Fermi levels in p-n junction under voltage

[smallphi]

I understand that for a semiconductor p-n junction under zero voltage, the Fermi level is constant throughout the junction because that is the condition for a system in thermodynamic equilibrium.

I also understand that p-n junction under applied voltage (forward or reverse) is not a system in thermal equilibrium because there is a nonzero total current that obviously tries to revert the system back to equilibrium. Then I don't understand how come away from the junction we still use two Fermi levels to describe the electron distribution in energies, one for the left side and one for the right. Most importantly how come those Fermi levels differ exactly by the applied voltage?

I've consulted 2 textbooks in semiconductor devices and Ashcroft-Mermin solid state textbook. They all state the difference in the Fermi levels must be equal to the applied voltage without any derivation as if that is something obvious.

Can someone enlighten me where that formula comes from?

 

[rbj]

maybe post this under the "Atomic, Solid State, Comp. Physics" forum.

semi-conductor physics is very tough, and i don't remember half of it.

 

[marcusl]

Think of the Fermi level as an energy level (it's often called that and we use E_f to denote it). If you apply a potential across the junction, you impose an energy difference between the two sides (that's the definition of a voltage or potential drop). The Fermi levels simply shift to reflect that.

By the way, you will sometimes see the Fermi level called "chemical potential," which is technically the correct term for a semiconductor.

 

[smallphi]

It's not so simple. In equilibrium (junction with no external voltage) there is a potential difference across the junction (the so called built-in potential). The conduction bands on both sides of junction differ by that potential difference but the Fermi levels are at the same level. On the other hand, the external potential difference is ZERO in equilibrium case.

So the picture is that the electric potential difference across the junction is measured by the difference of the conduction bands on both sides of the junction. The battery voltage, which may not be the actual voltage applied on the semiconductor sample, is measured by difference in the Fermi levels and is NOT equal to the electric potential difference across the junction.

That are the things I can't comprehend:

1. Why the Fermi levels measure the battery voltage but the conduction bands don't?

2. Why the conduction bands measure the potential difference across the junction but the Fermi levels don't?

3. Why the battery voltage is NOT equal to the potential difference across the junction? Where is the missing voltage applied ?

4. Is the potential difference on the part of semiconductor away from the junction zero? The band edges and the Fermi levels are drawn horizontal there, no bending, and that makes me think it must be zero. If so then how come there is a current flowing there?

 

[marcusl]

The conduction and valence bands on the two sides are shifted relative to Ef because of the different dopings there. That is, Ef is not in the center of the band gap for a bulk material but is higher or lower for n and p materials. When Ef is equalized as the sides are joined, charges move to either side because of the relative shifts in the energy of the conduction band edge Ec, until space charge balances it out. This is where thinking in terms of the chemical potential mu is helpful, because mu describes diffusion along a concentration gradient just as in a chemistry problem.

The built in potential essentially arises, then, from the difference in work functions of the two materials. The "actual" potential across the junction is V_applied - V_builtin.
Your textbook probably covers this; if not, see any of the excellent semiconductor books out there like Sze, Physics of Semiconductor Devices.
Section 4.2 in this link also covers it briefly
http://ece-www.colorado.edu/~bart/book/book/index.html"

You can't see V_builtin directly because as soon as you apply contacts and attach a meter, additional potentials arise at those interfaces that cancel it and prevent any current flow.

 

[smallphi]

I know what the textbooks say. I have a whole tower of them on my desk right now. The problem is it doesn't make sense at all. The most ridiculous statements go like this:

A: The applied potential V appears mostly across the transition junction region because it has way bigger resistance compared to the neutral semiconductor regions away from it.

B: The electrostatic potential across the junction is Vb - V (Vb is the built-in).

Now statements A and B obviously contradict each other. If the total electrostatic potential difference on the sample is indeed V then crossing the whole sample we have V = -(Vb - V) + Vx, so the missing potential is Vx = Vb and it must be applied somewhere. Statement A, claims it can't be on the neutral bulk of the semiconductor because the voltage drop on that is much smaller than V and Vb supposedly. Statement B claims it is not on the junction region. So where is Waldo ?

