# P-N junction voltage under polarization

All the books that deal with the p-n junction under applied bias assume that the same equations used for the equilibrium case (no bias) can be used for the biased case provided that the junction voltage $V_{ic}$ is substituted bt $V_{ic}-V$, where $V$ is the applied bias. Although this can seem plausible I don't see any reason for why the junction voltage drop should be the $V_{ic}-V$, i.e., why are these voltages simply added?

Is there any reason on the ground of fundamental electrostatic / thermodynamics?

Thanks.

• The things you claim don't imply that they obey a superposition principle. You could only say that they obey superposition principle if the relevant calculated quantities were linear in the voltages, e.g. for current $I(V_1+V_2)=I(V_1)+I(V_2)$. This is surely not true for semiconductors. Here, the things are nonlinear. One only adds a bias and the bias means - it is defined - as effectively reducing the voltage. Jul 22 '14 at 9:00
• Biasing = "connecting a battery of V volts to the ends of the p-n junction". Why applying V volts at the ends of the junction should result in adding that V volts to the voltaje drop of the deplection región is not obvious at all. That's why I said "sort of" superposition, but you can overlook that phrase, is irrelevant for the question. Essentially your comment says that the voltages are added by definition of "addition". My question is why should they be added. Jul 22 '14 at 11:22

The built-in potential, $V_{ic}$, is obtained by analyzing the equilibrium situation, where drift and diffusion currents cancel each other out in the depletion region. This built-in potential is necessary for current not to flow in the system.

When you apply an external voltage you expect some current to flow, so the built-in potential must be altered, such that a net current can flow in the system. The question is, why is the change in the built-in potential exactly that of the applied voltage?

The reason is that nowhere else in the system can you have such a potential drop. The non-depleted regions of the p and n sides have such large carrier concentrations that a very small (negligible) electric field on them is enough to drive a current across the system. The negligible electric field results in a negligible potential drop across the non-depleted p and n regions.

Put another way, the depletion region has much less carrier concentration, and therefore presents the largest resistance to current flow in the system. The electric field associated with the voltage $V$ adds to the built-in field in the depletion region, such that the proper amount of current can flow.

Of course, the better way to approach this problem is to start out with the equations that govern transport in semiconductors (drift, diffusion, generation+recombination, etc.) and treat the applied voltage as a difference between quasi-Fermi levels on the two ends of the junction.

The numerical solution to the set of differential equations will show that the results of the depletion model (for the depletion region width) are a good approximation if the built-in potential $V_{ic}$ is replaced by

$$V_{ic} - V$$

In introductory textbooks, this result is simply quoted as if it was obvious. It turns out is not that obvious, but it is a pretty good approximation for this situation.

• Thanks for your answer Pedro. Do you know any text which treats this topic with the equations instead of assuming it from scratch? Dec 1 '14 at 7:14
• I have found the discussion in Sah's book is very good (it is in section 531). Even though the differential equations can be set up and solved, an analytical formula is not available (perhaps not what you wanted to hear). Nevertheless, the thickness of the space-charge region in the depletion model with a modified built-in potential is a good approximation. Dec 1 '14 at 16:33