3 ms·

Say an NPN BJT is in conduction mode with: Vbe = 0.7v, Ib = 1 mA, and Ic = 10 mA. In practice, Vce might be 0.2v or so, depending on the transistor. How can

Say an NPN BJT is in conduction mode with: Vbe = 0.7v, Ib = 1 mA, and Ic = 10 mA. In practice, Vce might be 0.2v or so, depending on the transistor. How can Vce < Vbe, when C-B-E essentially form a series circuit? I liked this video, but I don't think this model can explain this observed behavior.

gw99 4y ago
Using that simplified model, it's better to think of it as a simple diode between B-E and a completely separate current source between C-E. But it's way more complicated than that.
kragen 4y ago
As I understand it, and I'm not a quantum physicist so I may be wrong, electrons lose energy when they go from the emitter into the base, because they're moving from N material where their energy is high into P material where it's low, which is why you have that 0.7-volt drop. But, by the same token, electrons that move from the base into the collector gain energy. It's just that they're regaining the energy they lost when they crossed into the base in the first place. That is, the built-in potentials of the two diode junctions are in opposite directions, so in series combination those voltages mostly cancel out. Since the base-emitter current heats up the junction, but the collector-emitter current doesn't (or rather it heats it up to the tune of 0.2 milliwatts per milliamp rather than 0.7) it stands to reason that this collector-base transition must be sucking heat out of the junction when the electrons cross it. And I think this is actually how thermoelectric generators (TEGs) and Peltier coolers made out of semiconductors like bismuth telluride work. One problem with this account is that I think the band-bending diagram of a p–n junction has the electrons in the P material at a higher energy level, although I guess the relationship of the Fermi levels is opposite?
- exmadscientist 4y ago
  So I am a (former) quantum physicist (though the wrong kind, an experimental–elementary-particles type rather than a theoretical–condensed-matter type) and am now an electrical engineer, and I can say with confidence, bipolar junction transistors are hard to really understand. I don't think I've ever seen a single explanation that goes through it all in an understandable way. In particular, factors that have to be considered are the built-in potentials of the space charge layers (aka depletion zones), and the different doping profile of the collector versus emitter. The doping is very important, and is often neglected. But without it, transistors would basically operate equally well in forward mode as in reverse, and they definitely don't do that. These two things have a big effect on the Fermi/band-bend diagram and probably help clear up your last point. I barely feel like I actually understand these things on a good day, and I pay the bills with them....
  - Koshkin 4y ago
    > hard to really understand Just like QM itself, I guess. So, “shut up an calculate” all over again?
    - exmadscientist 4y ago
      They're different kinds of complexity, though. QM isn't too bad. The basic ideas are alien but straightforward enough. It's hard to extend small examples to larger systems, though, and that plus the innate weirdness really messes with your intuition. BJTs are complicated devices built from multiple simple-enough effects all happening at the same time. Each piece isn't too bad, but you've got to track them all and combine them into an overall function. QFT (quantum field theory, such as QED) is basically all the bad parts of both of those: the basics aren't too bad, but it's difficult to scale, there's a lot going on, and it's hard to develop intuition. QCD then makes the basic calculations intractable, which is why it's still poorly understood to this day (in my opinion). So, yeah, in conclusion, complexity is complex? Not so profound, I guess.
      - kragen 4y ago
        Yet Shockley was able to make quantitative predictions about, say, solar cell performance that are still considered useful today. And I don't think kT shows up en the Ebers–Moll model just by chance.
  - kragen 4y ago
    Btw I think that typically not just the doping but also the geometry of the emitter and collector are very different.

gw99 4y ago

Using that simplified model, it's better to think of it as a simple diode between B-E and a completely separate current source between C-E. But it's way more complicated than that.

kragen 4y ago

As I understand it, and I'm not a quantum physicist so I may be wrong, electrons lose energy when they go from the emitter into the base, because they're moving from N material where their energy is high into P material where it's low, which is why you have that 0.7-volt drop. But, by the same token, electrons that move from the base into the collector gain energy. It's just that they're regaining the energy they lost when they crossed into the base in the first place. That is, the built-in potentials of the two diode junctions are in opposite directions, so in series combination those voltages mostly cancel out. Since the base-emitter current heats up the junction, but the collector-emitter current doesn't (or rather it heats it up to the tune of 0.2 milliwatts per milliamp rather than 0.7) it stands to reason that this collector-base transition must be sucking heat out of the junction when the electrons cross it. And I think this is actually how thermoelectric generators (TEGs) and Peltier coolers made out of semiconductors like bismuth telluride work. One problem with this account is that I think the band-bending diagram of a p–n junction has the electrons in the P material at a higher energy level, although I guess the relationship of the Fermi levels is opposite?

exmadscientist 4y ago
So I am a (former) quantum physicist (though the wrong kind, an experimental–elementary-particles type rather than a theoretical–condensed-matter type) and am now an electrical engineer, and I can say with confidence, bipolar junction transistors are hard to really understand. I don't think I've ever seen a single explanation that goes through it all in an understandable way. In particular, factors that have to be considered are the built-in potentials of the space charge layers (aka depletion zones), and the different doping profile of the collector versus emitter. The doping is very important, and is often neglected. But without it, transistors would basically operate equally well in forward mode as in reverse, and they definitely don't do that. These two things have a big effect on the Fermi/band-bend diagram and probably help clear up your last point. I barely feel like I actually understand these things on a good day, and I pay the bills with them....
- Koshkin 4y ago
  > hard to really understand Just like QM itself, I guess. So, “shut up an calculate” all over again?
  - exmadscientist 4y ago
    They're different kinds of complexity, though. QM isn't too bad. The basic ideas are alien but straightforward enough. It's hard to extend small examples to larger systems, though, and that plus the innate weirdness really messes with your intuition. BJTs are complicated devices built from multiple simple-enough effects all happening at the same time. Each piece isn't too bad, but you've got to track them all and combine them into an overall function. QFT (quantum field theory, such as QED) is basically all the bad parts of both of those: the basics aren't too bad, but it's difficult to scale, there's a lot going on, and it's hard to develop intuition. QCD then makes the basic calculations intractable, which is why it's still poorly understood to this day (in my opinion). So, yeah, in conclusion, complexity is complex? Not so profound, I guess.
    - kragen 4y ago
      Yet Shockley was able to make quantitative predictions about, say, solar cell performance that are still considered useful today. And I don't think kT shows up en the Ebers–Moll model just by chance.
- kragen 4y ago
  Btw I think that typically not just the doping but also the geometry of the emitter and collector are very different.