Electrostatic potentials have to add up, doesn't matter if you can or cannot measure them.

 

[marcusl]

It's hard to interpret A from that small snip. I imgaine it's just saying that you can ignore ohmic drop in the bulk regions since the resistance there is low due to heavy doping. In other words, all points in a good conductor are at the same potential. Accordingly, all the drops are at the junction.

B is correct. Imagine joining a piece of n to a piece of p material (abrupt junction). There is a large concentration gradient, so electrons leave the n side, leaving behind a positively charged depletion zone there. The same occurs with holes from the p side. Together, these create a space charge field across the junction that balances out the original flow. The junction is now in equilibrium, and the electric field in the junction is triangular as shown in sect. 4.3 of the link I posted, and also presumably in your textbooks. Since \vec{E}=-\vec{\nabla}\phi, or E(x)=-\frac{d\phi}{dx} in this 1-D problem, we can find phi by integrating -E resulting in a smooth curve that is parabolic on each side. The potential difference across the junction (that is, the difference between values of \phi measured well away from the junction) is the built in potential \phi_i. It's around 0.7V in silicon.

The junction is not a battery, however. The built-in potential arises from the depletion zones, which in turn arose from carrier diffusion that balanced the charge or carrier gradient between n and p when the junction was (virtually) created. That diffusion brought the chemical potential to a constant level throughout the material (indeed, it was mu that caused the carriers to move in the first place if you want to think of it that way), so equilibrium is reached. Connecting the two sides via a wire cannot support a current flow because the chemical potential is constant throughout the entire circuit including at each semiconductor/metal contact, which has its own fields, potentials, etc. And if there's no current flow, there's no emf or potential driving the flow, hence you can't directly see the built-in potential.

Finally, any externally applied voltage gets added onto/subtracted from \phi_i.

 

[smallphi]

I am talking about forward biased junction not junction in equilibrium. What you said can be found in any textbook but it doesn't address the question of the missing potential drop in forward bias. This is question 3 from my list above.

Just because you can't measure the built-in voltage doesn't mean it's should be treated as 'special voltage that is actually not there'! It's created by static total charge distribution and as such it obeys laws of electrostatics: the total potential difference between points A and B is the same irrespective of the path you use to sum it up. So if it's V between the terminals of the battery, and there are negligible potential drops on wires, on contacts between semiconductor and wires and on bulk semiconductor away from junction, then it must be V on the junction which contradicts the statement it's Vb - V.

The only resolution I see is that actually the potential drop on the semiconductor sample is Vb-V. And the missing potential drop is between the sample and the metalic contacts - a statement I've never seen discussed in any textbook.

Finally, any externally applied voltage gets added onto/subtracted phi ...

That is not obvious at all. Forward biased junction is a system totally different from junction in equilibrium. The extra voltage V-Vb could be distributed between the p-n junction and the semiconductor-wire junctions. Assuming the potential change V is applied on the junction only is a bit of a stretch that needs deeper explanation. Otherwise it's like solving the problem by assuming the solution.

 

[marcusl]

It seems we agree with the description of the static system, so now we apply bias. The forward biased case is more complex in details as you say (away from equilibrium we use quasi-Fermi levels, for instance) but conceptually the picture is still simple. When you look at the static band diagram, the “potential hill” (different height on left and right) reflects the built-in potential discussed above, about 0.7V for silicon. Suppose you apply an external potential of 0.2V in the forward direction, that is, positive applied to the p side. This has the effect of flattening the diagram, reducing the barrier that a carrier must overcome to move from one side to the other. A hole, for instance, now “sees” a barrier of 0.5V. The simple addition really holds here, and there is no "missing" potential because we assume that the other parts of the circuit have low resistance. At 1 or 10 mA, for instance, the ohmic drop across the wires and extrinsic (heavily doped) bulk regions can simply be ignored. If the applied potential is \phi then the potential barrier seen by a carrier near the junction is \phi_i-\phi.

A small current begins to flow, at room temperature anyway, because a tiny fraction of carriers have enough thermal energy to cross the barrier. Flow remains small, however, until phi applied approaches 0.7V where the barrier flattens enough for carriers to flood across. The junction “turns on” in the language of the engineer, and in this regime the applied voltage changes little regardless of the current.

