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The diagram of a BJT implies that the two p-n junctions are identical. However, I do know that they are not since we cannot connect the transistor the other way around i.e swap emitter and collector, and still make it work.

So how do the collector and emitter side differ (in both npn and pnp) such that it prevents the emitter and collector from being swapped?

EDIT: This is the diagram I have been seeing for a very long time. enter image description here

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    \$\begingroup\$ In the old days, they were identical. I'm not sure just how far back that was. But for a small time. Even when I was first starting to learn books I read then said as much. My later experiments confirmed it with some devices I could barely afford to try out. Soon (early 60's?) pretty much everyone tumbled to the idea of varied doping to enhance \$\beta\$ at the cost of lowering the reverse voltage the base-emitter can withstand. The performance gains are worth the asymmetry and the old symmetric BJTs are long gone. You can, and people still do, swap E and C for certain specialized reasons. \$\endgroup\$
    – jonk
    Nov 26, 2022 at 0:00
  • \$\begingroup\$ The diagram you are likely looking at is a highly simplified affair. It's just to get the basics across. If you look at a physical cross-section of a modern lateral (or vertical) BJT and read about the doping profiles used you can very readily see that today there's nothing even close to see with respect to the 1D simplified diagram you are probably examining. And if you ever try to understand these things in 3D (full 3D integrals and edge effects) you'll see that a BJT is a rather complicated affair. It takes a while to study and only some take the time to fully apprehend all their details. \$\endgroup\$
    – jonk
    Nov 26, 2022 at 0:08
  • \$\begingroup\$ Which diagram? There's lots out there. If you refer to something, please cite it or post a copy of it or both. Certainly this diagram of a BJT doesn't show a same-sized collector and emitter. \$\endgroup\$
    – TimWescott
    Nov 26, 2022 at 0:38
  • \$\begingroup\$ I have added the diagram. \$\endgroup\$
    – gyuunyuu
    Nov 26, 2022 at 15:57
  • \$\begingroup\$ I remember reading long ago that an inverted BJT could have lower noise. On the other hand, I also remember that the source didn't back this statement in any way. \$\endgroup\$
    – fraxinus
    Nov 26, 2022 at 18:34

2 Answers 2

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Regarding swapping, it's not "prevented", just discouraged by the characteristics.

Back in the days of symmetrical (typically germanium alloy junction) types, it didn't make much difference; perhaps thermally (heat dissipates in the collector junction, which is bonded to the case, so is slightly closer).

Specialty types ("muting" or analog-switch transistors) have been made, although still asymmetrical, but with modest voltage ratings (VEB > 10V?), hFE (inverted hFE > 50, say?) and characterized in terms of RCE(on).

A word about VCE(sat) first. Normally, this is limited by the built-in potential due to the doping profile of the collector region and junction. While you might think it's not possible to saturate below VBE, that's actually not the case: effectively, charges are sucked across the collector junction and so the junction's built-in potential cancels out most of that voltage. What doesn't cancel out, is the potential due to doping concentration, and because emitter is typically higher doped, VCE(sat) consequently has a small minimum value -- some 10s of mV typically.

In inverted mode, the built-in potential works for you, so although hFE might be poor, at least saturation voltage can be quite small. There's no free lunch of course, you can't get negative ~mV of saturation -- but it can get very close to zero at least. (That's actually not so clear of a conclusion, mind; it's not that a negative C-E voltage drop would come out of nowhere, there is a power source here: base current. As it turns out, base current doesn't help, or the voltage drop across bulk resistance dominates over the built-in potential anyway. There is, in fact, one operating condition where negative voltages can be generated by a transistor: E-B avalanche, which puts C-B slightly negative, as if through photovoltaic means -- indeed an avalanched junction does emit light, albeit very feebly in silicon, so this is indeed quite an obscure (and not very useful) characteristic!)

Anyway, through a combination of geometry and doping, such muting transistors are made to give reasonably low and symmetrical VCE(sat), so that it looks more or less resistive. (Rohm 2SD2704K is a current production example, though it only has the high VEB and doesn't specify other characteristics. Also an interesting example in that, despite the high VEB, forward hFE is extremely high; I'm guessing the base is extremely thin, which will reduce fT (seems to be the case) and also exaggerate Early effect. Unfortunately they don't give collector curves or h parameters that would show the Early effect.)

