Electronic – BJT operation and conservation of charge

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.

You are using the wrong topology for what you are trying to achieve. Here is a better circuit:

First, we figure out R2. This is what sets the LED current. You want 700 mA. D1 drops 1.4 V, and let's say the transistor saturates to 300 mV. That leaves 3.3 V across the resistor. (3.3 V)/(700 mA) = 4.7 Ω. Note that this resistor will dissipate (3.3 V)(700 mA) = 2.3 W.

Now for the base resistor. You need 700 mA collector current, so you need at least that divided by the transistor gain as base current. Let's say the minimum guaranteed gain of the transistor at 700 mA is 50. (700 mA)/50 = 14 mA minimum required base current. Figure the B-E drop will be 750 mV, so that leaves 4.25 V across R1. (4.25 V)/(14 mA) = 304 Ω. So the largest common value that guarantees everything within spec is 270 Ω.