The depletion region is formed at the junction due to diffusion which creates a thin width of negative ions on the p side and positive ions on the n side whose electric field restricts further diffusion of the free carriers from the n side to the p side.
A forward bias of this junction will allow the free carriers from the n side to overcome this electric field and recombine with the hoes on the p side.
However, this seems to me to be the same scenario as how the depletion region was formed in the first place. When the recombination occurs due to forward bias, then the p side forms even more negative ions and the n side forms even more positive ions which would then increase the electric field and therefore the forward bias applied (typically .7V) would not be sufficient anymore and no more current would flow. What am I understanding incorrectly?
Best Answer
A few corrections that should hopefully clear things up.
I would phrase this as "Diffusion of mobile charge carriers leaves behind non-mobile dopant ions."
I would phrase this as "Whose electric field establishes a drift current in opposition to the diffusion current. At equilibrium these currents exactly cancel out. "
I mostly agree with this statement, but I would prefer to say "A forward bias of the junction will allow the diffusion current component to exceed the drift current component resulting in a net current"
I'm not following you at this point, which is probably where your confusion lies. First, note that those positive and negative ions are formed during fabrication and remain undisturbed throughout device operation. A chunk of charge neutral doped semiconductor has an amount of ionized dopants and free charge carriers that perfectly cancel out resulting in a net zero charge.
What you are describing sounds more like a diode in reverse bias. In reverse bias, charge carriers are extracted from the device and the depletion region gets wider, and continues to prevent a net current through the device.
In forward bias the depletion region width decreases due to the addition of charge carriers from outside the device. At a basic level, we assume that recombination is negligible in a diode. However, I think you are misunderstanding what recombination in a semiconductor device is.
Recombination is when an electron and a hole meet up and, well... recombine. When this happens they are effectively gone. Recombination is not what happens when extra holes enter the depletion region on the p side of the junction and electrically cancel out a negatively charged acceptor ion.