BJT in Active/Linear mode -- How can it still conduct current?

Question

So I've read quite a few books on how BJT's work and all of them seem to gloss over how a BJT in the linear mode with the base/emitter junction forward biased and the base/collector junction reversed biased is still able to conduct current from emitter to collector.

In this situation, the base/emitter PN junction is biased in such a way that the depletion region is very small or negligible. However, the PN junction of the base/collector region has its depletion region widened by the reverse biasing; which makes you think at first glance that no current should be allowed to conduct.

Here is my take on the device theory of why we still see (large) current from emitter to collector for an NPN BJT in the Active/Linear mode. Please correct me if my theory is wrong or has some misconceptions:

Since the base is very thin in relation to the collector and emitter, and it is lightly doped to a P-type material, 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. When the mass amounts of electrons in the emitter get pushed into the base by the emf of the base/emitter battery, the vast majority of them have no holes to combine with. Now we have a situation where we have a high entropy of electrons in the base which begin to defuse into the base/collector depletion region; effectively doping and shrinking the base/collector depletion region into N-type material. When enough electrons diffuse, the depletion region all but disappears into n-type material, allowing the positively biased n-type collector region to sweep those diffused electrons out of the collector terminal. Since the collector is moderately doped to N-type, the diffused electrons are able to pass right out the positive-polarity collector terminal.

So, am I right about entropy playing a role in this by diffusing the base/collector depletion region with electrons to effectively dope it into n-type material?

Thanks!

Answer 1

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.

Answer 2

Base-width modulation is indeed an effect, but only a second-order one. Transistor gain (beta) occurs because the transit time tt of injected electrons across the base is much less than the recombination time tn of said elections within the base, so most of the injected electrons reach the reverse-biased base-collector junction and get swept across it (exiting as collector current) instead of exiting as base current. Numerically, beta = tn/tt.