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Carrier flow in BJT when saturated

  1. Oct 24, 2014 #1
    I would like to know how carriers flow in a bipolar junction transistor when working in saturation mode.

    Normally, electrons would flow from the emitter to the base. Then, from the base, some recombine with holes nad leave, forming the base current, and the others (the majority) diffuse in the collector and are swept because of Vcc (forming the collector current). But when saturated, the collector base junction is forward biased, so some electrons should be moving from the collector to the base. Is this right? Or there are so many electrons in the base and because the base is so thin they diffuse into the collector and are still "attracted" by the "positive" potential of Vcc (even though the normal electric field between the base and the collector would push electrons the opposite way if it were for a wire instead of a pn junction)?

    What is the correct flow of charge carriers when the transistor is saturated?
  2. jcsd
  3. Oct 24, 2014 #2
    First of all let's address your statement:
    "Normally, electrons would flow from the emitter to the base. Then, from the base, some recombine with holes nad [sic] leave, forming the base current, ---"
    This small number of electrons recombining in the base forms only a tiny fraction of the base current, called the "transport component". Nearly all base current, at low frequency, is due to holes injected from base to emitter when b-e junction is biased by an external source, and is called the "injection component". If 0 electrons recombined in the base in transit from emitter to collector, the base current is barely changed. The injection current is nearly all the base current. The part you described is the transport current, and is negligible. I just thought we would clear that up.

    In saturation, the base-collector junction is forward biased, so that base-collector current becomes substantial. Holes from base enter collector, and electrons from collector enter base. The collector current has 2 main parts, the electrons coming in from the emitter via the base, and the electrons leaving collector towards the base. They are opposite in direction.

    This is why once a bjt has entered saturation mode, increasing base current has little influence on collector current. The increased electron collection from the emitter is countered by increased electron emission by the collector into the base. As base current increases, the emitter current increases while the collector current increases only slightly.

    Did I help at all?

  4. Oct 25, 2014 #3
    Where are the electrons that go from the collector to the base coming from? Are electrons injected by the emitter into the base, then diffuse into the collector, and some of them are "attracted" by the collector's lead potential and some are swept back into the base, again? Thank you.
  5. Oct 27, 2014 #4
    Electrons in the collector are majority carriers, they are abundant since collector is n-type material. When bjt saturates, b-c junction is forward biased as well as b-e junction. The signal source driving the base provides positive (hole) current into the base, and an E field that sources holes, and sinks electrons. The Vcc rail provides an E field that sources holes into the collector, and sinks electrons.

    Regarding your question on where electrons come from that go from collector to base, Drs. Ebers & Moll described this in 1954 by modeling the bjt as 2 bjt devices, one right-side-up, the other upside-down. The b-c collector junction when forward biased behaves like a b-e junction forward biased. The collector region is the "emitter' of the upside down bjt. This upside down emitter current is opposite in direction to the right side up collector current.

    Do electrons migrated from emitter through base into collector return to the base? Statistically some of them do. When an electron enters an n type material, it is a majority carrier with long lifetime and high mobility. It can survive a long time and eventually be drawn into the base by the b-c E field. Some electrons from the emitter ending up in the collector survive the trip through the entire collector and are sinked by the Vcc rail.

    Did I help?

  6. Oct 27, 2014 #5
    So if the potential of the base is increased, more electrons from the emitter go into the base (like in the forward active region), but now electrons that arrived from the emitter in the collector, instead of leaving through the collector lead, they go back into the base, because the base collector junction is forward biased. It that it? Thank you for your help!
  7. Oct 27, 2014 #6
    In part yes, but please remember that electrons arriving in the collector from the emitter undergo 4 possible scenarios.
    1) Some go back into the base and exit the base lead because the b-c junction is forward biased as you just stated. Or:
    2) Once the electron has left the collector to enter the base (again), it may encounter recombination with an atom that is positive ionized (excess hole), or recombine with an acceptor atom (excess hole), thus ionizing it negatively. Eventually another electron exits the base lead preserving charge neutrality in the base region. The base is p type material, and an electron is a minority carrier. Its mobility is low because there are an abundance of holes in the base, providing a probability of recombination greater than that for n type material.
    3) Some travel through the collector region and exit the collector lead. In the collector, an n type material (for an npn bjt), has an excess of negative carriers, i.e. electrons. Thus an electron drifting/diffusing through the collector has a lower chance of recombining with a hole, than it would if the material were p type. In the n type collector, the electron is a majority carrier. Still:
    4) Electron-hole pairs (ehp) are constantly being generated by lattice vibrations (thermal energy) in both the n type and p type materials. When an electron is separated from its parent atom, a hole remains, and the atom is positively ionized. Thus the electron moving through the collector can recombine with said ion. Eventually another electron exits the collector lead, and charge neutrality is kept.
    I hope this makes things easier to understand. This stuff is not trivial, and it baffles even the best minds.

  8. Oct 27, 2014 #7
    Thank you for your time and help!
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