Relation between built in potential and doping
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.