You still don’t directly access \phi_i, but its effects are indirectly seen: there is little flow at small forward bias, increasing exponentially and becoming large as \phi=\phi_i is approached. This is very different than for a regular conductor where current flow is directly proportional to impressed potential.

 

[smallphi]

My questions are not answered by citing the textbooks. In fact, what you have said so far doesn't even address my questions.

 

[Reality_Patrol]

This is a good question, it's something I've really never seen a good explanation of in standard semiconductor texts. It is explained in electrochemical texts but with respect to electrochemical applications.

I think your problem is you haven't been taught about the electrochemical potential - which is really the driving force in electrochemical systems. Roughly speaking it's the sum of the chemical potential (think Fermi level for electrons/holes in unbiased junctions) and electrostatic potential (uhh, the externally applied voltage, which is zero with no bias).

Equilibrium thermodynamics shows the electrochemical potential difference is driven to zero. So, if the unbiased junction has a chemical potential difference equal to the difference in the fermi levels, then a biased junction in equilibrium (constant current) can only be produced via an applied electrostatic potential "equal and opposite" to original chemical potential difference.

Hope this helps.

 

[smallphi]

No it doesn't help.

You can combine the electrostatic potential in whatever potentials you want but there is a simple law of electrostatics that can't be broken when the circuit is in steady state: electrostatic drop around a closed loop must be zero. So involving electrochemical potential doesn't resolve the missing electrostatic drop in the closed circuit. How come the total electrostatic drop between the battery terminals is V but the drop on the junction is (V-Vb). Apparently there is electrostatic drop Vb, created by accumulated charge somewhere in the circuit and my guess was it is at the contacts between the semiconductor and the metal wires. Not a single textbook discusses that and they should cause that missing potential drop is the first thing all beginners notice.

The existence of that electrostatic potential drop on the contacts raises the second question. How come the total increase on electrostatic potential given by the battery voltage V is applied as an increase entirely on the junction. How come the contacts didn't get any increase? After all the voltage on the contacts Vb is bigger than V or V-Vb so they have more charge there. How come only the junction responds and not the contacts.

The fake 'explanation' that it is 'simply addition of voltage' doesn't pass. Imagine a capacitor charged to some voltage Vc. Now imagine that you connect it to a battery of voltage V. Following the fake logic of 'voltage addition' now the capacitor voltage must be V + Vc. Systems do NOT respond in that way. When the applied voltage changes the system reconfigures, it doesn't simply 'add voltage only at its special point'.

 

[Reality_Patrol]

No it doesn't help.

Sorry. You do have some genuinely valid questions, but you're also really confused on some points. This is a complex subject. The texts, as I said above, do not offer entirely valid explanations. On this point you are correct. So, if your objective here is to pass a course the most practical advice I can give is to study the texts, memorize their explanation, and spit that back out on any exams. If your objective is to really understand a PN junction the bad news is you will have to learn about several other subjects as well. I will give a little more detail below just in case your interested.

The "fermi level" is just a name given to a quantity generally referred to as the chemical potential. It is NOT a field quantity, an exact "position" can not be assigned to it. It can't be "measured" using voltmeters or oscilloscopes. It comes from statistical thermodynamics. Every statistical system has one. Systems whose chemical potential is a "fermi level" are those that obey Fermi-Dirac statistics. There are other types of statistics and if a system behaves in one of those ways it is not appropriate to call the chemical potential a fermi level. If a system has more than one component (phase), like a chemical mixture or PN junction, then each phase has it's own chemical potential. Thus we can talk about differences in chemical potentials. These are what drive chemical reactions in purely chemical systems, i.e. reactions occur until the chemical potentials are equal (equilibrium).

