Relation between built in potential and doping

What is the relationship between the built in potential and the doping concentration of a pn junction diode ? I could only find the relationship between the depletion region width and the doping concentration.

• The Fermi level. $$eV =| E_{f_n} - E_{f_p}|$$ Commented Aug 17, 2016 at 0:44
• @TTV where did you find the relationship between the depletion region width and the doping concentration? can you post a link please? Commented Feb 16, 2017 at 17:45

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.

• Sorry,I think you missed 'ln'. Thank you anyway. You helped me know the relation. By
– TVV
Commented Aug 18, 2016 at 0:22