The question is in the title, to clarify: I know that in a P-N-junction, at the boundary, there is going to be a voltage drop in thermal equilibrium. I also know that we can't measure this voltage drop by measuring the potential difference over the whole semiconductor device: If the diode is in thermal equilibrium, and we clamp a voltmeter to the endpoints of the diode, we will measure zero voltage.

But has at anytime somebody anyhow actually observed the voltage drop inside the diode, right at the pn-junction? Or is it just assumed that diode's work this way, because the model yields fine results and can explain the current-voltage behaviour of the diode?

What I imagined was a measurement that gives specific values of the potential inside the diode, at specific points in the diode, shown graphically in an $\Phi$ vs $x$ Diagramm, where x is the spatial distance in the junction.

  • 2
    $\begingroup$ Possible duplicate of Pn junction voltage drop? $\endgroup$ – Rol Feb 26 '17 at 9:35
  • $\begingroup$ Well ... no. The question you mention deals with why there is no net current, although there is a drop. I just want to know if the drop has been measured. Has somebody ever measured the actual potential (modulo constant) inside the diode, right at the junction? $\endgroup$ – Quantumwhisp Feb 26 '17 at 11:07
  • $\begingroup$ Like the open-circuit voltage of a solar cell (which is just a diode)? Yes. $\endgroup$ – Jon Custer Feb 27 '17 at 14:03

You can measure the voltage drop across by measuring the voltage required to make a current flow in the forward biased direction.

If you measure the current as a function of voltage for a forward bias you get something like:

V-I curve

(image is from this question)

The knee voltage ($0.65 - 0.7$V in this case) is a measurement of the potential across the junction.

  • $\begingroup$ But there is no measurement of the actual potential inside the diode? Like a function $\phi(x)$ ? $\endgroup$ – Quantumwhisp Feb 26 '17 at 18:44

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