi level cannot describe this illuminated steady state.