A simple transistor question - Page 1
Forum section: electronics / beginners

author: Sredni
author_rank: Frequent Contributor
timestamp: May 25, 2022, 09:17:50 pm
content: The purpose of this post is twofold.Bipolar transistors are arguably the hardest transistor type a beginner has to study; with FETs and MOSFETs the carrier transport is unipolar and the 'pinched hose' model provides a reasonably intuitive simple model to explain how they can provide amplification. But BJTs are a little more complicated due to the interplay between drift and diffusion currents and the role played by both carrier types.Is it possible to use the doubts a student is exposing in a seemingly naive question to teach them how a bipolar junction transistor actually works as an amplifier without delving too much (or at all, if possible) into semiconductor theory?Here is the question:QuoteMy question to the users of this forum are two:1) First, can you understand what the student's doubt(s) is/are?2) Can you answer this question in a simple but meaningful manner so that the student will learn how a transistor in that configuration can be used to amplify power and/or adapt impedances?I already have an answer in mind, but I will add it later in order not to put boundaries on your creativity and imagination.Let's say this is a small didactic experiment.

author: Sredni
author_rank: Frequent Contributor
timestamp: May 25, 2022, 09:19:25 pm
content: Analysis of the questionFirst, there is an error in one of the formulas (maybe introduced with the formatting of the original question), but I see that everybody here has automatically intepreted what the student meant, nevertheless (I wrote this part before Kleinstein spotted it). He wrote "V2 = VE - VB" instead of VC - VB and there is also a question of signs (VC - VB or VB - VC?) that might be discussed but from the actual question asked it seems clear that the student is talking about absolute values.I have to disagree with the conclusion reached by rfeecs that we cannot say V2 > V1, because the drawing, as ugly as it is, shows that the battery that is reverse biasing the collector junction is bigger than the one forward biasing the emitter junction. This is the classical circuit configuration used in several textbooks to explain how transistor works and it is implied that the transistor is in active mode. If VBE is at least 0.6V, then |VBC| is at least 0.6V. Now, this might seem to void the question of any meaning, since if we already know (from the batteries sizes) that the output voltage is in absolute value greater than the input voltage the question seems already answered. But it's not. What the student is grappling with is how is it possible that in the input mesh we have a current IE that gives rise to a voltage V1 and in the output mesh we have a smaller current IC that (never mind the sign for the moment) gives rise to a voltage V2 that is bigger than V1.And this brings us to the interesting point of the 'resistances' Rpn and Rnp. A diode is indeed a resistor, albeit a nonlinear one as correctly noted by TeslaCoil (one reference to rule them all: Chua, Desoer, Kuh, "Linear and Nonlinear Circuits"). We are used to deal with the dynamic - differential - resistance of the device because our way to analyse nonlinear circuits is by linearizing them in the neighbour of their operating points, but this is not the case: here we are considering a large signal behaviour and the resistance (the value of the ratio V/I) happens to be a nonlinear function of the voltage (or the current).The coexistence of these two similar but completely different concepts of resistance is nothing new: we do this for hFE and hfe when the relation between IB and IC is not linear, and in physics for phase and group velocity when the dispersion relation is not linear.https://i.postimg.cc/9MBpqFn2/Nonlinear-relations.jpgfig. nonlinear relationships and global and differential quantitiesSo, the resistances the student is referring to are the values of the ratio V/I for a given voltage (or current). From the point of view of the input mesh we see a small voltage V1 associated with a current IE, while in the output mesh we see another voltage V2 associated with a current IC. And the student is puzzled: if IE>IC how can we be sure that V2 > V1?In my view, there are several opportunities here: explain why IE > IC is actually IE ≈ IC, and then what is the mechanism that allows basically the same current to be associated with different voltages. I would even go further and add a source resistor to 'program' IE and a load resistor to 'pour the programmed current into' and show that I can get power amplification (in this configuration, it's voltage amplification).Many interesting answers so far, and I am grateful to all participants. Some of the answers are quite similar to the one I had already written. free_electron makes a valid point about the doping asymmetry of the BJT but I feel that is a bit premature to discuss at this student's level. That could be the subject of a follow-up question by the student.I will post my answer shortly (it contains parts that have already been put forward), after producing a couple of pictures. And then I will explain - in a separate post so that it can be moved if necessary - the second purpose of my question here (even if I have just been threatened in PM to "mind my own