In systems that have an external FIELD applied to them thermodynamics shows that the a new quantity is needed, i.e. the so-called electrochemical potential I mentioned above, which is more generally the sum of the chemical potential and external potential. It is what drives processes until all are equal. Here the external potential is a FIELD, and one can talk about how it is distributed spatially across the system - as some of your questions point to. But THERMODYNAMICS doesn't allow such treatment. This is a limit introduced due to the lack of more robust thermodynamic field theory. In thermodynamic treatments, each phase (component, i.e. P or N semiconductor) are modeled as if the external potential is constant across the phase (no spatial variation). Thus, the measurable external potential drop must be treated as if it occurs entirely across the phase interface (PN junction).

So, what the texts are really trying to say is that whenever an external potential difference is applied to a PN junction, an equal and opposite difference in chemical potential (fermi levels) develops. Again, this is because the system is trying to get back to equilibrium (zero electrochemical potential difference). Such an explanation would require a solid background in statistical physics, electromagnetics and general thermodynamics - something most students simply don't possesses at the time they look at this material. In short it's not really practical to give an accurate, consistent, simple explanation.

 

[smallphi]

I appreciate your trials, but like the previous poster, you are NOT addressing my questions:

1. Where is the missing electrostatic potential drop in p-n junction under forward/reverse bias?

2. The external voltage is not 'applied on the p-n junction'. It's applied on a circuit of two metal-semiconductor junctions and the p-n junction. Why would the p-n junction alone get the total increase in external voltage but the metal-semicondor junctions get no change at all?

The electrochemical potential cannot prevent the electrostatic potential obey the laws of electrostatics. The very notion of electrostatic potential is not defined if it doesn't obey them - if it jumps across a closed loop then you will have multi-valued function with several values at the same point. The equalizing of electrochemical potential in equilibrium leads to certain charge distribution in the system. Now you can imagine that you make a copy of that charge distribution but using charges that are frozen in place by some external forces (you can imagine a daemon holding each charge in place). Are you trying to tell me that the electrostatic potential drop around any closed loop in that system would not be zero?

And by the way, forward bias in p-n junction is NOT equilibrium situation, there is nonzero total current. They even have to describe electrons and holes with two different quasi-Fermi potentials inside the transition region. The Fermi levels match only for junction in equilibrium with no applied external voltage.

If you claim that the Fermi levels on the usual diagrams of forward biased junction are the 'chemical' potentials not the electrochemical ones, then adding the electrostatic potential energy represented by the edge of the conduction band to them will lead to even bigger jump between the p and n side because the fermi levels and the conduction band jump by the same amount in the SAME direction, they don't compensate each other. So no matter how you define the electrochemical potential, like the fermi levels or fermi levels plus conduction band, it jumps for forward or reverse biased p-n junction.

 

[Reality_Patrol]

I appreciate your trials, but like the previous poster, you are NOT addressing my questions:

Do you appreciate it? I don't get the impression you realize how convoluted some of your questions really are. It's tough to answer some because they're confusing multiple concepts simultaneously. Also, I wasn't trying to answer your questions in my last reply. I was replying to your last post and trying to providing some general guidance. I used caps in places where I was trying to highlight places that concepts from different theories were being introduced.

1. Where is the missing electrostatic potential drop in p-n junction under forward/reverse bias?

There simply isn't a "missing" electrostatic potential drop under either bias.

2. The external voltage is not 'applied on the p-n junction'. It's applied on a circuit of two metal-semiconductor junctions and the p-n junction. Why would the p-n junction alone get the total increase in external voltage but the metal-semicondor junctions get no change at all?

Each metal-semi junction has a very small drop, compared to the drop across the semi-semi junction. This drop is sometimes called a contact potential. A reasonable estimate would be around 1% of the total applied voltage at each metal-semi junction. For this reason these drops are typically neglected in simple explanations and first, sometimes even, second order mathematical models.

In the forward bias case, when current flows, there is also some drop through each semi region, due to the ohmic loss of the materials. This is also usually neglected, because it is likewise usually quite small. I can't give a useful % for that though it depends on to many different factors.

This is why simple models treat the "applied voltage" as if it were dropped entirely across the junction. It's a very good approximation.

I will also point out this "drop distribution" I've just described is really a description of how the electrostatic field behaves spatially. This has nothing to do with chemical potentials (fermi levels). Again, chemical potentials are merely numbers calculated from a statistical ensemble, nothing to do with spatial distributions in any way.

