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A transistor is a three terminal device. One terminal is called emitter, one collector and in between them is base. Now, during biasing the junction between emitter and base is made forward biased and the junction between collector and base is made reverse biased.

My question is that if one of the junction of a transistor is reverse biased, how does the transistor allow current to flow through it because the reverse biased junction (diode) doesn't allow the current to flow through it?

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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.

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Your question is fundamentally how a bipolar junction transistor (BJT) works. That is too complicated to get into here, and there is much written about it out there.

Very briefly and with a lot of hand waving: The collector junction is reversed biased as you say, which is why a BJT is off when nothing else is done to it. The special property of a BJT as apposed to just two separate junctions, like to diodes wired together, is that running a little current thru the base causes the collector junction to "leak" considerably more current. This is possible because the depletion regions of the two junctions interact. A BJT is NOT just two diodes, but one integrated device.

Again, there is much written out there about BJTs. Go look it up. Duplicating all that here would be lengthy and pointless.

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  • \$\begingroup\$ It is not correct that the "depletion regions of the two junctions overlap" in forward-active operation. See my answer for slightly more detail. \$\endgroup\$ – The Photon Jan 8 '13 at 17:06
  • \$\begingroup\$ @TheP: Maybe they don't overlap in all modes of operation, but that's really what makes a transistor different from two back to back diodes. The base has to be "thin". When the base is thick enough so that the depletion regions of the two diodes don't interact anymore, then you don't have a transistor anymore, just two diodes. \$\endgroup\$ – Olin Lathrop Jan 8 '13 at 18:49
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    \$\begingroup\$ They interact, but they don't overlap. When they overlap, the transistor action stops working. \$\endgroup\$ – The Photon Jan 8 '13 at 18:50
  • \$\begingroup\$ @TheP: OK, I'm not going to split hairs. I have edited the answer accordingly. \$\endgroup\$ – Olin Lathrop Jan 8 '13 at 18:54

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