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.
2
-
2\$\begingroup\$ The Fermi level. $$eV =| E_{f_n} - E_{f_p}|$$ \$\endgroup\$– Tom CarpenterCommented Aug 17, 2016 at 0:44
-
\$\begingroup\$ @TTV where did you find the relationship between the depletion region width and the doping concentration? can you post a link please? \$\endgroup\$– Mr.RobotCommented Feb 16, 2017 at 17:45
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
1
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.
-
\$\begingroup\$ Sorry,I think you missed 'ln'. Thank you anyway. You helped me know the relation. By \$\endgroup\$– TVVCommented Aug 18, 2016 at 0:22