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We all know that when the transistor is in its linear or active region, collector current does not increase further. At this point, the base collector junction is reverse bias. But how come there is still a collectr current flowing given the fact that the B-C junction is reversed biased.

Isnt it when a diode is in reverse bias, it blocks current, but how come in this setup above, there is still collector current?? Also why does the current doesn't increase any further?

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Your question is about the fundamental operation of a BJT. Surely there is much written about that out there.

Briefly, C-B-E is a sandwich of three semiconductors of opposite polarity doping. However, what makes a BJT more than just two diodes in series pointing in opposite directions is that the base region is so thin that the depletion region of each junction extends to the other junction. The collector is still within reach of the emitter, if it weren't for the base region in between with all its carriers depleted. A little bit of externally applied base current injects carriers into the base region, which now allows current to flow accross it between collector and emitter.

Due to a whole bunch of semiconductor physics you should look up elsewhere, a few carriers in the base go a long way. This is where the transistor gets its gain from. You inject a few carriers into the base (provide a small base current) and that lets a lot of carriers conduct (allows a larger current to flow) accross the otherwise depleted base region between collector and emitter.

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  • \$\begingroup\$ i get it,, so, the overlapping feature causes the flow of current. :) \$\endgroup\$ – WantIt Sep 13 '12 at 11:42
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A BJT is two junctions in close proximity to each other — the base region is very thin. The B-C junction can carry current because charge carriers flowing through the B-E junction (electrons or holes) drift all the way through the base region into the B-C junction. The collector current is limited by the number of charge carriers made available by this action, and is pretty much completely independent of collector voltage.

The number of carriers (current) through the B-E junction is controlled by the voltage across it (VBE), using the standard diode equation. The emitter current gets divided between the collector and the base; the ratio between them is called the current-transfer ratio, or hFE. If VCE is more than a few volts, most of the emitter current flows through the collector and a tiny fraction (a few percent) flows through the base. As VCE drops towards zero volts, however, a larger fraction of the current flows through the base.

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