Now, if you have a higher concentration of dopants, you end up with more diffusion pressure which means the electric field must be bigger in order to oppose it, meaning that the depletion/space charge region must be wider. However, what I just said is incorrect, but i don't understand where my reasoning is wrong.
If you understand that there's a built-in voltage, then you should also know that distance and path is unimportant to a voltage. Voltages are scalar. 5V over 1 billion miles is the same as 5V over 2um. Now the electric fields associated with those two voltage are entirely different. Now the reason that the depletion region gets smaller with higher dopants is that there is more static charges exposed in the depletion region. Think of it like this, with a low dopant level I'd have to take 3mm of space to free "20" charges, but if they were a higher density, I'd only have to take 3um to free "20" charges. Remember it's the exposed charges in the depletion region that create the field, not the charges in the drift region.
Now again, if you have a higher electric field applied you should get proportionally less diffusion current meaning that the width of the depletion region can be smaller and still fully oppose the diffusion current. This is also incorrect reasoning, but where did i go wrong?
You went wrong when you thought about the field inside the depletion region effecting things outside the depletion region directly. The field is stuck there. It can't directly effect anything. It does have some secondary effects though. For instance the wider the depletion region, the easier it is to diffuse across the non-depletion areas. This is because there is less distance to the end of the device when there's more length in the field. We use this all the time in BJT's. The other things that a field will do is create a build up of charges on it's edges (You know, all those ones you ripped out to create the depletion region, they want back to where they were). These "free charge" (free because they are able to move, unlike in the depletion region where the charges are held in the atoms of the crystal lattice) areas are actually what is driving the diffusion in the other areas. The free charges are more concentrated by the depletion region than in other places. Again, they want back, but the built in field removes them every time they try so they're stuck at the edge. This creates the "High to Low pressure" that drives diffusion. The higher the field, the more charges you ripped away that want to go back that the field at the boarder resists them, the more concentrated the free charges are, the more diffusion there is. All linked. I'll Try to get picture in here. It really is difficult to see without picture.
Edit: I've found a good Youtube video with all the particles present and a very decent explanation. I hope this helps.
The distance that minority carriers can move across the other material is called the "diffusion distance". The time that it takes until the minority carrier disappears is called the "minority-carrier recombination lifetime".
The distance / time that minority carriers have is dependent on the number of recombination sites in the base material. Recombination sites are crystal defects.
If your majority carrier has better mobility than your minority carrier, you want to create more recombination sites close to the junction, so that your injected minority carriers are quickly converted to majority carriers.
Since electron mobility is better than hole mobility, you probably want your
"diffusion distance" and "carrier recombination lifetime" to be shorter for holes on the N side, and to be longer for electrons on the P side.
Best Answer
What you have to understand is holes don't actually exist, electrons are the only charge carriers in a semi-conductor but there are 2 types of electrons 1) free electrons and 2) bounded electrons.The "Electron flow" that is usually referred to in literature refers to the flow of free electrons and the "Hole flow" is the flow due to bounded electrons.
If a bounded electron moves in a particular direction it leaves a hole at the position it was previously at and it removes the hole at its new position.So it is equivalent to think of bounded electron flow as hole flow.
To understand the process better lets let b - bounded electron , h - hole and take the following illustration as a row of bounded electrons and a hole in a semi-conductor.
When there is a potential difference across semiconductor the bounded electrons will be attracted towards the positive terminal and so the bounded electron to the right of the hole will move left, this causes a hole to "move" right, the same process will repeat when the hole has moved to its new position, it will then seem as though the hole is the one moving.
$$ \text{+ b h b b b b -} \\ \text{+ b b h b b b -} \\ \text{+ b b b h b b -} \\ \text{+ b b b b h b -} \\ $$
Now with this knowledge in mind we can proceed to analyze a reverse biased semiconductor.Consider the following a negativily biased p-region (excess holes) next to a depletion region (all electrons bounded).
When a negative bias is applied on the p-type material holes will move left making the depletion region get bigger as it can be seen in this illustration.
originally: $$ \text{ b h b b b h} \hspace{0.3cm} \text{b b b b} \\ \text{ b b h b h b} \hspace{0.3cm} \text{b b b b} \\ \text{ b b b h b h} \hspace{0.3cm} \text{b b b b} \\ \text{ b b b b h b} \hspace{0.3cm} \text{b b b b} \\ $$
then $$ \text{- h b b b h b} \hspace{0.3cm} \text{b b b b} \\ \text{- b h b h b b} \hspace{0.3cm} \text{b b b b} \\ \text{- b b h b b b} \hspace{0.3cm} \text{b b b b} \\ \text{- b b b h b b} \hspace{0.3cm} \text{b b b b} \\ $$
then $$ \text{- b b b h b b} \hspace{0.3cm} \text{b b b b} \\ \text{- h b b h b b} \hspace{0.3cm} \text{b b b b} \\ \text{- b h b b b b} \hspace{0.3cm} \text{b b b b } \\ \text{- b b h b b b} \hspace{0.3cm} \text{b b b b } \\ $$
note how the depletion region has now increased by 2 columns, this is a simplified illustration of what happens.Hope the diagrams were clear enough, I know they are very crude.