I am trying to get an intuitive grasp of the derivation of the ideal diode equation and I am struggling with a couple assumptions although the math itself looks pretty straightfoward. I am self-stuyding this from Pierret's book. My major problem is with the boundary conditions at the contacts. It says that
$$
\Delta n_p (x \rightarrow -\infty) =0
$$
When one can assume the diode in question is a wide-base one, meaning whose contacts are several diffusion lengths or more from the edges of the depletion region. This means we can consider them to be infintely far from the depletion region (p beeing on the left side). Yet current in the quasi-neutral p region is diffusion current which is determined by
$$
J_N =qD_N \frac{d \Delta n_p }{dx}
$$
Since the electrical field is nul by assumption.
Now doesn't the first equation imply the diffusion current to be 0 in the diode. The way I understand it under forward bias the electrons from the n region should cross the now lower potential barrier through diffusion and become minority carriers in the p region where they end up reaching the ohmic contact and eventually the cathode of the battery. Am i missing something here?
Best Answer
No there is an electric field.
There is a balance in the depletion region, a built in E field
(from the n and p junction, that you can change with an external bias voltage)
that drives charge carriers one way,
and diffusion the other.
For me the current due to diffusion is harder to understand.