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I have always argued to myself that drift and diffusion components of the current though a p-n-junction cancel for each type of carrier because any electron diffusing from n into p will sooner or later require the same electron to drift back. (And analogously for holes).

However, it now seems to me that the generation of EHPs (electron-hole-pairs) profoundly contributes to the drift current:

"The supply of minority carriers on each side of the junction required to participate in the drift component of current is generated by thermal excitation of electron-hole pairs." (In "Solid-State-Electronic Devices" , Streetman and Banerjee)

In thermal equilibrium, that implies an equal amount of recombination events.

Anyway, one can imagine an EHP being generated on the p-side, with the electron drifting and the hole diffusing to the n-side, where thy recombine. In that scenario, there is no hole drift and no electron diffusion, but hole diffusion and electron drift.

So, how in earth does one dare to argue that "the drift and diffusion components must cancel for each type of carrier"?

I feel like I am missing the real argument.

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    $\begingroup$ Your first sentence is nonsense. If drift and diffusion components cancel each other, the current through p-n junction is zero and diode has infinite resistivity. Which contradicts the facts. It seems that your question is incomplete: you mean no bias, or some other conditions? $\endgroup$
    – Misha
    Commented Jan 25, 2013 at 7:47
  • $\begingroup$ Well, the current though an UNbiased p-n-junction IS zero. I don't understand your problem. Resistancy can't really be defined when there is no bias, so I wouldn't argue with that. $\endgroup$
    – Konstantin Schubert
    Commented Jan 25, 2013 at 23:02
  • $\begingroup$ I see no mention of zero bias in the question. And even for zero bias photoexcited carriers generated inside p-n junction will give you non-zero total current (which means that components are not compensated). That's how photodiode works. $\endgroup$
    – Misha
    Commented Jan 26, 2013 at 7:10

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The concentration of electrons ($n$) and holes ($p$) in a semiconductor is determined by the Fermi-Dirac distribution function. As you correctly mentioned, the Electron-Hole Pair (EHP) generation and recombination rates are equal in thermal equilibrium. As a result, when the semiconductor is in thermal equilibrium, the concentrations $n = n_0$ and $p = p_0$ are constant (not necessarily equal). If for some reason these quantities are (say) depleted (i.e. $n < n_0$ and $p < p_0$) then the rate of EHP generation will exceed the rate of recombination until $n$ and $p$ are back to $n_0$ and $p_0$ respectively. In other words, the system returns to thermal equilibrium.

Now, in your analysis you are forgetting that there is EHP generation occurring on the n-side as well. When a EHP is generated on the n-side, the hole drifts and electron diffuses to the p-side. The current due to the electrons drifting to the n-side must cancel the current due to electrons diffusing to the p-side. Similarly, the current due to the holes diffusing to the n-side must cancel the holes drifting to the p-side. If they did not cancel each other perfectly, there would be an accumulation or depletion of holes or electrons on the n- or p-side depending on which current is dominant. When I say accumulation or depletion, I mean deviation of $n$ and $p$ from $n_0$ and $p_0$ respectively. And as I mentioned in the previous paragraph, deviation of $n$ or $p$ from $n_0$ and $p_0$ will cause the EHP generation and recombination rates to adjust such that the system returns to $n_0$ and $p_0$.

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  • $\begingroup$ I can't follow your logic. As you say, "If they did not cancel each other perfectly, there would be an accumulation or depletion of holes or electrons on the n- or p-side[...](which) will cause the EHP generation and recombination rates to adjust". In other words: Wouldn't it be possible for the electron drift and the hole diffusion to exceed the currents in the other direction, while the EHP generation on the p-side and the EHP recombination on the n-side exceed their respective counterparts. The equilibrium would still be given. $\endgroup$
    – Konstantin Schubert
    Commented Nov 26, 2012 at 22:04
  • $\begingroup$ The problem with that argument is that EHP generations occur in pairs whereas recombinations do not occur at the same rate for both species. For example, say $x$ electron-hole pairs are generated on the p-side. Then they enter the n-side. The electrons that drifted over to the p-side see a very small quantity of holes (since holes are in minority on the n-side) but the diffused hole sees a lot of electrons. As a result, the recombinations for both species does not occur at an equal rate. The probability of recombination is proportional to the product of the electron and hole concentrations $\endgroup$
    – NanoPhys
    Commented Nov 26, 2012 at 22:15
  • $\begingroup$ Are you sure about that: "the recombinations for both species does not occur at an equal rate". I don't quite agree. The number of electrons recombining on the n-side is still equal to the number of holes recombining there. Simply because recombination involves an electron and a hole. Obviously, for an individual electron, the recombination probability is lower, but since there are more electrons, the expectation value will be unaffected. $\endgroup$
    – Konstantin Schubert
    Commented Nov 26, 2012 at 22:22
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    $\begingroup$ Yes, I'm sure. I explained the reason in the previous post too (i.e. statistical probability of recombination). Hopefully this will clarify the misunderstanding: I was only talking about the recombination of the electrons and holes which traveled from the p-side over to the n-side. There are additional electrons and holes already present on the n-side. Those additional electrons and holes are participating in EHP recombinations independently. The "total" generation and recombination rates (which include carriers crossing and not crossing the junction) are equal on the n- and p- sides. $\endgroup$
    – NanoPhys
    Commented Nov 26, 2012 at 22:51
  • $\begingroup$ Thank you for explaining, I think that I understand now what you said in your first comment: The electron in the EHP that crossed from p to n is less likely to recombine than the hole that it came with. The latter will likely recombine with a different electron of the conduction band. However I still don't see how this affects the argument I made in my first comment. That is, I don't see why "The 'total' generation and recombination rates [..] are equal on the n- and p- sides "(respectively). $\endgroup$
    – Konstantin Schubert
    Commented Nov 26, 2012 at 23:13

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