For steady state optical excitation, we can write the hole diffusion equation as Dp d²δp/dx²=δp/τp - gop Assume that a long p^+ - n diode is uniformly illuminated by an optical signal, resulting in gop EHP/cm³-s. Calculate the hole diffusion current Ip(xn) and evaluate it at xn = 0. Compare

Question

For steady state optical excitation, we can write the hole diffusion equation as

[latex]D_p \frac{d^2δp}{dx^2}=\frac{δp}{τ_p} - g_{op}[/latex]

Assume that a long p[latex]^+[/latex] - n diode is uniformly illuminated by an optical signal, resulting in [latex]g_{op}[/latex] EHP/cm³ -s. Calculate the hole diffusion current [latex]I_p(x_n)[/latex] and evaluate it at [latex]x_n = 0[/latex]. Compare the result with Eq. (8–2) evaluated for a p[latex]^+[/latex] -n junction.

(8-2): [latex]I=I_{th}(E^{qV/kT}-1)-I_{op} \ \quad \quad \quad I=qA\Big(\frac{L_p}{ au_P}p_n+\frac{L_n}{ au_n}n_p\Big)(e^{qV/kT}-1)-qAg_{op}(L_p+L_n+W)[/latex]

Accepted Answer

[latex]\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} [/latex]

(a) [latex]\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}[/latex]

(b) [latex]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} [/latex]

The last equation corresponds to Eq. (8–2) for [latex]n_p \ll p_n[/latex], except that the component due to generation on the p side is not included.

Question 8.1: For steady state optical excitation, we can write the hole d...

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