The electrochemical potential cannot prevent the electrostatic potential obey the laws of electrostatics. The very notion of electrostatic potential is not defined if it doesn't obey them - if it jumps across a closed loop then you will have multi-valued function with several values at the same point. The equalizing of electrochemical potential in equilibrium leads to certain charge distribution in the system. Now you can imagine that you make a copy of that charge distribution but using charges that are frozen in place by some external forces (you can imagine a daemon holding each charge in place). Are you trying to tell me that the electrostatic potential drop around any closed loop in that system would not be zero?

Not at all. This is an example of a poorly phrased question founded on a faulty claim. Again, there is no "missing" electrostatic potential drop.

And by the way, forward bias in p-n junction is NOT equilibrium situation, there is nonzero total current. They even have to describe electrons and holes with two different quasi-Fermi potentials inside the transition region. The Fermi levels match only for junction in equilibrium with no applied external voltage.

You are correct, when steady current flows in the forward bias case the system is described thermodynamically as "near equilibrium". This is exactly the type of state where the electrochemical potential gradient is used to describe system dynamics.

If you claim that the Fermi levels on the usual diagrams of forward biased junction are the 'chemical' potentials not the electrochemical ones, then adding the electrostatic potential energy represented by the edge of the conduction band to them will lead to even bigger jump between the p and n side because the fermi levels and the conduction band jump by the same amount in the SAME direction, they don't compensate each other. So no matter how you define the electrochemical potential, like the fermi levels or fermi levels plus conduction band, it jumps for forward or reverse biased p-n junction.

I claim "fermi level" is merely a name generally used to describe the chemical potential of systems who obey fermi statistics. The texts you are looking at may be doing something non-standard in their terminology. I have no way of knowing so I will have to leave my answer at this. I hope all this hard work pays off for you. This will be my last reply. I don't have enough time to type all this out!

Good luck.

 

[smallphi]

Direct citation from "Solid state electronic devices" Ben Streetman, page 152:

"Indeed the contact potential cannot be measured by placing a voltmeter across the device, because new contact potentials are formed at each probe, just canceling V0 (the built-in potential".

Potentials canceling V0 are NOT small compared to V0. Those are the missing electrostatic potentials for a closed circuit with zero external voltage. The same potentials have to appear in circuit with battery to balance the electrostatic potential change around a loop to zero.

 

[smallphi]

I finally found an explanation in William Shockley's "Electrons and holes in semiconductors", 7th edition, 1959, section 4.3 On the nature of metal semiconductor contacts:

Although the potential at the surface of the semiconductor is independent of the work function of the metal which makes the contact, the application of voltage to the point contact produces equal changes in the potential inside the semiconductor.

When the metal is in contact [with the semiconductor], however, the surface states exchange electrons very easily with the metal and maintain the same potential in respect to the metal as they would under equilibrium conditions. Thus the surface of the semiconductor comes to a potential in respect to the metal which is independent of the work function of the metal; and this difference in potential then remains fixed so that the surface of the semiconductor follows the potential changes applied to the metal.

So the missing electrostatic potential drops are indeed between the semiconductor and the metal probes. The 'surface states' explain why those potential drops remain rigidly constant when the potential of the metal changes, producing exactly the same potential change on the semiconductor. That explains why the p-n junction voltage changes by V.

The book gives a reference to article discussing the surface states:

J. Bardeen, "Surface states and rectification at metal semiconductor contact", Phys. Rev. 71, 717-727 (1947)

 

[marcusl]

Interesting! We also have
"You can't see V_builtin directly because as soon as you apply contacts and attach a meter, additional potentials arise at those interfaces that cancel it and prevent any current flow."
earlier in this thread, and also
"Connecting the two sides via a wire cannot support a current flow because the chemical potential is constant throughout the entire circuit including at each semiconductor/metal contact, which has its own fields, potentials, etc."

Seriously, glad you understand it now.

