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| BJT NPN: Apparently Ohm's Law Does Not Apply |
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| T3sl4co1l:
Ahh; that took a moment to figure out, but I see it is actually very clear. You didn't follow schematic conventions, but the stuff in the corner provides a definition. "Math, in my schematics?!" :P FYI, it's perfectly okay to label components with say SI units; and it can help to label them with designators as well, so they're easy to refer to in discussion. Instead of "the base resistor" (meaning what, the fixed or variable one?) one can just say R1 or R2 or whatever. Another FYI, you can't embed PDFs as images; also the inline embed function is disabled right now (for... reasons), but you can embed an image-image (JPG or PNG would be fine for these) by using [ img ] tags around the attachment link (submit post, then copy link, then edit). As for the problem at hand -- it seems you stopped short of asking the question? -- The question I was expecting, was whether Vce(sat) is ohmic. FWIW, it generally is, though there's an offset in the forward direction (it's no lower than about 20mV for most parts), while in reverse ("inverted", swap C and E), it can be very low indeed but hFE sucks instead (typically 5-10, though some parts have over 100). Vcb, is a red herring -- but why, requires some explanation. And note that, the B-C junction is still a diode as usual, so if anything, as collector voltage falls (to the near-zero range), current will flow from base to collector. That current may ultimately flow out the emitter, but the key is that internal Ib seems to fall; this is sensible externally as Vbe(sat) being lower than Vbe(on) for the same Ib. This can be harnessed more directly, by wiring a schottky diode in parallel (B to C). The schottky has a lower Vf than the PN junction, so when Vce < 0.3V or so, base current is shunted. This is an application of negative feedback, using the transistor's high gain to solve the equilibrium of Ic (whatever load current happens to be in this scenario) at Vbc = Vf. This arrangement has the effect of raising Vce(sat), without incurring stored charge; the transistor can turn off much faster. (Stored charge is essentially diode reverse recovery. In saturation, the base has excess charge carriers, which must be removed, by waiting for them to recombine (~10µs typically for this part), or by applying a negative base current to shunt them out. Only then will Vbe and Ic fall.) As for the reason why Vce can be so low, it's physics. You would think (and, I also thought for a long time, a long long time ago :) ) that, because the collector chunk is stacked on top of the base chunk, well, how could it ever be anything else, it must be flowing current into the base for Vce < Vbe, right? How could Vce(sat) ever be below ~0.6V? In what way is the emitter common, obviously the base is in the middle, only Vbc and Vbe should matter, right? The explanation is found in semiconductor physics. The PN junction has a built-in potential, meaning any electrons/holes flowing across it, automatically gain or lose the corresponding amount of energy (i.e., a Vbe, so about as much energy or ~0.6eV). Electrons emitted by the emitter, diffuse into and across the base (and holes vice-versa). They're drawn by applied current, against the N-P junction potential -- hence Vbe is ~0.6V. Because the base is very thin (those block diagrams are NOT drawn to scale, by a huge margin!), a lot of them spill over into the B-C depletion region as well. There, they get sucked up by the built-in potential (and any additional collector voltage, if applicable), and are able to leave the collector terminal. Because the E-B and B-C junctions are opposite, they largely cancel out, and Vce(sat) can be very low. It's not quite zero, because there's a smaller component to the potential, due to doping concentration. A doping gradient, for example, has a built-in potential (on the order of some 10s mV). Typically the emitter is most heavily doped, and collector lightest, and this works against the electron flow. Thus, Vce(sat,min) is typically some 10s of mV. Transistors can be constructed with more symmetrical junctions, or with creative doping gradients, to help reduce this. Among other improvements, we have "low Vce(sat)" transistors, which really do quite well as the title suggests. Surprisingly, they also have quite good inverted hFE; even though they seem to have quite ordinary doping levels (namely, E-B breakdown is the typical ~7V). Notice the effect doesn't quite work in reverse, of course. We might have a transistor with, whatever it is say 30mV of built-in potential from C to E. If we simply swap C and E, we still have an NPN transistor, we can still apply base current and get "collector" current flow, right? As mentioned, inverted hFE is typically pretty low, but it also happens that Vce(sat) is merely very low -- it can be single mV -- but not negative. There are other effects of course, which prevent us from having a free lunch at the expense of that built-in potential. Physics, amirite?... (I forget what the detailed explanation of that effect is; something about metallic contacts to those different doped regions, maybe? Whatever it is, it happens that the BIPs add up in one direction, but don't perfectly add up the other, so of course you don't have a free energy source.) About saturation. Note that, at low Vce, the B-C junction isn't very depleted anymore. In linear operation (Vce >> Vce(sat)), it is quite attractive (wide depletion region, strong electric field pulling charges across). In saturation, it's kind of just whatever, mainly the built-in potential and almost no depletion region at all. The consequence is, at low Vce, less current reaches the collector, even when Ib is large (or, indeed, more Ib might flow in response, if we've applied a constant Vbe). Another way to put this is, simply, hFE falls at low Vce -- which makes perfect sense, you can't draw any more current than is available at Vce ~ 0, so any extra input you give, has to be wasted. It can't pull negative, that would generate free energy. (Well, it could be powered by the base, but that can't be any more improvement than Vbe / hFE -- and there's no mechanism for that to happen, anyway.) About current flow. BJTs exhibit electron and hole flow, stimulated by current flow through PN junctions -- called minority carrier flow (minority meaning for example, electrons are the predominant charge carrier in the P-type (hole-doped) base). Both types of charge carriers are ultimately absorbed (recombined) at the contacts. Recombination can simply happen over enough distance (10s of µm), where charge carriers are replaced with the local (doped) variety; it can happen over short distances in heavily doped regions ("n+" or "p+" on schematics), or instantaneously on contact with metals (which are effectively "100% doped" with electrons). Metal contacts are Schottky junctions, and come in two flavors: ohmic and rectifying. Obviously, ohmic is required for most purposes (connecting to transistors and such). These usually involve strong (n+/p+) doping, then aluminum metal. Schottky diodes intentionally (ab)use the opposite effect, typically using platinum metal (or related metals or compounds) over lightly N-doped silicon. The metal being extraordinarily abundant in electrons (majority current carriers), conduction can be essentially instantaneous, no recovery time as suffered by PN junction diodes. Thus they are used for high frequency switching and radio applications. Tim |
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