There are other types that have surprisingly good inverted behavior, given they aren't made for it (in the way muting transistors are). Typically, low-VCE(sat) types (example: Nexperia PBSS303NX) also have good inverted characteristics, presumably a consequence of the push for low VCE(sat) in general. It's been a while since I measured, but I recall something like 75-150 inverted hFE on one of these types. Curiously, the SPICE models do not usually reflect this -- presumably they sacrifice some inverted accuracy (they're not specified in that region, who cares, right?..) for better forward accuracy.

So, not that the characteristic is all that useful -- it's the same ~7V VEBO that most general-purpose transistors have -- but interesting, anyway.

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  • \$\begingroup\$ Nice that you included the photovoltaic part. I've experienced it and it is definitely different on different BJTs. Some I couldn't get a detection from. But I now need to get some of the Rohm 2SD2704K and subject them to testing. Sounds like it may be fun/interesting. I didn't know about the category of 'muting transistors.' Completely ignorant. Now I know. So now I'm curious. Thanks! \$\endgroup\$
    – jonk
    Nov 26, 2022 at 9:32
  • \$\begingroup\$ Models may be inaccurate because the parameter isn't controlled. I've seen dramatic differences in the inverted beta between different batches of 2N2484. \$\endgroup\$
    – John Doty
    Nov 26, 2022 at 13:20
  • \$\begingroup\$ @JohnDoty Fair point -- and I only measured a few of the one type, too! \$\endgroup\$ Nov 26, 2022 at 13:44
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In almost all bipolar junction transistors (bjt's) the emitter is much more heavily doped than the collector, and this difference is responsible for the asymmetry of bjt.

The reasons why the emitter is usually much more heavily doped than the collector, is that the design goals for the emitter and collector are different. The most common design goal for the emitter, is to make it efficient at transferring its majority carriers to the base (where they become minority carriers). The most common design goal for the collector is that the maximum \$V_{CE}\$ of the transistor is sufficient for practical applications. These two different design goals lead to the use of different doping concentrations in the emitter and in the collector.

The ratio between the doping concentrations on the two sides of a pn junction affects how many majority carriers, in each direction, cross over the junction to become minority carriers. In a transistor the emitter is more heavily doped than the base (as well as of the collector). It is an objective of most transistor designs to efficiently have the majority carriers in the emitter cross over the emitter-base junction and become minority carriers in the base. However, any base-majority carriers that cross over the emitter base junction and become minority carriers in the emitter, will recombine with majority carriers in the emitter. This will reduce the number of majority carriers in the emitter that arrive at the emitter-base junction and cross over it. Therefore, reducing the number of majority carriers in the base that cross over the junction and become minority carriers in the emitter is a design objective. That design objective is met by making the emitter more heavily doped than the base.

Another design objective of bjt's is to have a relatively high reverse break-down voltage for the base-collector junction. The reverse break-down voltage for the base-collector junction limits the maximum collector to emitter voltage. If the collector had the same doping level as the emitter, it would have a similar reverse break-down voltage, which is often quite low, perhaps in the vicinity of 5V. So, the collector is less heavily doped than the emitter to give the transistor a higher maximum collector to emitter voltage.

This is why a transistor is almost always asymmetric. However, it is not exactly true that you cannot operate a transistor with the emitter and collector reversed. You can, but it's performance in terms of "beta" (the ratio between collector and base currents) and in terms of maximum operating voltage, will both suffer. There are however, some specialty cases where transistors are operated in this "reverse active region".

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  • \$\begingroup\$ What reason could one have for operating the transistor in reverse active region when the beta is so much lower? \$\endgroup\$
    – gyuunyuu
    Nov 30, 2022 at 22:58
  • \$\begingroup\$ @Quantum0xE7 per Tim Williams answer, exploiting the extremely low \$V_{CEsat}\$ might be a motive for using a transistor in the reverse active region. \$\endgroup\$ Dec 1, 2022 at 15:22
  • \$\begingroup\$ Yes but could you give an application where this makes sense? \$\endgroup\$
    – gyuunyuu
    Dec 4, 2022 at 14:18

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