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In "Principles of transistor circuits, 8th edition" (top of p26) it is said that once holes have been injected in the base region by a forward biased base-emitter junction (which I understand) of a PNP transistor, they diffuse towards the collector and are then swept across.

However, 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?

Edit after the answers: After drawing a diagram it is much clearer. The potential barrier creates a field reinforced by the reverse bias, but that field goes against holes going towards the base - it's actually accelerating holes from the base to the collector. enter image description here

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  • \$\begingroup\$ I've seen answers on SE that say "the base is thin so carriers end up on the depletion layer and hence are swept across", which I find counter-intuitive. \$\endgroup\$ – Mister Mystère Sep 17 '14 at 16:41
  • \$\begingroup\$ The CB junction is reverse biased. So Collector is aat lower potential than base. Holes are attracted to negative polarity, they move from higher to lower potential. So where is the barrier ? I think you overlooked the fact that holes are in minority in the base for PNP. \$\endgroup\$ – Plutonium smuggler Sep 17 '14 at 17:04
  • \$\begingroup\$ I understand that they would, if they weren't stopped by the potential barrier (which is reinforced by the CB reverse bias) that I regard as a wall. I know I'm wrong, I just don't know where. \$\endgroup\$ – Mister Mystère Sep 17 '14 at 17:07
  • \$\begingroup\$ There is a barrier at CB junction, but its for majority carriers. Holes are minority carriers in the base ( bcoz base is negative). Think about it. Holes always move from higher to lower potential. Base is at higher potential than collector because of reverse bias. So the Electric field is directed from base to collector. The barrier is for holes in collector trying to move towards base, not for holes in base trying to move towards collector. \$\endgroup\$ – Plutonium smuggler Sep 17 '14 at 17:14
  • \$\begingroup\$ Even after Alfred Centauri's answer it was still unclear so I've drawn a diagram (shown) and then it hit me. Stupid mistake, in my head what affected minority carriers had to affect majority carriers but they are of different polarities. Thanks. \$\endgroup\$ – Mister Mystère Sep 18 '14 at 9:29
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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.

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  • \$\begingroup\$ Thanks, stupidly enough in my head what affected minority carriers had to affect majority carriers but after drawing the diagram in my edited post, it hit me that obviously from their different polarities it wasn't the case. \$\endgroup\$ – Mister Mystère Sep 18 '14 at 9:33

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