Electronic – Why can current flow through the reverse biased base-collector junction (N-P junction) in a BJT with a forward biased base-emitter junction

First, lets look at why there's only very small current in a reverse-biased pn-junction diode. The junction doesn't block all current when reverse biased. The electric field in the junction opposes the majority carrier current whether forward-biased or reverse-biased, but quickly sweeps any available minority carriers (electrons in the p-region, holes in the n-region) across it. In forward bias, minority carriers are being continuously injected from the contacts, so there is a sustained current. In reverse bias conditions, there's very few minority carriers available, so the junction carries all the available carriers away in a very short time, and there's no more carriers available to sustain a current.

So what happens in a forward-active BJT is that the forward biased base-emitter junction creates a large number of minority carriers in the base region. The (reverse-biased) collector-base junction then has no problem "finding" carriers to create a current, and so you can have a large collector current.

It is not correct that the depletion regions of the two junctions overlap. If that happens, you have a condition called "punch-through" where there is no gain in the device.

I found a slide-set that gives a very quick explanation of BJT operation here. In particular note that current in the depletion regions is mainly caused by "drift" (carriers being pushed around by electric fields), but in the bulk regions it is mainly caused by diffusion --- that is simply carriers randomly moving around, so that the net movement is from areas of high concentration to areas of low concentration. Finally, remember that the important currents are the minority-carrier currents.

Edit

My explanation of forward biased operation was not right. Let's try again: Whether the junction is forward or reverse biased, the electric field in the depletion region (the area right around the junction) opposes "majority" carriers crossing the junction and encourages minority carriers. In forward bias, the size of this barrier is reduced to the point where some fraction of the majority carriers have enough thermal energy to overcome the barrier. But anyway, the operation in reverse bias is more important to answering your question.

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.