 

[Reality_Patrol]

I finally found an explanation in William Shockley's "Electrons and holes in semiconductors", 7th edition, 1959, section 4.3 On the nature of metal semiconductor contacts:

So the missing electrostatic potential drops are indeed between the semiconductor and the metal probes.

smallphi,

In case you're still interested I have a reference that explains the PN junction in electrochemical terms: "Solid-State Physics, An Introduction to Principles of Material Science, 2nd Ed, 1995", see sections 12.6.1 (equilibrium) and 12.6.2 (biased).

In particular note eq 12.48, where the "electric field" is introduced/defined. If you follow closely, you'll see it's NOT an electrostatic field. It's a type of non-conservative field usually called a "concentration potential" in electrochemistry. You may or may not know yet that there are 2 types of electric fields in general, conservative (e.g. eletrostatic) and non-conservative (e.g. transformer, motional and diffusional). Nonconservative fields don't have to sum to zero around a loop.

I hope this information doesn't cause you too much grief. You're working very hard to understand something that's not being explained correctly in the texts you're studying. Your solution above, re contact potentials, is also incorrect. But who am I, right?

I like your spirit so when I came across this section in this text today I remembered this thread. If nothing else I think you'll find in interesting and hopefully credible. I'll share something with you as a final comment: personally I think even this description is incomplete physically, though less so than the ones you've seen.

 

[smallphi]

I was talking about the electrostatic field created by a static electrical charge distribution. There may be charges drifting and diffusing around but the total charge density at every point of the circuit is constant with time after the stationary state is reached (I am assuming DC circuit). The field of the static total charge is always conservative and its potential drops have to be zero across a closed loop - it comes from Maxwell equations and there is no way around it. If there are other forces in the problem, they do not change the fact that the electrostatic force is conservative. The built-in potential is a real electrostatic potential - they calculate it from the charge distribution.

 

[Reality_Patrol]

I was talking about the electrostatic field created by a static electrical charge distribution. There may be charges drifting and diffusing around but the total charge density at every point of the circuit is constant with time after the stationary state is reached (I am assuming DC circuit).

Yes, electrostatic fields are created by charge distributions. However, other types of fields can be present as well even in the stationary state.

The field of the static total charge is always conservative and its potential drops have to be zero across a closed loop - it comes from Maxwell equations and there is no way around it.

Yes, but this is the case only when an external voltage is applied across the junction. The external voltage is the electrostatic potential and it sums to zero around the loop (circuit) as per Maxwell equations.

If there are other forces in the problem, they do not change the fact that the electrostatic force is conservative.

Correct, but you are ignoring those other forces in your model.

The built-in potential is a real electrostatic potential - they calculate it from the charge distribution.

Electrostatic potentials are calculated from charge distributions. However, the built-in potential is calculated from carrier gradients - not the same thing. The resolution: the built-in potential is not electrostatic, it's not conservative. See the reference I provided above, it's authors are highly competent researchers and the text is standard around the world in solid state physics courses.

 

[smallphi]

The built in potential is calculated by setting the diffusion current equal and opposite to the drift current at each point of the semiconductor. The drift current is proportional to the average electric field which is the gradient of the ELECTROSTATIC potential V. The built-in potential is the difference of V on both sides of the depletion region so both V and the built-in potential are electrostatic.

The presence of other forces in a system of static charges doesn't cancel the requirement of conservative electric field. If you hold two charges stationary in your hands applying external forces against their repulsion or attraction, that doesn't mean the produced electrostatic field is not conservative.

 

[Reality_Patrol]

smallphi,

I think I'm going to drop out of the conversation after this reply. It appears from your last answer that you now accept diffusion processes are present, whereas earlier you made no mention of those. That's why I initially pointed out the need for the electrochemical potential in a more accurate explanation. On the other hand you now seem to be arguing that Vb must be electrostatic, whereas earlier you said it couldn't be. Perhaps you still think that contact potentials are the resolution. I will only point out if they are you now have a new problem to resolve: diffusion potentials.

You're smart enough to take it from here. Good luck and a few final comments =>

The built in potential is calculated by setting the diffusion current equal and opposite to the drift current at each point of the semiconductor.

No, it's calculated from carrier concentration densities on both sides of the junction - see eq 12.42. This is a concentration potential and is frequently encountered in electrochemical applications.

