densité de porteurs en excès dans le volume de la diode

L'équation différentielle à résoudre est :$\frac{d^2\widehat{p}}{{\mathit{dx}}^2}-\frac{\widehat{p}}{L_p}=0$ ,

La solution est de la forme : $\widehat{p}(x)=A\exp \left(-\frac{x}{L_p}\right)+B\exp \left(\frac{x}{L_p}\right)$.

$\begin{array}{cc}\begin{array}{c}\text{en}x=0\mathrm{:}\widehat{p}={\widehat{p}}_0\\ \text{en}x=W_N\mathrm{:}\widehat{p}(W_N)=0\end{array}\}& \Rightarrow 0=A\exp \left(-\frac{W_N}{L_p}\right)+B\exp \left(\frac{W_N}{L_p}\right)\Rightarrow {\widehat{p}}_0=A+B.\end{array}$Zoom

$\Rightarrow 0={\widehat{p}}_0\exp \left(-\frac{W_N}{L_p}\right)+B\left[\exp \left(\frac{W_N}{L_p}\right)-\exp \left(-\frac{W_N}{L_p}\right)\right]\Rightarrow B=-{\widehat{p}}_0\frac{\exp \left(-\frac{W_N}{L_p}\right)}{\exp \left(\frac{W_N}{L_p}\right)-\exp \left(-\frac{W_N}{L_p}\right)}.$Zoom

et

$A={\widehat{p}}_0\left[1-\frac{\exp \left(-\frac{W_N}{L_p}\right)}{\exp \left(\frac{W_N}{L_p}\right)-\exp \left(-\frac{W_N}{L_p}\right)}\right]={\widehat{p}}_0\frac{\exp \left(\frac{W_N}{L_p}\right)}{\exp \left(\frac{W_N}{L_p}\right)-\exp \left(-\frac{W_N}{L_p}\right)}.$Zoom

$\widehat{p}(x)={\widehat{p}}_0\frac{\exp \left(\frac{W_N-x}{L_p}\right)-\exp \left(-\frac{W_N-x}{L_p}\right)}{\exp \left(\frac{W_N}{L_p}\right)-\exp \left(-\frac{W_N}{L_p}\right)}$Zoom

Comme $\exp (a)-\exp (-a)=2\text{sh}(a)$ , on peut écrire  : $\widehat{p}(x)=\widehat{p_0}\frac{\text{sh}\left(\frac{W_N-x}{L_p}\right)}{\text{sh}\left(\frac{W_N}{L_p}\right)}$.