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.
Any Bipolar junction transistor (BJT) constitutes :
- a heavily doped emitter
- a lightly doped and small size base
- a moderately doped collector
So according to your question, the emitter should have been doped to a concentration of 10^23/cm^3, the base doped to 10^16/cm^3 and the collector doped to a concentration of 10^18/cm^3.
Now your assumption about the emitter being connected to a P region, the base to a N region and the collector to a P region is correct.
And keep in mind that a Si atom has 4 valence electrons and that hole concentration will be in the order: emitter > collector > base and reverse for electron concentration.The calculations are quite easy to do.
Best Answer
No, you're incorrect.
The Doping doping levels are:
E: highest doping level
B: less than E
C: weakest doping, less than B
It needs to be like that because the ratio of B/E doping sets the beta
The collector has the weakest doping level so that the depletion layer will be large so that it can handle a large reverse voltage. Remember that in the active region the BC junction is reverse biased !
The thickness of the base must be small, the base must be thin because the carriers coming from the emitter should be have already passed through the base and enter the collector region before they realize:
Oh, oops ! Too late to recombine in the base.
And, Oh, blast now we're pulled to the collector contact (because that will have a large positive voltage, for an NPN that is).
That is what makes the base current small and forces the carriers to go to the collector instead of recombining in the base.
The depletion region in the base will be smaller than the depletion region in the collector because the base has a higher doping level ! Your assumption that the base has a lighter doping than the collector is not true. It is the collector that has the lightest doping.