Diffusion current
When a p-n junction is formed, a diffusion phenomena causes electrons from the n-doped region to diffuse to the p-doped region. At the same time (even if it's an abstraction) holes diffuse from the p-type region to the n-type one. The atoms that lose a carrier (electron or hole) become ions, which means that instead of being neutral, they have a positive or negative net charge. This happens because the ideal equilibrium would have the same concentration of mobile carriers equal all over the region.
Ohmic current
However, this diffusion causes the growth of a region, populated by ions, called depletion region, because all atoms have lost their carrier. These ions, as we said, are electrically charged, and cause an electric field directed from the n-region to the p-region, pushing carriers in the opposite way than diffusion. Therefore an equilibrium is reached in which the current (movement of carriers) caused by diffusion is perfectly balanced by the current caused by the electric field (ohmic current).
Effect of biasing
Applying a potential to the junction causes a perturbation on this equilibrium, making one of the currents dominant on the other. Reverse biasing the junction causes the ohmic current to prevail, while forward biasing increases the diffusion current.
Now, the diffusion current is a much stronger phenomena, from which derives the exponential growth of the forward bias current with the bias voltage. Ohmic current, on the other side, is much weaker, and saturates quite soon (neglecting avalanche effect) because the width of the depletion region (which determines the resistivity) is proportional to the reverse bias voltage.
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.
Best Answer
I'll just steal a diagram from the web so we know what we're talking about:
(Image source: circuitstoday.com)
In forward bias, holes are injected from the p-region to the n-region and then diffuse from left to right. That means current is flowing from left to right in the n-region.
Electrons are injected from the n-region to the p-region and diffuse from right to left. Since electrons have negative charge, this means conventional current is flowing from left to right, the same as it is in the n-region.
In each region the injected minority carriers recombine with the much more numerous majority carriers. A tiny average drift of the majority carriers to "fill in" where a tiny fraction of them have recombined is enough to maintain a continuous current through the whole structure.