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Bjt: Base transit time

  1. Aug 29, 2009 #1
    Hi to you all,
    I'm reading about semiconductor devices on Muller-Kamins' "Device Electronics for Integrated Circuits" and I have some problem understanding the base transit time of a bipolar transistor.
    I read that when the npn bjt is forward biased the emitter injects electron in the base. In the case of high injection, the majority population of the base (holes) increase at the emitter-base junction "because of charge neutrality", Muller says.

    -So, this is my first question: why it is necessary the hypotesis of charge neutrality? It's maybe for th hypotesis of completely depletion of the space charge region, so that the voltage is 'seen' only in the depleted region(s, if we count the base-collector) and in the other parts of the system the voltage (and so the field) is veeeery small?

    -Second question: from where this addictional charge comes? From the base lead?

    -Third question: why they arrive?? That's because the injected electrons create a field that attracts the holes so that they're arranged in that triangular shape? In this case the total charge is 0, but, Muller says, the electrons see the opposite field caused by holes that helps them and so they can move faster than only with the diffusion-method. That's it? Otherwise, can anyone explain me what happened to the fields and to the carrier concentration during the minority injection?

    Thank you very much [and sorry for my english].
  2. jcsd
  3. Aug 31, 2009 #2
    1) Charge neutrality is a must or the bjt would cease to function. If a large concentration of electrons were to build up in the base region, electron emission from the emitter would be greatly reduced to the point of the device being non-functional. When an electron transits into the base, and then recombines with a hole in the base, another hole replaces it. Otherwise, a build-up of electrons would result in a field that prevents electrons emitted from the emitter from transiting through the base and onward to the collector.

    2) Yes, the additional charge does indeed come from the base lead.

    3) I don't understand your question. "Why they arrive" is vague. What is "they", and where are they "arriving". If you be more specific I can help. BR.

  4. Aug 31, 2009 #3
    Thanks for your answers..!
    3) The subject is the additional holes: why the additional holes arrive in the base? I mean, I can't see who-cause-what.. When the injected electrons enter in the base, they change somehow the original field (made by the small voltage drop in the base region) in the base? And this field modification attracts the additional holes from the lead?

  5. Sep 1, 2009 #4
    For npn devices, the base region, being doped with p-type material, has an abundance of free holes, h+. With an external bias, the holes will drift from base to emitter. Then the holes will recombine once in the emitter, with free electrons, e-. The emitter being n-type material, has an abundance of electrons. The same field which moves e- from emitter to base, also moves h+ from base to emitter. Once the electrons enter the base region, they encounter a strong force due to the electric field associated w/ the reverse biased b-c (base-collector) junction. Most of the e- continue through the base and arrive at the collector. These e- were majority carriers in their original native emitter region, being n-type, and have high mobility. As they pass through the base region, they are minority carriers and have low mobility, as the base is p-type. Upon arrival in the collector they are majority carriers again w/ high mobility as the collector is n-type.

    En route to the collector, while in the base region, a small fraction of e- do not make it. They recombine with holes in the base. The base is p-type, w/ lots of free holes available.

    So charge neutrality is maintained as follows. Let's say for purpose of computation, the collector current is 160.2 uA (microamp), and hFE (beta) is 100. The emitter current is then 161.802 uA, and the base current is 1.602 uA. This literally means that in 1 femto second (1 fsec = 1e-15 second), that 1010 electrons enter the emitter lead.

    Of the 1010 electrons that enter the emitter lead, 10 of them do not make it to the collector. Of the 10, around 9 of the e- entering the emitter lead recombine in at the edge of the emitter region near the base, with holes, h+, injected into the emitter from the base. These 9 e- entering the emitter are neutralized by 9 h+ entering the base and injected from base to emitter. Of course, outside the silicon, it is usually stated that holes do not enter the base lead, but that electrons exit the base lead.

    The remaining 1 of the 10 e- transits through the emitter, into the base, where it recombines with a hole in the base region. An extra hole enters the base through the base lead to neutralize this charge.

    The remaining 1000 e- survive the trip all the way to the collector. Thus 1010 e- enter the emitter lead, 10 e- exit the base lead, and 1000 e- exit the collector lead. Charge balance exists.

    For a pnp, the polarites are reversed. Does this help?

  6. Sep 1, 2009 #5
    Thanks for your numeric example, that was very clear!
    But [ :) ] my problem is a little bit different.. I'll try to explain my 'vision', that maybe isn't correct.
    I see the current in the bjt composed by many contributes:
    -the 'ideal' contribute is the linking current, the current of electrons from emitter to collector;
    -the 'second-order things': base current due to recombination in the base region and due to the holes injected in emitter [and then there is also recombination in the space-charge region, but let's skip this].
    Now, for the existence of the linking current, I see (in my vision) that there has to be a sort of zone [like in the MOS] of electrons that permits the flow of current: in the bjt, this zone is the triangle in the base region [like here: http://bmf.ece.queensu.ca/mediawiki...t_carrier_conc.png/350px-Bjt_carrier_conc.png ]. This region, in steady state, is constant at every moment, and "upon it" the minority carrier can flow. [like a bridge......]
    Besides, there are the contributes of the 'second-order things', that aren't represented by this density. With them you illustrated to me the charge neutrality during the current flow.

    My problem is in the shape of the majority carrier in base [http://bmf.ece.queensu.ca/mediawiki...carrier_conc.png/350px-Bjt_carrier_conc.png]: as you said before, there has to be the charge neutrality in base, otherwise the bjt won't work. But... how these holes become like the picture? In my vision I think of three components of holes: two of them feel the field applied by the voltage and are the recombination current and the injectec current of electrons. And then, the third part, moved again by the field, arrive to the base-emitter junction and there she has to stop because she meet the opposite field in the space-charge region of the emitter-base junction. So this part of holes reorganizes in the triangular shape to create a new field that eliminates the field applied.... but I think is really confusing and confused.... Moreover, in the case of high injection from the emitter, the field created by the holes is 'seen' by the electrons injected and so they are accelerated [drift-aided bjt]. And also here, I have some difficulty thinking of a field that was created by holes to contest the field applied that is felt by the electrons..

    I don't know if I made my problem(s) clear [problems about bjt; about my oratory and my english I think they are clear since my first pos..]. Thanks again!

    Last edited by a moderator: Apr 24, 2017
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