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.
Electronic – Relation between built in potential and doping
diodes
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Best Answer
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.