(What do you mean by "can be measured directly", Thomas? Well, I think one will measure exactly zero as the voltage resulting from a p-n junction that is kept in the dark.)

Andrea, the key points here are the meaning of "thermodynamic equilibrium" and "band energy". In the bulk regions far away from the p-n junction, nothing changes, so the Fermi energy is close to either band (according to the doping type). At the junction, however, there is a space-charge region with an electric field, and therefore a change in the electrostatic potential occurs.

In thermodynamic equilibrium, the Fermi energy is the same throughout the device. The energy represented by the bands includes the electrostatic energy. Therefore, since the latter changes across the p-n junction, so do the band energies. Their total change in energy, which amounts to eU_bi, is in accordance with the concentration ratio of the minority and majority carriers (considering the same carrier type at both sides of the junction), as can be expressed by the Boltzmann factor (for not-too-small built-in voltages U_bi):

C_min / C_maj = exp(–eU_bi / kT).

Hi Andrea, the voltage drop you refer to is the built-in voltage of the pn-junction, which is the basis of every diode and of course CAN be measured directly with a voltmeter attached (assuming Ohmic and neglecting Schottky contacts). The point really is that the Fermi levels on either side of the junction need to be the same in thermodynamic equilibrium, and that on the n [p] side it is closer to the conduction [valence] band edge due to the doping.    

(What do you mean by "can be measured directly", Thomas? Well, I think one will measure exactly zero as the voltage resulting from a p-n junction that is kept in the dark.)

Andrea, the key points here are the meaning of "thermodynamic equilibrium" and "band energy". In the bulk regions far away from the p-n junction, nothing changes, so the Fermi energy is close to either band (according to the doping type). At the junction, however, there is a space-charge region with an electric field, and therefore a change in the electrostatic potential occurs.

In thermodynamic equilibrium, the Fermi energy is the same throughout the device. The energy represented by the bands includes the electrostatic energy. Therefore, since the latter changes across the p-n junction, so do the band energies. Their total change in energy, which amounts to eU_bi, is in accordance with the concentration ratio of the minority and majority carriers (considering the same carrier type at both sides of the junction), as can be expressed by the Boltzmann factor (for not-too-small built-in voltages U_bi):

C_min / C_maj = exp(–eU_bi / kT).

Hi Andrea, I think Jan-Martins answer tries to explain where the built-in voltage of the pn junction comes from  in more detail, and he is correct in that the electric field is restricted to where the space charges are at the pn-junction, the built-in potential, however, which can be obtained by integration, really extends up to the surfaces and can be measured with a voltmeter as the applied voltage (bias) needed for the diode to conduct. This bias needs to at least compensate the built-in potential, which is ~0.6V for most silicon diodes. This also works for any diode and even in the dark! I attach a pdf file of one of my lectures where I explain this step-by-step. Thomas

Well, I think the measurement you described, Thomas, is rather optimistic -- at least for a silicon diode, where the built-in voltage can amount to roughly 0.9 V (at least for silicon solar cells this is a typical value). Applying such a voltage directly to the device results in an enormous current which most likely destroys the device. On the other hand, the characteristic value of 0.6 V for a silicon diode is well-known, but this is the threshold voltage and not the built-in voltage.

Therefore, without destroying the device, I think one cannot perform a direct measurement of U_bi; instead, one rather determines the current--voltage characteristic from which U_bi can be obtained (for known temperature and known diode ideality factor).

Further, I think there is a slight mistake on your transparency number 14 where it says "exponential increase for V_d > V_b-i": The current always increases exponentially (well, except for the smallest voltages), which can be seen by plotting the current data logarithmically.

Moreover, the value of the threshold voltage depends on the absolute scaling of the linear current axis; on a logarithmically scaled current axis a threshold voltage doesn't make much sense.

Thank you Thomas and Jan-Martin for your fast and on-point replies. Sorry for me being late. I read all of them and I think that what I was missing is:

Therefore, since the latter changes across the p-n junction, so do the band energies. Their total change in energy, which amounts to eU_bi, is in accordance with the concentration ratio of the minority and majority carriers [...].

I didn't study quantum mechanics so I do not expect the concept of Fermi Level to be crystal clear to me. Finally: speaking of electrons can I just say that the conduction band in the n-type semiconductor is "lower" than the p-type's one because on the n-side there are "less electrons"?

Excuse me for my not-precise language, I'm just trying to figure out a possible easy way to get the idea and remember it. Hopefully one day I will dive deeply in these concepts.

Thanks again.

