BJT NPN: Apparently Ohm's Law Does Not Apply - Page 1
Forum section: electronics / general

author: mbennett555
author_rank: Contributor
timestamp: August 22, 2021, 07:18:38 pm
content: Hello to everyone. FYI, this is my first post to this forum.I believe I know a bit about electronics, and specifically bipolar junction transistors. But I'm confused about something. It's something that's confused me since I was at university years ago (I have a BS degree in Electrical Engineering). It has to do with the voltages and currents observed when a BJT NPN is in saturation, with its emitter connected to ground.There are 2 images in this post, and they're described below.• Image 1 shows the circuit diagram.• Image 2 shows: the measurements I made, in a table (see the bottom-most row); and a zoomed-in view of a portion the circuit, namely the NPN transistor.I do not understand why, as shown in Image 2, VCB= -572 mV. According to Ohm's Law, that means there must be a current that flows from B to C, over some resistance. But we can see from the measurements that there is a current IC= IE= 5 mA that flows from C to E. How can there be a current that flows from B to C? (Also as shown on Image 2, there is a current IB= 50 uA that flows into the base. I believe this current can be neglected, since its value is 1/100 of IC.)Circuit_BJT NPN_1-NPN_202108_B1.pdf(68.09 kB - downloaded 122 times.)Circuit_BJT NPN_1-NPN_Measurements_202108_B1.pdf(1909.23 kB - downloaded 133 times.)Circuit_BJT NPN_1-NPN_P01= CircuitDiag_P02= Measurements_202108_B1.pdf(1973.39 kB - downloaded 105 times.)

author: bob91343
author_rank: Super Contributor
timestamp: August 22, 2021, 07:33:18 pm
content: I didn't download the files but when you draw reverse current through the BE junction there will be some laser operation and thus the collector voltage goes negative. Or something like that. Bob Pease explained it.

author: amyk
author_rank: Super Contributor
timestamp: August 22, 2021, 07:51:59 pm
content: The one-sentence answer is "because a BJT is not ohmic".

author: magic
author_rank: Super Contributor
timestamp: August 22, 2021, 08:04:19 pm
content: Quote from: mbennett555 on August 22, 2021, 07:18:38 pmPrecisely according to Ohm's law,ifthere is a straight metal connection between B and C then it has a resistance and a voltage across those point causes a proportional current to flow.In practice that's a PN diode so Ohm's law doesn't quite apply. But you are right, the BC junction becomes forward biased in saturation and steals base current away from the BE junction so a transistor can't saturate down to zero or negative Vce.As for how current flows from 50mV collector to the emitter despite the 600mV base in between, that's because BE junction voltage attracts charge carriers (here electrons) from the emitter to the base layer which is thin enough that 99% of electrons ram into the collector before they get a chance to exit through the base terminal. A BJT with the sort of proportions that you have drawn wouldn't work very well (like, maybe β=1 if you are lucky).

author: antenna
author_rank: Frequent Contributor
timestamp: August 22, 2021, 08:11:29 pm
content: I don't think there is a current flowing from the collector to the base lead, aside from the minor current your volt meter is using to read the potential difference across the junction. If you put your voltmeter across a battery that has nothing else connected to it, you still can read the voltage, because that voltage is now flowing through the 1MΩ resistor inside the voltmeter (ohms law satisfied). When the transistor is saturated, the B-E junction becomes forward biased and becomes conductive, and in the process, the holes in the p-type material are filled with electrons. Now, the p-type material at the base has an excess amount of electrons filling the holes in the B-C depletion region and the voltage across C-E can pull them through. There is a current between B and C, but the electrons are not coming from the base lead, they are coming from the emitter and going straight through the forward-biased junction to the P material at the base connection and filling the depletion region holes. I've read somewhere, that the reason you cannot make a transistor with discrete diodes is because the electrons that are being attracted to the p-type material by the voltage at the base lead will not have the momentum to make it down the wire into the next diode junction to fill that junctions holes, which is why they need to be sandwiched together with a very thin p layer.This video has a great explanation of the inner workings of a transistor if you are interested in how the semiconductor material behaves.https://youtu.be/J4oO7PT_nzQ

author: Kleinstein
author_rank: Super Contributor
timestamp: August 22, 2021, 08:45:00 pm
content: Quite often the CE saturation can be reduced if the transitor is used with the C and E terminals swapped. This only works for low voltage (e.g. < 5 V) with most types.It is One can not really say there is some current flowing from base to collector, as the electron are all the same and there is stil a net current flow from the collector. It is still a picture to imagine why the the gain drops quite a bit.Ohms law is is not V = R * I !The point is that for a metal I/V is constant if the temperature stays constant. So one has a linear voltage / current relation, that allows to define a resistance. Ohms law itself needs no resistance.

author: mbennett555
author_rank: Contributor
timestamp: August 22, 2021, 10:06:23 pm
content: To "antenna": Thanks very much for the excellent explanation, and the link to the video.

author: mbennett555
author_rank: Contributor
timestamp: August 22, 2021, 10:09:11 pm
content: To Bob91343: In your reply earlier today, you said: "... when you draw reverse current through the BE junction there will be some laser operation and thus the collector voltage goes negative. Or something like that. Bob Pease explained it." What do you mean by "laser operation"? Is that a joke?

author: bob91343
author_rank: Super Contributor
timestamp: August 22, 2021, 11:10:37 pm
content: Not a joke. When you put reverse current through the B-E junction, there is an emission that causes the collector to go negative. I am not fully clear on this but I remember trying it and it works. As I said, Bob Pease described the situation, so maybe you can google it.https://hackaday.com/2021/06/10/an-explanation-of-a-classic-semiconductor-riddle/

author: David Hess
author_rank: Super Contributor
timestamp: August 23, 2021, 03:10:56 am
content: He means the breakdown of the base-emitter junction results in emission of infrared photons which hit the base-collector junction forward biasing it.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 23, 2021, 05:48:55 am
content: 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?!"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