The electron states in the band gap are localised, whereas the states which contribute to the bands are not.
The electron states in a solid are not simple, there's a lot of non-trivial quantum mechanics going on. The electron states in a free atom are localised around the atom - the electrons in those states can't leave the atom without a lot of energy, so they can't conduct anything. When you pack lots of atoms together, the surrounding electron states overlap and mix. Which states mix with each other is dictated by the energy: similar energies means more mixing. You end up with a new set of states which extend over the whole block of material. If the material has a periodic lattice, these electron states group together into bands.
Every state in a band has some velocity (called a Fermi velocity) associated with it, and an electron in that state can be thought of as moving through the material with that velocity. The Fermi velocity of electrons in the conduction band is very large, but because the electrons are all going in different directions, there is no net current. An applied electric field moves some electrons from states which were going in one direction, to states going in the other. In a metal, one of the bands is part full, so there are plenty of nearby states to move electrons into. In a semiconductor, there is a gap between a full band and an empty one so it's much harder to push electrons into the higher band.
When dopants are added, they don't form a nice periodic lattice and they are much more spread out than the silicon atoms that host them. This means that the electron states around the dopant can't mix with states from other dopants to form a band. Since the energy levels of the dopant states are different from the silicon states, they don't mix (much) with them either. Instead, the electron states are localised around the dopant, much like the states around the free atom. An electron in that state can't conduct in the way one in a band can. It either has to jump up into the band, or jump to another nearby dopant. The former happens in semiconductors, the later is known as incoherent transport, and appears in some other materials.
I'm not sure how well I've explained this, but if you don't get a clear answer here, you could try the physics stack exchange. This definitely feels more like condensed matter physics than electrical engineering!
I don't know how you missed the first formula for the built in voltage that I can find.
$$
V_{bi} = V_t\ln(\frac{p_nn_p}{n_i^2})
$$
$$ p_n = \frac{n_i^2}{n_n}$$
$$ n_p = \frac{n_i^2}{p_p}$$
and last but not least:
$$
n_n = N_D - N_A
$$
with Nd and Na being the donor / acceptor doping in the n-region
$$
p_p = N_A - N_D
$$
with Na and Nd being the acceptor / donor doping in the p-region
Assuming you know algebra you can easily express the built in voltage in terms of the acceptor and donor concentrations.
$$V_0 = V_t \cdot ln\Big(\frac{N_d N_a}{n_i^2}\Big)$$
This equation is what I missed.
Best Answer
No --- the built-in potential is equal to the difference of the Fermi levels in bulk semiconductor N and P, so for a given material it basically depends on the doping of the P and N zones. See the figure in In band diagram, why the Fermi energy (EF) is constant along the device?
Think about it --- if you were right, all the diodes of a given material would have the same built-in potential.
So the answer is that (using the simply common model of the diode):
$$ V_0 = \frac{k_BT}{q_e} log (\frac{N_AN_D}{n_i^2}) $$
with the usual meaning of the symbols. Notice though that in the term \$n_i^2\$ lie a dependency to the band-gap of the material, too.
(And be sure to not mistake the built-in potential with the threshold voltage, they are completely different beasts).