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In a PN junction, under no bias, when the depletion region forms an electric current is also formed in the direction from N side to P side. This electric field, also has a potential difference, obviously, which is known as the barrier potential. This barrier potential is required to be overcome by the majority carriers on both sides in order to cross the depletion region in order to diffuse, and hence also provides a force opposite to the diffusion of the majority charge carriers.

Now, my question lies in the drift current that follows:

When the electric field is induced, due to the uncovered ions on the both sides it resists its diffusion to the other side. Now, about the minority carriers that are formed in the P and N region due to thermal energy, if they manage to reach to the edge of the depletion region. (Provided it doesn't recombine with any majority charge carriers present in the said region.)

With holes in the N region at the edge of the depletion region near the N region, there's uncovered positively charged ions. Will they not be repelling the holes back into the N region? Why do the holes still drift to the N region?

Is barrier potential just another name for junction built-in voltage?

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2 Answers 2

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In a PN junction, under no bias, when the depletion region forms, an electric current is also formed, in the direction, from N side to P side.

It would be more correct to say that under no bias conditions, there are TWO currents formed, a drift current and a diffusion current, and they exactly cancel out. Or it would be correct to say that under no bias conditions there is NO net current.

This electric field, also has a potential difference, obviously, which is known as the barrier potential.

By "this electric field" I think you obviously mean the electric field that forms spontaneously when N and P type semiconductors form a junction, i.e. the "built-in" electric field. You refer to this as "the barrier potential", and I think it is important to stop here. I think the term "barrier" is perhaps misleading. While a PN junction can be, in some cases, a "barrier" to net current flow, it is, in no sense a "barrier" to the movement of individual carriers. It is not as though an electron, traveling somewhere near 100 km / sec comes up to the depletion region and sees it as an insurmountable barrier and suddenly turns around. Rather, individual carriers can and do cross the depletion region regularly, but if they cross with equal frequency in both directions, there is no net current.

This barrier potential is required to be overcome by the majority carriers on both sides in order to cross the depletion region in order to diffuse,

No. The diffusion current density depends only on the gradient of the carrier density. Carriers do not need to "overcome" some "barrier" in order to diffuse. The diffusion current is affected by the electric field only in the following way. The electric field controls the width of the depletion region. The wider the depletion region, the lower the slope of the density function, i.e. the lower the density gradient. Hence the lower the diffusion current. Conversely, the more narrow the depletion region, the steeper the carrier density function, the higher the gradient, and the higher the diffusion current.

provides a force opposite to the diffusion of the majority charge carriers.

The electric field does provide a "force", and that "force" is responsible for the drift current that flows in the opposite direction of the diffusion current. But the "force" doesn't stop the diffusion. It creates a counter-current, making the net current zero in the no-bias case. The electric potential difference across the depletion region affects the width of the depletion region, which affects the carrier density gradient, which affects the diffusion current.

When the electric field is induced, due to the uncovered ions on the both sides, it resists it's diffusion to the other side.

"it resists it's diffusion"? I think you are trying to say that the electric field "resists" the diffusion of carriers. But, as I am trying to explain, the individual carriers still diffuse. They diffuse at a lower rate the wider the depletion region.

Now, about the minority carriers that are formed in the P and N region due to thermal energy, if they manage to reach to the edge of the Depletion region[,] will they not be repelling the holes back into the N region? Why do the holes still drift to the N region?

Carriers will still diffuse across the junction, AND they will still drift back. It isn't an either or.

Also, is barrier potential just another name for junction built-in voltage?

I think they are generally used interchangeably, but I'm not an expert in the precise use of physics language. A physicist might draw a distinction that I am unaware of.

Addendum:

So from what you're saying is: The magnitude of the diffusion current works to manage the width of the depletion region which in turn affects the magnitude of the electric field induced in that region. And that induced electric field regulates the magnitude of the drift current which cancels out the diffusion current in equilibrium. (from comment in chat)

I think the actual physics goes like this.

The applied voltage modifies the width of the depletion region. The wider the depletion region, the smaller the slope in the carrier concentration curves, and therefore the smaller the diffusion current.

Also, the wider the depletion region, the greater the distance for the voltage drop. The electric field in the depletion region depends on both the voltage drop across the depletion region, and the width of the depletion region. In negative bias, and also forward bias with low applied voltage, the change in width and change in voltage across the drop work in tandem to keep the electric field, and hence the drift current, fairly constant.

