calculating the contact potential of a p-n junction with donor energy level?

Question

A p-n junction is such that the acceptor and donor dopant levels are 0.16 eV from the nearest band edge. The doping level on both sides is \${10}^{16} \text{atoms}/\text{cm}^3\$. What is the built-in potential for this junction?

How do I find the intrinsic carrier concentration for this? I know that the equation to calculate the built-in potential is

$$V_{bi}=\frac{k\, T}{q} \ln \frac{N_a\,N_d}{n_i^{\,2}}.$$

I have the acceptor concentration N_a and the donor concentration N_d, but the bit about the donor dopant level (\${10}^{16}\$) is throwing me off.

Answer 1

Deep donors and acceptors are not fully ionized at ordinary temperatures. OP needs to evaluate the Fermi level positions in neutral n and p material for this nontypical case. Then qVbi=(Efneutraln-Efneutralp).

Look at the usual textbooks to see where the quoted Vbi equation comes from.