As far as I understood, when we connect P and N materials, electrons from N drift to P, thus create negative ions in P and positive ions in N.

=> electric field (barrier potential) must depend on an amount of electrons recombined with holes in P region.

=> The more electrons recombined, the more ions created, the more barrier potential.

BUT! Silicon has 14 electrons and 4 on the 3d level. Germanium has 32 electrons and 4 on the 4th level.

=> at 25C degrees, electrons in N region in Germanium must have more energy than electrons in Silicon, because less energy required to remove electron from an atom => more electrons can drift to P region, thus barrier potential for Germanium should be higher, however

Vbp Germanium = 0.3 V Vbp Silicon = 0.7 V


P.S. I am talking about P doped and N doped silicon compared to the junction between P doped and N doped germanium

  • $\begingroup$ Are you asking about a Si-Ge junction, or about the junction between P doped and N doped silicon compared to the junction between P doped and N doped germanium? $\endgroup$ Apr 1, 2014 at 14:58
  • $\begingroup$ P doped and N doped silicon compared to the junction between P doped and N doped germanium $\endgroup$
    – Herfox
    Apr 1, 2014 at 14:59
  • $\begingroup$ What is your question exactly? Can you make your question really clear and simple? $\endgroup$
    – boyfarrell
    Apr 4, 2014 at 5:50

1 Answer 1


The question appears confusing because the requester has confused at least three different things.

The first confusion: the submitter appears to believe that it should be much easier to ionize a Ge atom, since it has so many more electrons and they are stuffed into higher levels. However, the well they are in is also much deeper, since the Ge nucleus has 32 protons rather than 14 for Si. When all is said and done, the first ionization energies are actually pretty close (8.15eV for Si, 7.89 for Ge).

The second confusion: does the band gap of a crystal have anything to do with the energy required to remove an electron from a free atom? No, since the band gap arises from overlap and degeneracy splitting of atomic levels, and how those levels get filled in the crystal. While it is true that, going down the column IV diamond cubic semiconductors, both the first ionization (free atom) and the band gap (diamond cubic crystal) monotonically decrease from C (diamond) to Sn (in diamond cubic form), that is pretty much coincidence. And note that while Ge has a band gap of 0.66eV at RT, GaAs (elements on either side of Ge) has a band gap of 1.42eV, bigger than Si.

Finally, n and p regions are made by doping the lattice, where the dopant has electrons (or holes) very near to the band edges such that they will be free carriers at room temperature. It is these carriers, very easily promoted into "free" states to move, that are responsible for the electrical properties of the pn junction.


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