I would like to solve Poisson's Equation for a space charge region (fixed and free charges) that is located within a device of semiconductor material.

Unfortunately I do not have any boundary conditions available for the surface of the device.

I am now wondering if in my case I can work with the general solution to Poisson's Equation:

$$\phi(r) = \frac{1}{4\pi\epsilon_s} \int \frac{\rho(r')}{|r-r'|} d^3 r'$$

This equation is a solution to Poisson's Equation for the boundary condition $\phi=0 \;\text{for}\;|r - r'| \to \infty$.

I would believe that this boundary condition must also hold true for any charge carrier in my device (-> Coulomb's Law). Is this correct?

Is it a problem that the boundary condition is outside my volume of interest?

| cite | improve this question | | | | |

The Poisson equation inside the (homogeneous) semiconductor is

$\Delta \phi = - \frac{\rho}{\epsilon_0 \epsilon_r}$

whereas outside it, the relavite permittivity $\epsilon_r$ is different, e.g., if the material is sitting in vacuum

$\Delta \phi = - \frac{\rho}{\epsilon_0}$

The solution you propose does not fulfill both equations simultaneously. So, the answer you gave is correct only outside the device, where it fulfills Poisson's equation with your boundary condition as we already know.

But you want the potential inside the device. As you can tell, the only difference between the equations is a division by the relative permittivity at the RHS. In this case you should use the same formula but with an extra factor $\epsilon_r$ in the denominator. At last, you should not forget to add a constant, which pops up from the integration that you can use to make sure that the potential curve stays continuous at the boundary.

| cite | improve this answer | | | | |
  • $\begingroup$ Actually, I am only interested in finding solutions to Poisson's Equation within the (homogeneous) device. Can the fact, that the device sits in vacuum, then be ignored? Still, within my confined device volume for any charge it should be true V = 0 for |r - r'| -> Infinity. This boundary condition leads to the "general solution" stated above. Does that make sense? $\endgroup$ – Sweetheart Aug 27 '15 at 6:54
  • $\begingroup$ I find contradictory your sentence about solving for a region of interest even when the boundary is far away. In other words, if you are given a boundary, you have to find a path to it in order to use it. In that sense you cannot neglect the vacuum region, because you have to integrate somehow from your device to infinity, where your boundary is. But in general this boundary at infinity is used only to fix a dummy constant, it has no physical relevance. $\endgroup$ – Rol Aug 27 '15 at 18:31
  • $\begingroup$ I can see your point. However, I also think that for every single charge carrier in the material Coulomb's Law must be true. Coulomb's Law requires for V = 0 for |r - r'| -> Infinity, although infinity in my case is outside the material. Now, if I have a cloud of charges, the superposition principle applies. Then also V = 0 for |r - r'| -> Infinity should apply. I can also see a logic in this reasoning. $\endgroup$ – Sweetheart Aug 28 '15 at 6:25
  • $\begingroup$ @Torro That's right. Using Coulomb's law and superposition is the way to reason the problem, I was wondering why did you consider Poisson's equation. After all, you are going to use your semiconductor device in an electric circuit, thus you are only interested in potential differences, no need to assume $V=0$ at infinity. $\endgroup$ – Rol Aug 28 '15 at 7:39
  • $\begingroup$ The contradiction remains: Coulomb's Law can be derived from Poisson's Equation assuming V = 0 for |r - r'| -> Infinity. Thus, I would think that in my case (charge inside semiconductor) it should also be allowed to solve Poisson's Equation with this boundary condition. $\endgroup$ – Sweetheart Aug 28 '15 at 11:01

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.