I have seen several diagrams and explanations depicting formation of Schottky Barriers. Most of them focus on the workfunction and electron affinity to predict what will happen when the semiconductor and metal are put together (accumulation layer, depletion layer, inversion layer). However, I haven't been able to find an explanation as to what happens when the Fermi level of the metal is lower than the valence band maximum of a semiconductor before contact (i.e. before equilibrium or charge transfer). I am assuming that an accumulation layer will form at the interface. However, I am not sure whether such thing is possible. Can a metal have a fermi level that is lower than the valence band maximum of a semiconductor?
It seems to be possible. For example platinum has a larger work function than the sum of the electron affinity and band gap of Si, which is about 5.17eV. Therefore, according to this picture, this should lead to a degenerate inversion layer (n-type Si) or accumulation layer (p-type Si) in Si at the interface. The actual position of the Fermi level at the interface and thus the Schottky barrier in equilibrium is, however, often not given by the work function difference of semiconductor, but depends on chemical factors at the interface. See, e.g., "The physics and chemistry of the Schottky barrier height", by R.T. Tung, APPLIED PHYSICS REVIEWS 1, 011304 (2014)