No, it is the other way round! You obviously missed the minus sign in the Boltzmann factor. Have a close look at that formula and think again; then, you will much easier understand the rest of my explanation. The general principle (to be remembered by heart) is that the higher the energy, the lower the probability to find particles there. This holds independent of the underlying statistics (Fermi or Boltzmann). Since on the p-type side their band energy is higher, there are less electrons than on the n-type side.

(Note the difference between "probability" and "concentration" or "total number": In the most simple case, the latter follows from the former by multiplying with the effective density of states. For further reading, have a look at "2.2 Basic Semiconductor Physics" of http://www.tf.uni-kiel.de/matwis/amat/semi_en/)

Another important thing to remember by heart is the following: On the n-type side there are much more electrons (with "much more" being measured in several decades), because this is what "n-type" means -- that the negatively charged carriers are the majority ones. On the p-type side the electrons are the minority carriers, with a (typically very small) concentration following from the "law of mass action" (cf. https://en.wikipedia.org/wiki/Law_of_mass_action#In_semiconductor_physics) as

nmin = (ni)2 / NA, with the (strongly temperature dependent) intrinsic carrier concentration ni and the acceptor doping copncentration NA.

Mr. Wagner I'm really sorry (again) of me being this late. I read your answer a few times and I do believe that I didn't undestood because the formula you proposed ion my textbook (link below) is in the form:

n3/n2 = exp[q/kT * (P3 - P2)]                           (4.1.9) p. 178

That's for the analysis of a so called graded impurity distribution. Here P3 - P2 is the potential difference, not the potential energy difference. Therefore, I would say that: "Free carriers are more numerous in regions of "high" potential or (equivalently) in regions of "low" electrostatic potential energy."

Am I right? I really hope so and thanks again for your replies.

http://eu.wiley.com/WileyCDA/WileyTitle/productCd-0471593982.html

Wait: Please do not stick to that special book of yours, nor to that special case of a "graded impurity distribution", but take some steps back first: Before looking at a p--n junction you have to understand the general concept of doping a semiconductor, making it either n-type or p-type. Further, you have to understand the concept of majority and minority carriers in a doped semiconductor and how their concentrations are related by the law of mass action.

You need to go through all this because in your key statement -- that free carriers are more numerous in regions of "high" potential or (equivalently) in regions of "low" electrostatic potential energy -- you only speak of "free carriers" without differentiating between electrons and holes, and also not between majority and minority carriers. But without these differentiations your key statement doesn't make any sense, because all of them -- electrons, holes, majorities, minorities -- are free carriers; however, their behavior is completely different!

In case you did not find the section "2.2 Basic Semiconductor Physics" of the script I mentioned, here's the direct link: http://www.tf.uni-kiel.de/matwis/amat/semi_en/kap_2/backbone/r2_2_1.html

You may also have a look at the following related discussion: https://www.researchgate.net/post/Concentration_of_electrons_in_the_P-type_region_of_semi-conductor

May be a suitable starting point to be be able to draw the energy band diagram of a pn junction under thermal equilibrium under no bias condition is to be able to draw the energy band diagram of the p-type and n-type alone with reference to vacuum or the free electron level. This energy level diagram contains in addition to the reference energy level, the cond ution band edge EC and the valence band edge Ev and the Fermi level which is the level with 50 percent occupation probability with electrons. So, for p-material it is closer to the valence band and and for the n type material it is closer to the conduction band.

My question to you what will happen when we bring the two materials into intimate contact down to the atomic level?

The whole structure will go to thermal equilibrium by exchanging mobile charge carriers by diffusion from the regions of higher concentration to the adjacent region of lower concentration and by exchanging heat. Since every side of the contact was originally neutral when majority carrier flow to the other side it leaves behind it its neutralizing charges. This leads to the formation of an electric dipole layer around the metallurgical junction.

This electric dipole grows until the electric field associated with it balances the diffusion of the majority charges and the equilibrium is reached where the diffusion flow is counter acted by an opposite field flow. The dipole layer results in an associated built in electric field and potential difference called the contact difference of potential phi. And so, the p-side will be at lower potential than the n side by the contact difference of potential. Or the p-side will be at electron potential energy than the n-side

This can be also understood by the balance between the diffusion and the drift means leveling out the Fermi level over all the material of the pn junction that through out the n and p sides. This results in the shift of the band edges such as the the conduction band edge in the p-side becomes higher with the contact difference of potential from that of the n-side.

For more information and explanation of the formation of p-n junctions please see the text book: Book Electronic Devices

You will find answers for your question

Best wishes

mathematically, you need to solve Poisson's equation (2nd order differential equation) for the potential where the space charge density is given by the stationary ions of the dopants after their holes and electrons have annihilated by recombination. This is the dipole Abdelhalim correctly refers to above.