The drift current is proportional to the average electric field

Yes, the drift current is driven by an electrostatic electric field - but what force drives the diffusion currents?

which is the gradient of the ELECTROSTATIC potential V. The built-in potential is the difference of V on both sides of the depletion region

You're confused on V: in this treatment it's an "energy potential" function, i.e. it's not an electromagnetic field quantity (potential). It's more like the potential functions used in Schrodingers equation.

so both V and the built-in potential are electrostatic.

No. See previous comments.

The presence of other forces in a system of static charges doesn't cancel the requirement of conservative electric field. If you hold two charges stationary in your hands applying external forces against their repulsion or attraction, that doesn't mean the produced electrostatic field is not conservative.

Correct, I've never said otherwise.

 

[Defennder]

I don't want to interrupt a discussion here, certainly not one that has evolved into something increasingly technical, but are we assuming steady state conditions here? And how is it possible for electric fields to be non-conservative under steady state conditions?

And more importantly why isn't the electric field at any point along the device given by the gradient of the electric potential at that point? How does the "energy potential function" differ from the electric potential here?

Is V, as discussed and defined so far, the electric potential at any point in the device? This would imply that it is due to both the applied voltage bias and the built in potential Vb here.

 

[smallphi]

Yes the charge distribution is static and as such must obey the laws of electrostatics - the electrostatic potential is conservative. All textbooks I know use the value of the built-in potential to infer the charge distribution causing it using the usual Poisson equation of electrostatics. If Vb was not electrostatic potential, one is not allowed to use the electrostatic Poisson equation to infer the charge distribution, the depletion layer width and so on.

I never mentioned the diffusion current because the forces or processes that created the charge distribution and hold it in place are IRRELEVANT to the fact that the laws of electrostatics must be obeyed. Many problems in electrostatics simply start by stating the charge distribution not mentioning who created it and how and who holds it in place. That's why I gave the example of someone holding two charges in his hands. Whatever fancy notions one uses to derive the static charge distribution like electrochemical potential etc. that cannot change the fact that the electrostatic potential is conservative. Otherwise, the whole electrodynamics will collapse conceptually.

The diffusion current as any diffusion is a result of random thermal Brownian motion of the particles. If you imagine a membrane separating a high density region from low density, the probability for particles transferring from the high density to the lower is higher simply because there are more particles on the high side. Unless there is a force that maintains the density contrast, the density will equalize within the volume of the container. It is not impossible for the particles to clump in one corner of the container but the probability for that is vanishing and practically zero when the number of particles in the system is of the order of moles. Thus, the diffusion is driven by probability not by some 'force'. One can construct 'diffusion potentials' whose gradient is the 'diffusion force' that 'drives' the diffusion current but that has nothing to do with the fact the electrostatic potential of the final static charge distribution must be conserved.

The electrostatic potential is the part of the charge carrier energies due to the presence of excess non-neutralized charge but it is not the full energy. Even in noncharged semiconductor where the total charge is zero in any macroscopic volume of it, the electrons interact with each other and with the ion lattice which leads to certain quantum mechanical energies for them - the energy diagram. The electrostatic potential, when there is non-neutralized total charge present, shifts the whole energy diagram up or down. Electrons in semiconductors or metals are not free particles but we describe them as free but with 'effective mass' that is not the mass of free electron. The effective mass and the energy diagram capture and represent the interactions with the other electrons and the lattice.

 

[pepeburro]

The answer to this question is on page 79, Helpdesk 4.1 of Semiconductor Devices, by Dasgupta and DasGupta, where it is shown that we add the contact potentials with the metals you get zero.

http://books.google.com/books?id=Pl...&hl=en&sa=X&oi=book_result&resnum=1&ct=result

 

[Defennder]

This thread is quite old. Please don't reply to old threads unless you want to clarify something. And I don't know to which post your reply is directed.

 

[cabraham]

There is bulk resistance in the semiconductor material. When forward biased, the current through the bulk material resistance incurs a voltage drop. The voltage drop across the p-n junction depletion region is less than the terminal voltage. The difference is the I*R drop in the semiconductor material bulk resistance. Did I answer the OP question?