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I want to clear my conception of the depletion region in an unbiased PN junction.

Before the formation of a depletion region in a PN junction, in the P region there is an excess of holes, aka places that free electrons can occupy.

Similarly, in the N region there is an excess of free electrons that have no place to occupy.

Then the diffusion starts, where the electrons from the N region move to the P region and some recombine with the holes present there.

How do the holes move? How can I visualize the movement of holes, since it really isn't a "particle" or an "ion" per se?

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  • \$\begingroup\$ If you really want an accurate description of the concept of hole you need to resort to at least some basic QM. Electrons are not particles but somewhat in-between particle and wave. You need to understand the dispersion relation E vs k and how the curvature is related to the concept of effective mass. Then you need to consider that electron in the valence band near the gap behave as if they had negative mass, while this farther down in energy have positive mass. Then you express the current of the positive mass electrons in a nonfilled valence band in terms of the current of holes. \$\endgroup\$ Commented Apr 22, 2023 at 2:12
  • \$\begingroup\$ 5 minutes are not enough to get rid of typos and stupid autocorrect alterations. We'll we'll we'll, I'll leave it as it is. \$\endgroup\$ Commented Apr 22, 2023 at 2:20

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How do the holes move? How can I visualize the movement of holes, since it really isn't a "particle" or an "ion" per se?

The nuclei of the atoms are more or less locked into place in a crystal lattice.

Let us suppose that a hole moves from positively charged atom A to neutrally charged atom B in one physical "step". What is happening on a physics level is that an electron from neutrally charged atom B moves to A. This leaves B positively charged, and A becomes neutral.

How does this differ from a flow of free electrons?

The electrons that move when a hole moves are not "free electrons". They occupy a lower energy level (known as the valence band) than the "free electrons" which occupy a higher energy level (known as the conduction band). This may seem a rather academic point, but hopefully its significance will become more apparent in a moment.

When a free electron moves about in a semiconductor, it darts willy-nilly from location to another. The same electron moves about. On the other hand, when a hole moves about in a semiconductor, an electron jumps "backward" to fill the hole leaving a hole in the new location. Then a different electron from a different neutral atom leaves its home, to fill the hole, making its previous home a hole. Then a third electron leaves its home to fill the hole, leaving its prior home as a hole, and on and on.

So in the case of free electrons as carriers, one electron goes on a journey farther and farther from its original home. In the case of holes as carriers, it is many electrons, each making very small jumps, one after another, from one spot to another. Each of these individual electrons does not go far, but the hole goes on a journey farther and farther from its original home. If you could see individual electrons and atoms, electron current would look different from hole current. It would be like watching someone slide a scrabble tile over a board that was filled with scrabble tiles, vs having a board that was filled with scrabble tiles, except for one spot, and someone moving a tile at a time from a neighboring spot into the empty spot. I hope that helps to give you some intuition.

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  • \$\begingroup\$ I love your answer, but I just don't understand one thing, how do these "small steps" that the holes are taking, result to them diffusing from one side of the semiconductor to the other? That seems like a rather large distance, around 0.5 Micrometers. I've also gotten to know that the diffusion current is a rather discontinuous process and happens for a very short period of time, so how do the holes, or rather the electrons moving from a neutral atom to a positively charged atom result to it being diffused from P side to the N side? \$\endgroup\$
    – SubbSE
    Commented Apr 22, 2023 at 8:36
  • \$\begingroup\$ A silicon atom has a diameter of about 200 pm (picometers). Your distance of 500 nm (nanometers) = 0.5 um (micrometers) is about 2500 times greater. Although that seems large, the very, very, short period of time it takes for a hole to move one or a few atoms at a time makes the diffusion rate quite rapid. Electrons at room temperature are traveling somewhere around 100 km / sec. Much slower than light, but still incredibly fast. (Note the drift velocity of electricity is very slow in comparison. Somewhere on the order of millimeters per second. -- but that's another story.) \$\endgroup\$ Commented Apr 22, 2023 at 13:38
  • \$\begingroup\$ Thank you for your answer. I feel like to understand the movement of the holes, I'll need to look at things at a microscopic level and the understanding of the whole thing is much more complex than it appears at this macroscopic level. With that being said, if the "holes" are moving via the neighboring electron coming in to neutralize the initial positively charged atom, and that keeps going on... How does the hole diffuse from one side of the depletion region to the other? Isn't the boundary of the holes on the p side the positively charged electrons present in there? \$\endgroup\$
    – SubbSE
    Commented Apr 22, 2023 at 19:37
  • \$\begingroup\$ I'm not sure, but I'm guessing that you are thinking that there is a complete absence of carriers in the depletion zone, and that's why you are having trouble seeing how current can pass through the depletion zone. If that is what you are thinking, it incorrect. There are fewer carriers in the depletion zone, much closer to the number of carriers in "intrinsic" (i.e.non-doped) silicon. But carriers can still enter the depletion zone and cross it. Intrinsic silicon has a high resistivity because of the few carriers, but because the depletion region is short, the total resistance is smallish. \$\endgroup\$ Commented Apr 22, 2023 at 20:21
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If you call it a "hole", you do that very literally to treat it as a particle. That's kind of the point of calling it a hole.

Semiconductor modelling then proceeds to give holes effective masses, impulses… as only particles would have.

Note: it's a model of physical reality, fit do describe important properties of a semiconductor device. That doesn't mean you'll be able to isolate a hole and put it somewhere else – just as much as you can't pick up an electron. That particle model useful for explaining things like depletion regions, indirect semiconductors, drift speed…, but not for everything. Models have limited scope – but as soon as you call things "hole", then you're in that model where a hole is a particle for all practical aspects.

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