If the base-emitter junction of a BJT is forward biased, then current can flow through the reverse biased base-collector junction (N-P junction). This disagrees with my understanding of the PN junction, as I thought electrons cannot flow from the P-side to the N-side of the reverse biased junction, since there is a depletion region between them.

I understand why the current can flow through the forward biased base-emitter junction: the external voltage (positive connected to P-side, negative connected to N-side) creates an electric field from the N-side to the P-side, which cancels out the built in electric field caused by diffusion of carriers across the dissimilar materials. This collapses the depletion region.

However, in the reverse biased base-collector junction the external voltage will support the built in potential and cause a larger electric field (from the N-side to the P-side), which will stop positive charges flowing from N to P, and stop negative charges flowing from P to N.

But if you forward bias the base-emitter junction, and reverse bias the base-collector junction, electrons can still flow from the collector to the base, which is from P to N, which as I just explained in the previous paragraph should not be able to happen?

So what allows the electrons to flow through the reverse biased PN junction, as in the case of the collector-base junction of a BJT?

  • \$\begingroup\$ I think, your question touches the basics of BJT operation and is well documented/explained in each textbook dealing with BJT working principles. \$\endgroup\$ – LvW Aug 15 '14 at 12:15

In a nutshell, bipolar junction transistors work because of the physical geometry of the two junctions. The base layer is very thin, and the charge carriers that are flowing from the emitter to the base do not recombine right away — most of them pass right through the base altogether and enter the depletion region of the reverse-biased base-collector junction. Once this happens, the strong field in this region quickly sweeps them the rest of the way to the collector terminal, becoming the collector current.

  • \$\begingroup\$ Yes - and some of these carriers take the "short way" to the base node - and, thus, form the base current. One could add: unfortunately, because otherwise the BJT would have (like a FET) a high resistive input node (base). \$\endgroup\$ – LvW Aug 15 '14 at 12:51
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    \$\begingroup\$ I wouldn't say "unfortunately". Otherwise we would only have FET-like devices. BJTs continue being a basic building block of nowaday's electronics even if largely replaced by FET devices in many applications. \$\endgroup\$ – jose.angel.jimenez Sep 21 '14 at 8:41

In BJTs, charge-carriers are able to cross the reverse-biased CB junction because those charge-carriers have been placed in the "wrong" side.

  • Normal diode: the n-doped side is full of free electrons, while the p-doped side is full of mobile conduction-band vacancies (holes.) A forward bias will produce e-fields which force the two populations together, turning the diode "on." A reverse-bias voltage will pull them away from each other, forming a potential barrier and turning the diode "off."

  • Transistor CB diode: the p-doped base is full of free electrons! That's backwards. If they remain there for long, they'll quickly fall into holes, which ends up producing an EB base current. However, if first they wander too close to the CB junction because of thermal diffusion, they'll be gripped by the large e-field within the depletion layer, and violently attracted into the n-doped collector region. (Fast charge-carriers crossing a large potential, that heats up the transistor because of Ic x Vcb.)

(That example was for NPN. For a PNP transistor, we reverse the carrier polarities, where free holes have been deposited in the n-doped base, and they tend to wander too close to the CB junction and get violently flung into the p-doped collector region.)

Conclusion: diodes don't work normally if we somehow moved some of the n-side charge-carriers over into the p-side, or vice-versa. The extremely thin base-region in BJTs allows this to happen. In that case, a diode starts working backwards, and its reverse-bias voltage causes increased current, not decreased current. Yet at the same time it also works normally, with the reversed junction preventing the free electrons of the Collector region from flooding into the Base region.

Notice that the out-of-place carriers can easily pass through the depletion layer. The depletion layer was never actually an insulator; it doesn't pin charges in place like rubber or plastics do. Instead it was more like a vacuum region, with the intense e-field repelling the usual carriers, keeping them on their proper side.

But when we have out-of-place charge carriers in the diode, that e-field in the reverse-biased depletion layer no longer behaves like a potential barrier. Instead it becomes a particle-gun. (Heh, see, the NPN transistors can be viewed as triode vacuum-tubes! Depletion regions are the vacuum, and doped silicon forms the electrodes. Of course the Base must be very thin, because it acts like a perforated grid electrode!)

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    \$\begingroup\$ Your clarification that depletion region is more like "vacuum" rather than "insulator" was the key for me! I would also add a comment re: "the p-doped base is full of free electrons!". Not just the p-doped base is full of free electrons, it is also the p-doped side of CB depletion region is full of free electrons! The n-doped side of CB depletion region is attracting electrons so the electrons are now being attracted to collector by CB junction. \$\endgroup\$ – akhmed Jul 25 '17 at 9:39
  • \$\begingroup\$ I would also add that "the large e-field within the [CB] depletion layer" comment implies that CB depletion layer must be very thin for the e-field to reach the base. I assume this is why collectors are doped less than emitters. \$\endgroup\$ – akhmed Jul 25 '17 at 9:39
  • \$\begingroup\$ @akhmed note that the CB depletion layer is already inside the Base region, also it extends partly inside the Collector. It grows wider when CB voltage (collector supply) is higher. But, all the free charges inside the main Base region don't see the depletion zone's field, because the base is conductive, and behaves as an electrostatic shield. Only the carriers at the edge of the depletion zone 'see' the e-field, and are pulled into it. Random-walking carriers are grabbed up by collector, so they cannot random-walk back down into base again as usual. That's a diffusion current inside the base. \$\endgroup\$ – wbeaty Aug 23 '17 at 7:24

The Base Collector Junction is Reverse Biased which means that no current flows from the Collect to the3 Base. It doesn't mean current can't flow from the collector to the emitter. When The Base Emitter P-N Junction is forward Biased it ceases to act like a diode and is just a conductor and since the N-Doped Collector region is filled with electrons and the Baser-Emitter Depletion region no longer exists the Current Flows From Collector to Emitter. BUT there still is a reverse biased P-N junction between collector and Base so no Current flows from the Collector to the Base.


The main cause of collector current in case of bjt is the size of base region. The base region is very small in area in contrast to emitter and collector. Taking the npn transistor, we simply connect the emitter side to 'x' volt, base to 'y' volt and collector to 'z'volt, where x < y < z must be strictly followed to forward bias the emitter-base region and reverse bias the collector- base region. Since the emitter side is -ve with respect to the base, the electric field is established from base to emitter. So the major charge carriers(electrons) move towards the base. Now comes the core of transistor. Once the electron enters the base (which is very thin and lightly doped), very few electrons(5 out of 100) get the opportunity to recombine with holes. In addition to this, the electrons cant stay in base region for long time due to its small area. When you(electron) come running from ground(emitter) towards home(collector), you cant stay on the gate(base) for long time. You pass through the gate immediately. The same thing happens with electron. It passes to the collector-base depletion region immediately. So the charge carriers are electrons in the depletion region. The electric field established in this depletion region is directed from collector to base due to reverse bias. So the electron again tends to move towards the collector giving rise to collector current.enter image description here

If you simply search on web, you can see that the base is small in size like in the above one.


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