BJT Emitter to Collector current in linear mode, how?

Discussion in 'General Electronics Chat' started by SiegeX1, Mar 10, 2009.

  1. SiegeX1

    Thread Starter Member

    Mar 10, 2009
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    So I've read quite a few books on how BJT's work and all of them seem to gloss over how a BJT in the linear mode with the Base->Emitter junction forward biased and the base->collector junction *reversed* biased is still able to conduct current from emitter to collector.

    In this situation, the Base/Emitter PN junction is biased in such a way that the depletion region is very small or neglible. However, the PN junction of the Base/Collector region has its depletion region *widened* by the reverse biasing; which makes you think at first glance that no current should be allowed to conduct. Here is my take on the device theory of why we still see (large) current from emitter to collector for an NPN BJT. Please correct me if my theory is wrong or has some misconceptions:

    Since the base is very thin in relation to the collector and emitter, and it is lightly doped to a P-type material, there are not many holes available to be recombined with emitter electrons. The emitter on the other hand is a heavily doped N+ material with many,many electrons in the conduction band. When the mass amounts of electrons in the emitter get pushed into the base by the emf of the base->emitter battery, the vast majority of them have no holes to combine with. Now we have a situation where we have a high entropy of electrons in the base which begin to defuse into the base/collector depletion region; effectively doping and shrinking the base/collector depletion region into N-type material. When enough electrons diffuse, the depletion region all but disappears into n-type material, allowing the positively biased n-type collector region to sweep those difused electrons out of the collector terminal. Since the collector is moderatley doped to N-type, the difused electrons are able to pass right out the postive-polarity collector terminal.

    So, am I right about entropy playing a role in this by diffusing the base/collector depletion region with electrons to effectively dope it into n-type material?

    Thanks!
     
    Last edited: Mar 10, 2009
  2. beenthere

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  3. Wendy

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    One thing, I can't speak to your theory, but it sounds like you're thinking of a transistor as back to back diodes, which is false. The Base Collector junction is not "back biased".

    Here is an example where this logic breaks down. When a transistor is fully saturated, it drops between 0V to 0.2V across the Collector Emitter (depending on current and the transistor), even though the Base Emitter is dropping 0.6-0.7V.
     
  4. SiegeX1

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    Mar 10, 2009
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    Right, but Saturation is not the mode I am concerned about, as you said, the Base -> Collector PN junction is forward biased in Saturation.

    I'm strictly talking about the 'linear' or 'active' mode of the transistor where you still get current from emitter -> collector even though the base->collector PN junction is reversed biased, widening the depletion region. My theory is to try to explain the *why* behind a transistor not acting as a back-to-back diodes in this mode.
     
  5. Wendy

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    You can not separate the two, saturation is all the way on, while cutoff is off. Linear is between the two states. One model of a transistor is the base current causes a variable resistance, similar to a pot, between collector and emitter controlled by base current. The pot can swing end to end.

    Transistors are pretty complicated, though there are simiplifications that allow people to design with them without having a full understanding. There are lots of models out there that all work one way or another with varying degrees of success. I prefer the current models myself, although there are voltage models out there too.

    A lot of designs are meant to take the beta gain out of the equations, as long as the minimums are met. One of my favorite concepts is the transistor as a constant current source (common collector). By varying the voltage on the base you vary the current out the collector, which is one of the ways you can make a linear amplifier.
     
  6. SiegeX1

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    Mar 10, 2009
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    Well, actually I would say you *can* separate them. Although the active region is between cutoff and saturation, you can still distinguish the modes based upon the state of the two back-to-back diodes (or PN junctions). The fundamental difference between Active mode and Saturation is that in Saturation, you have now brought the collector voltage low enough to forward bias the base->collector diode basically creating a short from collector to emitter. Since the transistor can't do any better than a short between its two terminals, an increase in base current has little to no effect on the collector current.

    After reading the BJT chapter in the e-book (which I somehow glanced over, so thanks for the link!) it appears I was about half right. Entropy does play a role to diffuse the mass amounts of free electrons from the base into the base/collector depletion region, but I was wrong about it doping that material into N-type. Apparently, the electrons are never in the depletion region for very long as the electric potential of that region quickly sweeps them into the collector where they feel right at home in N-type material.

    I appreciate the discussion and the tips, great community here.

    Thanks
     
  7. Wendy

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    Mar 24, 2008
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    You do realize it goes both ways? NPN and PNP?
     
  8. SiegeX1

    Thread Starter Member

    Mar 10, 2009
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    Sure do, if you look at my original post I clearly stated my theory was in reference to an NPN so I don't have to make any generalities. Although the same process should apply, just with reversed polarities and reversed current.

     
  9. thatoneguy

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    I believe you have a bit of cross linking between BJT and MOSFET. From my understanding of your original post, it sounds kinda sorta similar to a Bi-FET.
     
  10. SiegeX1

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    Mar 10, 2009
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    Now that you mention it, the 'effective doping of the depletion region' part of my original post does sound alot like the inversion layer in a MOSFET that creates the thin channel to connect source to drain. I've never heard of a Bi-FET before, I'll have to look into that, thanks!
     
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