The lifetime of minority carriers ($\tau_\text{n}$ for electrons, $\tau_\text{p}$ for holes) represents the average time before recombination. But since an electrons must have a hole for recombination to take place, why not $\tau_\text{n} \equiv \tau_\text{p}$ by definition?


$$\text{Figure 1. Hole and Electron Bulk Lifetime for different Silicon Doping [3-4]}$$

"The Minority Carrier Lifetime in Silicon Wafer. Bulk and Surface recombination process."


1 Answer 1


Remember that the minority carrier lifetime is the hole lifetime in $n$ type semiconductors and the electron lifetime in $p$ type semiconductors.

In $n$ type conductors you have lots of electrons from donor gap states but only a few holes in the valence band. So the lifetime is the time for a hole to recombine with one of the large excess of electrons. In $p$ type conductors you have lots of holes due to electron promotion into acceptor states but only a few electrons in the conduction band. In this case the lifetime is the time for an electron to recombine with one of the large excess of holes. There's no obvious reason why the lifetimes should be the same.

If you look at the diagram you mention, note that the $x$ axes are not the same in the two graphs. The left graph shows donor state density on the $x$ axis while the right graph shows acceptor state density on the $x$ axis.

I wonder if you have implicitly assumed an intrinsic semiconductor, where the hole and electron densities are necessarily the same. In that case you would expect the lifetimes to be the same, and indeed if you look at the two graphs you'll see the low doping lifetime is around $10^{-3}$ seconds in both cases.

  • $\begingroup$ so the left and right graphs are not identical only because of the differences in effective mass? if yes, then I would expect the lower effective mass particle to have a shorter lifetime ($m_n^{*}<m_p^{*}\ \Rightarrow\ \tau_n < \tau_p$), however the opposite is shown in these graphs. $\endgroup$
    – Sparkler
    Oct 11, 2014 at 16:44

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