I am struggling with some approximation while deriving I-V characteristic of a pn junction. Let's consider a quasi-neutral region on the n side of pn junction.
There are four current components as follows:
Jndiff: majority diffusion current
Jndrift: majority drift current
Jpdiff: minority diffusion current
Jpdrift: minority drift current

Because there is no field in the neutral region then there can be no net charge at any point in this region. Then the excess majority carrier concentration should follow the decay of the excess minority carrier concentration. This results to Jndiff = Dn/Dp* Jpdiff.
I can see that Jndiff and Jpdiff is on the same order of magnitude. Also Jndiff is larger than Jpdiff because Dn > Dp.
The total current density:
Jtotal = Jndiff + Jndift + Jpdiff + Jpdrift
The approximation is as follows: Jndiff << Jndift so we can ignore Jndiff.
(similarly Jpdrift << Jpdiff, so we can ignore Jpdrift)
This results in Jtotal = Jndrift + Jpdiff.

However, what I am confused is that we ignore Jndiff and keep Jpdiff while Jndiff and Jpdiff are on the same order of magnitude (Jndiff is even larger than Jpdiff). If we ignore Jndiff then we should also ignore Jpdiff because Jpdiff is smaller than Jndiff. Could anyone please explain it?

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1 Answer 1


You are obviously referring to the Shockley model of the pn-junction where the recombination and generation is neglected in the depletion region so that the electron and hole currents across the depletion region each are constant. Thus the total current of the pn-junction must be the sum of the minority currents at the depletion region edges, hole current at the n-side, and electron current at the p-side edge.

The majority carrier diffusion and drift currents on the n- and p-side do exist, but, due to the electron and hole current continuity across the depletion zone, at the depletion edges they are identical to the minority diffusion and drift currents at the respective opposite depletion edge. Therefore it suffices to consider the minor current at the depletion edges to obtain the total pn-junction current.

  • $\begingroup$ Thanks a lot, freecharly. It makes some sense now. However, I am still not completely comfortable with it. Let's assume that there is no electric field in quasi-neutral region, how is it possible to have the drift current there? $\endgroup$
    – emnha
    Mar 11, 2017 at 13:29
  • 1
    $\begingroup$ When there is no electric field in the quasi-neutral region, then there is no drift current, only diffusion current. This is also the assumption for the pn-junction minority currents at low injection levels, which is usually part of the Shockley model. $\endgroup$
    – freecharly
    Mar 11, 2017 at 17:21
  • $\begingroup$ Hi, for an ideal wire there is also absolutely no electric field along the wire. However, it still carry some current (possibly drift current). Is this correct or something misunderstood here? $\endgroup$
    – emnha
    Mar 11, 2017 at 17:52
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    $\begingroup$ @anhnha - When you consider normal (metal) conductors, there is always an electric field (which might be very small) necessary for a current to flow. In superconductors, which have no resistance, you can have electrical current flow without electrical fields. $\endgroup$
    – freecharly
    Mar 11, 2017 at 18:01
  • 1
    $\begingroup$ @anhnha - In the depletion region you have usually both drift and diffusion current. $\endgroup$
    – freecharly
    Mar 11, 2017 at 18:34

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