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  • \$\begingroup\$ I have a question though, as we know, the drift current is comparatively very low to the diffusion current, so if we consider a case where the drift current is greater than the diffusion current, under no bias conditions. Will it be the case, that the depletion layer will expand due to the uncovering of more bound atoms on both sides of the junction and consequently, the electric field induced there will increase, "Limiting" the diffusion of majority carriers. \$\endgroup\$
    – SubbSE
    Commented Apr 23, 2023 at 17:52
  • \$\begingroup\$ Since the magnitude of the diffusion current has to be brought down to the equivalent of the drift current due to drifting of majority carriers and also movement of minority carriers, both of which contribute to the drift current. Also from my understanding, under equilibrium conditions, when diffusion current is equal to the drift current, diffusion of majority carriers still happens, but for each electron/hole that diffuses, there is an electron/hole that drifts back right? \$\endgroup\$
    – SubbSE
    Commented Apr 23, 2023 at 17:52
  • \$\begingroup\$ It's just that the induced electric field in the depletion region due to the diffusion of charge carriers would work as a "barrier" (as mentioned in my textbook and how the general terminology of "potential barrier" comes along), to limit, the diffusion of the majority carriers...Right? Do all the electrons have to be moving with 100 km/s speeds in the semiconductor? If that is the case then how does the electric field prove to do anything since the width of the depletion region is a few hundred picometers... \$\endgroup\$
    – SubbSE
    Commented Apr 23, 2023 at 17:57
  • \$\begingroup\$ "as we know, the drift current is comparatively very low to the diffusion current" ?? If there is no net current, then the drift current is equal to the diffusion current. Are you thinking of the velocities? "so if we consider a case where the drift current is greater than the diffusion current, under no bias conditions." <-- does not parse. \$\endgroup\$ Commented Apr 23, 2023 at 17:57
  • \$\begingroup\$ "under equilibrium conditions, when diffusion current is equal to the drift current, diffusion of majority carriers still happens, but for each electron/hole that diffuses, there is an electron/hole that drifts back right? " yes. Please note that comments are not for extended discussion, and if this turns into an extended discussion, a moderator will step in and move it to chat. \$\endgroup\$ Commented Apr 23, 2023 at 17:59
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In a PN junction, under no bias, when the depletion region forms an electric current is also formed in the direction from N side to P side. This electric field, also has a potential difference, obviously, which is known as the barrier potential. This barrier potential is required to be overcome by the majority carriers on both sides in order to cross the depletion region in order to diffuse, and hence also provides a force opposite to the diffusion of the majority charge carriers.

Firstly, the depletion region actually results from the diffusion current. A pn junction usually is meant with p-type semiconductor on lhs and conversely n-type on rhs; thus such diffusion current will be from p-type to n-type (the direction of electric current is conventionally the opposite one of electrons / parallel one of holes).

This electric field, also has a potential difference, obviously, which is known as the barrier potential.

Not really. I mean, the drift and diffusion models are macroscopic pictures for charge movement in the semiconductor.

Loosely stated, a barrier is usually meant in terms of energy in \$[eV]\$, to which you can associate a potential energy barrier height dividing by the elementary charge \$q\$ (E=qV). However, in the diode equation, there's no such barrier term. Furthermore, diffusion is purely driven by a concentration gradient: again there's no barrier term.

I guess you're mixing some current transport mechanisms of MS junctions and PN junctions. In the former, indeed, you can clearly link a current contribution to the different electronic affinities of the two materials (indeed defining a barrier height at the interface), which is called thermionic emission and it's a different beast (Bethe model); but then, also drift-diffusion is significant and must be accounted.

One for example does talk about barrier lowering effects (e.g. Schottky effect) in MS junctions, unlike PN.

When the electric field is induced, due to the uncovered ions on the both sides it resists its diffusion to the other side. Now, about the minority carriers that are formed in the P and N region due to thermal energy, if they manage to reach to the edge of the depletion region. (Provided it doesn't recombine with any majority charge carriers present in the said region.)

I must say this is quite confused. Majority carriers (e.g holes in p-region) move by drift there (recall their concentration is slightly increased by incoming minority carriers from the other side and charge balance must occurr), where the entire population is involved in the screening process; thermally generated minority carriers (although completely negligible) follow the same path of injected minority carriers, there's no reason they shouldn't.

With holes in the N region at the edge of the depletion region near the N region, there's uncovered positively charged ions. Will they not be repelling the holes back into the N region? Why do the holes still drift to the N region?

Injected minority carriers move by diffusion through the quasi-neutral region. Indeed, the field is very weak (actually 0 in the depletion approximation) since a tiny amount is needed by majority carriers, as said previsouly (recall drift current depends linearly on carriers' concentration).

You may want to go through some device physics device to clear up some confusion.

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