I need some help in numerically solving the nonlinear Poisson's equation for electrons in frequency domain. The steady-state equation is: \begin{equation} \nabla.(\epsilon\nabla\varphi) = q\left(n_i\exp\left(\dfrac{\varphi-\Phi_n}{V_\text{T}}\right) -N_D\right) \,. \end{equation}

In small-signal analysis, we have $\varphi = \varphi_{\text{DC}} + \delta\varphi$, with $\delta\varphi <<\varphi_{\text{DC}}$. This yields: \begin{equation} \nabla.\Big[\epsilon\nabla(\varphi_{\text{DC}} + \delta\varphi)\Big] = q\left(n_i\exp\left(\dfrac{\varphi_{\text{DC}} + \delta\varphi-\Phi_n}{V_\text{T}}\right) -N_D\right) \end{equation} which I can further simplify to: \begin{equation} \nabla.(\epsilon\nabla\varphi_{\text{DC}}) + \nabla.(\epsilon\nabla\delta\varphi) = q\left(n_i\exp\left(\dfrac{\varphi_{\text{DC}} + \delta\varphi-\Phi_n}{V_\text{T}}\right) -N_D\right) \,\,. \end{equation}

I do not know how to proceed from here. That is, how should I treat the exponential term on the RHS and what are the basic assumptions in doing so?

  • 2
    $\begingroup$ Honest question. Normally a Fourier transform would convert a function of time to a function of frequency, but Poisson doesn't have a time dependence. It is not obvious to me how this can be done in this case. $\endgroup$ – boyfarrell Jan 8 '16 at 21:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.