Why doped silicon does not undergo electrostatic discharge? Doped silicon is charge neutral overall, but since the extra added carriers are only weakly bounded (~45meV) they become delocalized. Since the concentration of silicon is 5-9 orders of magnitude higher than the concentration of dopants, the delocalized charges "spend most of their time" around silicon, such that locally speaking silicon becomes charged (?).
If so, why doped silicon does not discharge those extra carriers upon contact?
 A: The excess charge carriers in the conduction/valence band of Silicon (delocalized so that around silicon atoms there is a slight excess of local charge) are neutralized by the equal opposite charge of the randomly scattered dopants.
Thus, the total charge remains zero, and this is actually the only way that an infinite crystal can have a finite electrostatic energy density.
There will be current flow from doped Silicon to another material, as long as the Fermi-levels between the systems are not equal. However, very quickly there will a charge accumulation at the boundary, and this adjusts the Fermi levels (a dipole layer corresponds a step in the electrostatic potential). 
Just clarify further, all this is completely different from adding excess charge to the system. The system has to finite now for this to work. The extra charge (be it positive or negative) will now be located in partially occupied band, and therefore can move freely (like in metals). This happens, because the the response will be intraband like, with a finite effective mass of either valence or conduction band. Thus, this charge will be now located at the surface. This is the surface charge which would discharge upon touching.
A: Doping introduces allowed energy states within the band gap of the material, and these energy states are  very close to the energy band that corresponds to the dopant type.
From the Wikipedia article on doping in semiconductors:

[For example] electron donor impurities create states near the conduction band while electron acceptor impurities create states near the valence band.
The gap between these energy states and the nearest energy band is usually referred to as dopant-site bonding energy or E(b)and is relatively small.
For example, the   E(b)  for Boron  in Silicon bulk is 0.045 eV, compared with silicon's band gap of about 1.12 eV.
Because E(b) is so small, the room temperature is sufficient  to thermally ionize practically all of the dopant atoms and create free charge carriers in the conduction or valence bands.

From QuestionHub.net:

the delocalized charges "spend most of their time" around silicon, such that locally speaking silicon becomes charged (?).

So the above picture is not correct as one does not get silicon to be charged the charges are either in conduction band or valence band and can act only when some contact potential drives them to diffuse.
Another effect of Dopants  is that  they  shift  the energy bands relative to the Fermi level.
From Wikipedia again:

The energy band that corresponds with the dopant with the greatest concentration ends up closer to the Fermi level.

Therefore if one stacks  layers of materials with different properties it  leads to various electrical features  induced by band bending.

For example, the p-n junction's properties are due to the band bending in contacting regions of p-type and n-type material.

