I am not quite understand i-v character of PN-junction diode.

Here is the model in textbook. The PN junction diode can be divided into three regions. They are

  1. One depletion region near the PN interface, where no carrier but ionized atoms existed.
  2. Two quasineutral(QNR) regions extend from the boundary of depletion region to both ends of the diode.

This image is take from the post "Pn junction voltage drop?"

When positive voltage applied to this diode, current forms. In order to characterize the current, according to the model, the only thing one need to calculate is the minority diffusion within QNR. But, what about the majority carrier?

What is their concentration along the diode? Why don't we count their contribution to the diffusion?

While I guess that majority carriers also contribute diffusion by recombining with minority ones. In order to support diffusion (by maintaining concentration difference), excess carriers of both types must be injected into QNR.

Q1 Where do these injection carriers come from? Take the n-QNR for example, is it right to say that both electrons and holes are injected from p-QNR? If this is so, then carrier concentration in p-QNR would be lowered, and eventually, equals to zero, which I am sure is not the case.

Q2 Here near the interface between depletion region QNR, do the relationship n*p = ni^2 still hold? It seemed that it has been broken by both n and p injected.

Q1 and Q2 are quite clearly answered. With the explanation, I am now concerning with the number of carriers in this diode. Take the electrons for example, they are born in n-QNR by thermal excitation, then overcome the potential barrier in depletion region, and finally recombine (after some distance of diffusion) in p-QNR.

For each recombination, however, #electrons and #holes decreases by one. Since current flows continuously, a new electron-hole pair must be generated in n-QNR, otherwise, #electrons that can go through depletion regions would become zero eventually.

Q3 What about the holes being generated in n-QNR, where would they go? If they go through the depletion region (being accelerated) and become majority in p-QNR, are they risk of being recombined before going through the depletion region (in n-QNR)?

Q4 As bias increased, more current flows. Does it imply larger generation rate (in #/time-1 volume-1) of electron-hole-pair. If so, extra energy must be supplied (compare to the T.E case), since the #electrons with high K.E increase but the temperature remains. Where do these extra energy come from?

  • 1
    $\begingroup$ If, as you say, you like the answer to your Q1 and Q2, you should accept that post, and then ask Q3 and Q4 in a new question. $\endgroup$
    – Art Brown
    Commented Feb 1, 2015 at 9:12

1 Answer 1


A1: Under forward bias, the current-carrying carriers start out as thermally excited majority carriers (majority on the side they come from) with high kinetic energy. Because of their thermal excitation they have enough energy to overrun the potential barrier in the depletion region, and once they cross the depletion region they are now low kinetic energy minority carriers (on the far side). They then diffuse around a bit as minority carriers and then finally recombine.

A2: At equilibrium, $n_n n_p = n_i^2$ holds everywhere inside the semiconductor, both in the depletion region and outside of it. When forward bias is applied, $n_n n_p > n_i^2$ in the depletion region since extra carriers are being injected. When reverse bias is applied, $n_n n_p < n_i^2$ in the depletion region since carriers are being sucked out.

  • $\begingroup$ A clear story for the life circle of minority carriers. With your explanation, I am now concerning with the number of carriers in this diode, see Q3 and Q4 in the updated post. Thanks. $\endgroup$
    – user37528
    Commented Jan 20, 2014 at 9:08

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