marcus, your post is a little confusing to me! It seems like you are describing band bending on the metal side of the interface! There isn't a Debye length for a semiconductor (you need lots of free charges for debye screening, so the Debye depth applies to the metal side of the interface), and the charge density on the semiconductor side doesn't have an exponential decay profile - in fact, the profile is pretty close to the depletion profile you see in a regular PN junction.
jasum, you're partially correct in that if you have an n-type Schottky interface, the field on the SC side is primarily due to the ionized donors within the "depletion region". However, this additional potential doesn't make it "more easy to escape from bonding". What you describe would be seen as a reduction in the bandgap, not a bending of the bands. The bending is caused, as marcus explained, by the matching of chemical potentials across the interface. The bending profile, however, is a function of the total charge density, but typically, it goes approximately quadratically within the depletion region (assuming a nearly uniform density of ionized donors here).
savi, you want any "diode" to have the ability to rectify (conduct in one direction only). With a Schottky diode, you do that by putting the conduction band (of the SC) just above the Fermi energy (ofnthe metal). This way, a small positive bias voltage (exceeding the Schottky barrier height) results in a large conductivity, but any negative bias voltage (short of breakdown) leaves you with an essentially insulating barrier. That's what a diode needs to be able to do.