Animated Solution for Physics - Semiconductors: In a ....... biased $p$-$n$ junction, the net flow of holes is from the $n$ region to the $p$ region.

Imagine you are looking at a $p$-$n$ junction. On the left, we have the $p$-region, which is teeming with holes—these are the majority carriers. On the right, we have the $n$-region, where electrons rule as the majority carriers.

However, the story doesn't end there. Due to thermal excitation, a few electrons are generated in the $p$-region, and a few holes are generated in the $n$-region. These are known as minority carriers.

When we connect this junction to a battery, we apply a "bias."

If we connect the $p$-side to the positive terminal and the $n$-side to the negative terminal, we call it forward bias. In this state, the battery pushes the majority carriers across the junction. Holes from the $p$-region diffuse into the $n$-region, and electrons from the $n$-region diffuse into the $p$-region. This creates a large diffusion current.

But what happens if we flip the battery?

When the $p$-side is connected to the negative terminal and the $n$-side to the positive terminal, the junction is in reverse bias.

This configuration pulls the majority carriers away from the junction, widening the depletion region. More importantly, it strengthens the internal electric field, $\vec{E}$, which points from the $n$-region to the $p$-region.

Now, think about those minority carriers we mentioned earlier. The holes in the $n$-region are positively charged. The strong electric field $\vec{E}$ exerts a force on them, pushing them across the junction into the $p$-region. This movement of charge carriers under the influence of an electric field is called drift.

Because the electric field in reverse bias points from $n$ to $p$, it sweeps the minority holes from the $n$-region into the $p$-region.

Therefore, the net flow of holes from the $n$-region to the $p$-region occurs when the $p$-$n$ junction is reverse biased.