1
\$\begingroup\$

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.

\$\endgroup\$
  • 2
    \$\begingroup\$ The Fermi level. $$eV =| E_{f_n} - E_{f_p}|$$ \$\endgroup\$ – Tom Carpenter Aug 17 '16 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.Robot Feb 16 '17 at 17:45
3
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Sorry,I think you missed 'ln'. Thank you anyway. You helped me know the relation. By \$\endgroup\$ – TVV Aug 18 '16 at 0:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.