Question 4.3: As a numerical example, let us assume that 10^13 EHP/cm³ are...
As a numerical example, let us assume that 10^{13} EHP/cm³ are created optically every microsecond in a Si sample with n_0 = 10^{14} cm^{-3} and τ_n = τ_p = 2 μs. The steady state excess electron (or hole) concentration is then 2 × 10^{13} cm^{-3} from Eq. (4–14).
(4.14): \delta n=\delta p=g_{op}\tau_n
While the percentage change in the majority electron concentration is small, the minority carrier concentration changes from
p_0=n_i^2/n_0=(2.25\times 10^{20})/10^{14}=2.25\times 10^6cm^{-3} (equilibrium)
to
p=2\times 10^{13}cm^{-3} (steady state)
Note that the equilibrium equation n_0p_0=n_i^2 cannot be used with the subscripts removed; that is, np \neq n_i^2 when excess carriers are present. The steady state electron concentration is
n=n_0+\delta n=1.2\times 10^{14}=(1.5\times 10^{10})e^{(F_n-E_i)/0.0259}
where kT ≃ 0.0259 eV at room temperature. Thus the electron quasiFermi level position F_n – E_i is found from
F_n-E_i=0.0259\ln{(8\times 10^3)}=0.233\space eV
and F_n lies 0.233 eV above the intrinsic level. By a similar calculation, the hole quasi-Fermi level lies 0.186 eV below E_i (Fig. 4–11). In this example, the equilibrium Fermi level is 0.0259 ln(6.67 \times 10^3) = 0.228 eV above the intrinsic level.
