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I read a lot of answers but I still do not understand: the potential barrier is real (e.g.the left side of an otherwise open junction has higher potential than its right side) and why I cannot measure it by use of a voltmeter?

Often it is argued, that the two metal-semiconductors junctions from the tips forms another potential which exactly cancel out the barrier of the junction.

I believe that this is the case, but why cancellation is exactly?

Would it mean that when I connect two metallic caps on both ends of the junction without connecting them via a voltmeter, both metallic ends have now same potential? This sounds strange. Or is there a closed electrical loop needed? This is also somehow "absurd", because modern voltmeters are nearly ideal in terms of resistance and there is no current flowing.

I would like to learn, where exactly is my wrong interpretation of how things are going there. There must be some missing key fact in my considerations.

Btw, also the energetic argument is not really satisfying, because, although true, it doesn't explain details on a microscopic level.

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The electrons of the conduction band of the n-side of the junction can lower its energy by occupying the available states of the valence band in the p-side.

But if we are at the equilibrium situation, other electrons have already migrated, creating an E-field in the region. The work done (in the equilibrium state) against the E-field to migrate equals the energy decrease by change the quantum state. What means: the Fermi level is the same in both sides of the joint.

The excess of electron of the p-side of the joint are attracted by the ions of the n-side. It is like a spring hanging in the vertical position with a mass attached. There is an upward force due to the spring that is balanced by the gravitational force.

If a conductor (or a voltimeter) connects n and p terminals, nothing happens. The probability that an electron moves in one direction is the same as moving in the opposite one.

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  • $\begingroup$ >If a conductor (or a voltimeter) connects n and p terminals, nothing happens. Then we have the strange situation, that both probes of the voltmeter are at different electric potentials, but no voltage is displayed. How can that be? From the theory of pn junction it is derived, that there is a potential difference along the pn junction. $\endgroup$
    – MichaelW
    Commented Mar 22, 2021 at 19:29
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    $\begingroup$ Well, I though it was clear in my answer. If it is not, maybe it is better to wait for another one. $\endgroup$
    – Claudio Saspinski
    Commented Mar 22, 2021 at 22:42
  • $\begingroup$ What I see: there is a difference of electric potential directly on the left and right side of the junction. However, when I try to measure this difference, I get zero, which means, that between the + and - Input of the meter is zero voltage. Therefore, the voltage drop on the junction, although "real", is somehow hidden, because by contacting the junction with my tips another potential difference is created, which cancels out the voltage on the junction. As an electrical engineer I would claim, that there is no voltage drop at the junction, but I see, that this is not true. $\endgroup$
    – MichaelW
    Commented Mar 23, 2021 at 11:23
  • $\begingroup$ I think I understand now. Took a while to realize. Thanks $\endgroup$
    – MichaelW
    Commented Mar 23, 2021 at 14:47

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