Question 8.1: For steady state optical excitation, we can write the hole d...
For steady state optical excitation, we can write the hole diffusion equation as
D_p \frac{d^2δp}{dx^2}=\frac{δp}{τ_p} – g_{op}
Assume that a long p^+ – n diode is uniformly illuminated by an optical signal, resulting in g_{op} EHP/cm³ -s. Calculate the hole diffusion current I_p(x_n) and evaluate it at x_n = 0. Compare the result with Eq. (8–2) evaluated for a p^+ -n junction.
(8-2): I=I_{th}(E^{qV/kT}-1)-I_{op} \\ \quad \quad \quad I=qA\Big(\frac{L_p}{\tau_P}p_n+\frac{L_n}{\tau_n}n_p\Big)(e^{qV/kT}-1)-qAg_{op}(L_p+L_n+W)
\frac{d^2\delta p}{dx_n^2}=\frac{\delta p}{L_p^2}-\frac{g_{op}}{D_p}\\ \space \\ \delta p(x_n)=Be^{-x_n/L_p}+\frac{g_{op}L_p^2}{D_p} \\ \space \\ At \space x_n=0,\space \delta p(0)=\Delta p_n. \space Thus, \space B=\Delta p_n -\frac{g_{op}L_p^2}{D_p}
(a) \delta p(x_n)=[p_n(e^{qV/kT}-1)-g_{op}L_p^2/D_p]e^{-x_n/L_P}+g_{op}L_p^2/D_p \\ \space \\ \frac{d\delta p}{dx_n}=-\frac{1}{L_p}[\Delta p_n-g_{op}L_p^2/D_p]e^{-x_n/L_p}
(b) I_p(x_n)=-qAD_p\frac{d\delta p}{dx_n}=\frac{qAD_p}{L_p}[\Delta p_n-g_{op}L_p^2/D_p]e^{-x_n/L_p} \\ \space \\ I_p(x_n=0)=\frac{qAD_p}{L_p}p_n(e^{qV/kT}-1)-qAL_pg_{op}
The last equation corresponds to Eq. (8–2) for n_p \ll p_n, except that the component due to generation on the p side is not included.