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Potientials in npn transistor

  1. Nov 19, 2009 #1
    this query was previously in electrical engg; later i found that it actually have to be solved as a solid state physics one.

    (as in books) In an n-p-n transistor, emitter grounded, electrons enter the emitter from ground , where 90-98% come out of collector, and rest go out of the base.

    In saturation, Vbe=0.7-0.8 V. where as Vce=0.2-0.25 V.
    now consider electrons in p material; they have two routes; one out of the base & other out of collector.

    why does most of electrons go out of collector(0.2 V) inspite of having a greater (0.75 V) attracting voltage at base?

    please answer qualitatively.
     
  2. jcsd
  3. Nov 19, 2009 #2
    The b-e voltage is at roughly 0.7V, but the 0.2V you refer to is the c-e voltage. The b-c voltage is the difference, or 0.5V. The base is at higher potential than collector and emitter. The current in the base is the holes injected from base to emitter, plus the electrons injected from emitter to base which do not reach the collector, plus the holes injected into the collector, as well as electrons fron collector to base..

    Also, the base is very thin, while the collector is thick. When the electrons are emitted from the emitter, the enter the base but being so thin, they continue onward to the collector. The doping density of the emitter is much greater than that of the base so that Ie >> Ib.

    Did this help?

    Claude
     
  4. Nov 19, 2009 #3
    i already interpreted it before(base was thin), but not sure

    if in that case we can procede as followed:

    the area near base-edge will get accumilated by electrons due to high potientials.
    then as base is thin, and only some can pass, electrons get crowded, due to which potiential increases nearby.
    now the upcoming electrons will face these crowded electrons, having a repulsive force,
    moves towards collector.
    those repulsion causing electrons also get scattered , where at the same time some of the new upcoming electrons participate in repulsion action(where some get managed to go through).

    how did u find it?
     
  5. Nov 19, 2009 #4
    anyone has any modifications?
     
  6. Nov 20, 2009 #5
    No, I think you got it a little wrong.

    When electrons are emitted from the emitter and reach the base, they feel an enormous electric field that sweeps them out of base into the collector, because of the way the Collector-Base junction is biased.

    Think of it this way: Emitter-Base is like a separate diode which is forward biased so that a lot of carriers (say holes in a p-n-p structure) are emitted to the base.

    The Base-Collector junction is like a separate reverse biased diode, which normally wouldn't conduct because of the lack of carriers even though there is an enormous electric field.

    So if you connect these two, the junctions are coupled but relatively speaking, the base is MORE REPULSIVE for a carrier (hole or electron) THAN the collector. The way these junctions are coupled allows you to control the amount of EMITTER current that is injected BY MEANS of the base terminal.

    So, to get the qualitative picture right, see a band diagram and imagine the back-to-back diode picture.

    Does this help?
     
  7. Nov 20, 2009 #6
    when an electron passes across the junction of e-b i.e; n to p in npn
    it need energy.
    i'm very sure of this; it is definite.

    it cannot have even a bit of force by junction when it comes out of the junction width.

    coming out of the junction width means just entering the base.

    what repulsive force IN THE BASE are you mentioning about?
     
  8. Nov 20, 2009 #7
    Wrong. You need to learn how to read spatial band diagrams of solid state devices. Otherwise you won't go anywhere.
    Here is one:http://upload.wikimedia.org/wikiped...ds.svg/740px-Bjt_forward_active_bands.svg.png

    When an electron passes a junction it already HAS energy. It comes from the contact leads with a definite energy. It doesn't absorb ANY energy within the device (except due to scattering)

    What are you talking about? What about the depletion region which extends WAY over the base region into the collector region. This is the part where the electron feels this enormous force due to the reverse biased junction.


    Base-Collector region is REVERSE BIASED (do you know what that means?) therefore, there is a HUGE FORCE (typical: e*10^7 V/cm) to sweep the carriers to the collector while they are emitted.

    Electrons are FLYING PAST the base, because they are ALREADY kinetic, i.e, they already have MOMENTUM when they come into the base junction. Some of them are being recombined with the opposite carriers in the base ( which is NOT a good thing) that's why we make the base junction as small as possible. Once they fly THROUGH the base (BECAUSE OF THE FORWARD BIASED E-B JUNCTION) they are SWEPT into the collector (BECAUSE OF THE REVERSE BIASED B-C JUNCTION)

    The control of the collector current due to the base current is because of the COUPLING of these two back to back diodes.

    If you still don't understand, just consult to one of the infinitely many elementary solid state books on the issue.
     
  9. Nov 20, 2009 #8
    Very good. I was going to respond, but you've covered it well.

    Claude
     
  10. Nov 20, 2009 #9
    See thumbnail. Base is green, emitter is red, collector is black. Collector is 180 degrees out of phase with respect to base.
    Bob S
     

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