Electronic – Identifying parts of a transistor die

The first key, so they say, to understanding BJT behaviour is to understand that its driven by minority carrier behaviour. In an NPN device, that means that electrons in the p-type base region control the behaviour.

I think you captured that in your description, but most of the rest of what you wrote doesn't fit the usual way of describing the physics.

Since the base is very thin in relation to the collector and emitter, ... there are not many holes available to be recombined with emitter electrons. The emitter on the other hand is a heavily doped N+ material with many,many electrons in the conduction band.

This is the only part of what you wrote that makes sense. The forward bias on the b-e junction creates excess carriers in the base region. There are not enough holes to recombine with those electrons instantaneously, so the region of excess holes extends some distance from the beginning of the depletion region associated with the b-e junction. If it extends far enough, it will reach the opposite depletion region (for the c-b junction). Any electrons that get to that depletion region are quickly swept away by the electric field in the depletion region and that creates the collector current.

OK, so how is entropy involved?

A key point is that the spread of excess electrons away from the b-e junction is described by diffusion. And diffusion is, in some sense, a process that takes a low-entropy situation (a large number of particles segregated in one part of a volume) and turns it into a high-entropy situation (particles spread evenly across a volume).

So when you talk about "a high entropy of electrons", you actually have it backwards. Diffusion actually acts to increase entropy, not reduce it.

The idea that excess electrons are "effectively doping and shrinking the base/collector depletion region into N-type material" also doesn't make any sense. The excess carriers don't affect the extent of the c-b depletion region much. Electrons that reach the c-b depletion region are simply swept through by the electric field.

I thought the collector-base junction was reverse biased which means that the electric field created by the external bias adds to the potential barrier? How is it that holes spontaneously cross this huge potential barrier?

The built-in potential stops the diffusion current due to majority carriers diffusing from one side to the other. For example, electrons in the N material would diffuse to the P material (where they would recombine) if it weren't for the built-in potential. For the majority carriers, the built-potential is a barrier.

But in the N material of the base, holes are minority carriers and so, the built-in potential isn't a barrier at all. If a hole in the base exists long enough, it may be swept across the reverse biased base-collector junction by the electric field there and into the collector region.

When the base-emitter junction is 'on', lots of holes are injected into the thin and lightly doped base region so a large fraction of the injected holes exist long enough to 'slide down' the potential and into the collector region where they are majority carriers.