Imagine a PN junction in forward bias mode. The conventional current goes from the p-side to the n-side. However, as mobile holes move to the n-side, aren't there mobile electrons on the n-side which will combine with the mobile holes? Once they combine, they can no longer move. So how is it possible there is current?

Also, I have a second question. Because they recombine, doesn't that form a depletion region in forward-bias mode? The depletion region acts like an insulator which would completely block off current even in forward bias mode, right?

  • $\begingroup$ Yes, the recombine in the junction. Negative electrons flow in one way, positive holes flow in the other - current flows through the device. And, no, the depletion region does not act as an insulator. Go back and think how one forms in the first place. $\endgroup$
    – Jon Custer
    Dec 5 '15 at 23:47

So how is it possible there is current?

Mobile electrons are the (plentiful) majority charge carriers in the N type region whilst mobile holes are the (plentiful) majority carriers in the P type region

The fact that the holes that diffuse to the N type region recombine there does not take away from the fact that they crossed the depletion region contributing to the total current through that region.

Similarly, electrons that diffuse to the P type region recombine there but, as with the holes, they crossed the depletion region and contributed to the total current through that region.

Since the device is overall charge neutral, the (total) current through the depletion region is the current through the device (don't forget that there is an attached external circuit through which the current circulates).

Because they recombine, doesn't that form a depletion region in forward-bias mode?

I don't follow your reasoning; the depletion region forms at the junction on the, e.g., N type side since electrons diffuse to the P-type side leaving behind ions.

But when an external circuit is connected with the appropriate polarity, electrons flow from the external circuit into the N-type side which reduces the width of the depletion region.

  • $\begingroup$ I've understood your explanation for the second question but I'm still having trouble with the first question. Once all the electrons move from the n-side and combine with holes on the p-side, there are no longer any mobile current carriers in the pn junction. There is indeed an external circuit which moves electrons through the PN junction but how can these external electrons move through the PN junction if there are no mobile carriers in the PN junction? I thought a semiconductor is an insulator when it is not doped and the whole point of doping is to make it more like a conductor. $\endgroup$
    – Travis
    Dec 6 '15 at 18:47

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