Let's consider a PN junction which is at zero biased voltage. Once the junction is formed charges are diffused across the junction and form the depletion region and stops further diffusion. So no further diffusion takes place. But due to the created electric field minority carriers can drift across the depletion layer creating a drift current. Thing that cant understand, if diffusion has stopped after the depletion layer is formed, how come drift and diffusion current get equal. Is that because even the depletion layer is already formed diffusion contentiously occurs keeping depletion layer intact or same size? Can someone explain this scenario?
[Physics] How diffusion and drift current in a PN junction get equal
electronic-band-theoryelectronicssemiconductor-physics
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I think everyone must be right. (I'm certainly not going to contradict two-time physics Nobel Laureate John Bardeen.)
Schottky diodes have very short reverse-recovery times (switching from the conducting to the blocking state), so there can't be any significant minority-carrier currents, because it's extracting those minority carriers that slows down the switching of traditional pn diodes.
On the other hand, the classic Haynes-Shockley experiment uses a point rectifier specifically to inject minority carriers into a semiconductor. [The experiment measures various properties of these injected carriers and demonstrates that they are indeed minority carriers. Gibbons' "Semiconductor Electronics" even includes a detailed sample preparation procedure for a lab demonstration of what he calls the "Shockley-Haynes experiment".]
Shockley, in his Nobel lecture, includes a figure 2 depicting hole generation into n-type semiconductor, and states that
in a good emitter point it can be shown that more than 90 per cent of the current is carried by the process which injects holes into the semiconductor, and less than 10 per cent by the process which removes electrons.
I suspect the key is in the details of forming that "good" point, which could result in a completely different energy band structure at the interface than in a Schottky diode. However, that's only a suspicion, and so this post is not a satisfactory answer.
OK to get this all right you should look in a good semiconductor device book,
Maybe Ben Streetmans "Solid State Electronic Devices".
(But I'll wing it.)
To understand PN diodes we break the current up into two pieces.
The drift current due to the built in electric field in the depletion region.
and the diffusion current (about which you are asking.)
The built in E-field pushes the charges to one side, where they tend to pile up.
but now there is an excess of those carriers on that side.
Random thermal motion tends to cause this excess to diffuse away.
(if the concentration was equal everywhere there would still be random thermal motion, but no net current.)
In equilibrium (zero bias) these two currents have to cancel.
Diffusion under reverse bias: I'll have to think/ research this a bit more. The diffusion current is still just some concentration gradient and the thermal motion of carriers. There may be some slight increase because the applied E field should cause a higher concentration... but it also makes the depletion width bigger maybe the two things tend to cancel?
But as far as diffusion current, think of random thermal motion in the presence of a concentration gradient.
Best Answer
You answered it yourself. When the depletion region begins to widen, diffusion current decreases, but doesn't cease. Simultaneously drift current comes into picture due to electric field. At equillibrium, the value of depletion current and drift current becomes equal, hence the depletion region does not grow/shrink further.