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I have gone through a lot of questions but none of them ask how do the minority carriers approach the depletion layer in the first place.

When a p-n junction is formed, negative space charge accumulates at p side and positive space charge at n side. A minority carrier (say electron) from the periphery of the p side, when approaching the depletion region, should experience repulsion from the accumulation of negative space charge or ions in the depletion region, before it 'slides' down the potential barrier. The question is, minority carriers being so little in number, the probability of having thermal energy sufficient to overcome that repulsion is extremely low. So how is diffusion possible?

Is it because the space charge has no component of electric field away from the depletion region? Plus, why doesn't applied bias have any effect on the minority carrier current (according to the text books, this current is only a function of temperature). Shouldn't the reverse bias diffusion current be more than that during forward bias?

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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.

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  • $\begingroup$ Thanks for answering.. I realized that my question is a bit wrong. It should have been drift current rather which is a carried out by minority carriers. How do they approach the barrier? You see p side has electrons as minority carriers and therefore they must drift in order to cancel put the diffusion current. But doesn't the space charge build up in the dep region acts like a parallel plate capacitor? It should repel that electron away right?(cuz there's -ve space charge forming in the p region) $\endgroup$
    – Weezy
    Commented Jul 21, 2014 at 11:12
  • $\begingroup$ @eoshah, Yeah the charge build up (built in potential) is what causes the drift current, and the diffusion current balances that. (with zero applied bias voltage) It's a bit confusing the first time, but keep working on it. $\endgroup$
    – George Herold
    Commented Jul 25, 2014 at 14:16
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    $\begingroup$ In the field free region outside of the depletion layer, there is no excess charge - there is no field. So, a diffusing carrier can wander right up to the depletion layer edge. Now, a minority diffusing carrier has to not suffer recombination in the process... $\endgroup$
    – Jon Custer
    Commented Jan 24, 2017 at 0:34
  • $\begingroup$ First of all diffusion occurs due to the thermal energy of electrons, now due to diffusion depletion layer is formed, but as soon as it is formed, the positive charges on n-side pull back the electrons that had gone to n-side via diffusion, see this is the drift current everything is happening at the layer itself, there is kinda like a war going between thermal energy and forces due to the E field, this is going on continuously every moment but without an external battery, these two process are just in equilibrium.... $\endgroup$
    – Sarthak Sharma
    Commented Feb 28, 2017 at 10:29

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