business" if I am "not a teacher". LOLSo, here's my idea of answer. I modified my initial stance adding some more detail of the inner working, but I am not sure it is a good idea for a student at this level. Excessive detail and still to many simplifications (like my completely ignoring the electron current going from base to emitter) might lead to confusion.My take on the questionFirst: in this configuration and with the base junction directly biased and the collector junction reverse-biased, the transistor is in active mode. This means that the collector current is alpha times the emitter current, with alpha slightly less than one (typical values can be 0.95 - 0.99 and higher) and that means that even if the collector current is technically smaller than the emitter current, we can consider IC ≈ IE.From this, the reasoning of comparing Rpn IE with Rnp IC, when IE ≈ IC and Rnp >> Rpn resolves in favour of Rnp IC > Rpn IE. (It can be shown how we can add external resistors to the input and output mesh to exploit the fact that the same current can go through different resistances to provide amplification.)Here is the simulation of a fairly aggressive biasing using the simple default model of BJT in LTSpice. As you can see, even if they are mathematically different due to the relation IE = IB + IC, the input current IE and the output current IC are nearly the same - the differ by roughly 1%. For this example we can infer a value of alpha = IC/IE = 59.7/60.3 = 0.99.PNP BJT with ideal generatorsThe above explanation, though, is little more than a math trick because it does not actually explain why we can have basically the same current in both the input and output meshes when the resistances encountered can be so different.The 'coupling-decoupling' mechanismSo, why is IC nearly the same as IE? And why, whatbasicallyis the same current can have two different voltages associated with it?It has to do with the dual mechanism of transport (driftanddiffusion) inside a semiconducting material. Inside the copper conductors, the current is exclusively a drift current, where electrons are pushed along by the tiny (typically a few microvolt/m) electric field in copper. In the middle of the transistor, that is in the base, the current becomes a diffusion current that depends on the gradient in the concentration of carriers and not on the electric field.If we focus our attention of the main contributors to the currents in a PNP transistor which are holes (see footnote), theoversimplifiednitty-gritty details are as follows:Inside the transistor, the current in the emitter islargelya hole drift current, acted upon by the tiny electric field in the bulk of the emitter region; the large number of holes that overcome the weakened voltage of the first depletion region cross the EB junction and, once in the thin base, they become a diffusion current that depends on the concentration of carriers. In the base, these holes are minority carriers that diffuse independently of the electric field and almost all of them - due to the geometry and concentration profiles - reach the base-collector depletion region, ending up in the collector bulk as if they were a reverse saturation current for the reverse-biased BC junction. We can call 'alpha' the fraction of carriers that from the emitter cross the thin base without recombining and without 'exiting' the base terminal.In the bulk of the collector, the current becomes mostly a drift current again, this time subject toanother electric field- the one associated with the small (we're talking mV, here) collector bulk voltage. The now majority carriers end up being collected by the negative collector terminal thanks to the electrons that are pushed there by the external circuit.The large collector current is essentially an 'artificially inflated' saturation current for the reverse biased BC diode (in the VI characteristics, it is the nearly flat horizontal part of the exponential curve that sits right under the voltage axis) and, by being essentially independent on the base-collector voltage, it translate to the nearly flat horizontal output characteristics of the transistor.As a matter of fact, a BJT transistor can be seen as a reverse biased diode (the BC junction) whose reverse saturation current is dictated by the carriers injected by the emitter.Here are the input and output characteristics of the common base transistor, along with the vertical characteristics of the generators, seen as one-ports attached to the BJT terminals:Input and output chars with ideal voltage generators driveHow is this 'coupling-decoupling' modeled?The circuit under the characteristics is a large signal model for the PNP transistor drawn as a two-port: the input port sees a forward biased diode (the PN emitter junction), while the output port - with its horizontal characteristics at constant current (representing the reverse biased NP collector junction) - is modeled by a current controlled current generator.To reiterate:the fundamental mechanism that makes it possible to have a 'transistor action' is the existence of a 'bridge phenomenon' (diffusion) that is not ordinary conduction and that allows us to disjoint the effects of conduction current in the input mesh from the effects of (roughly the same) conduction current in the output mesh.The need for diffusion as the main transport phenomena in the middle of the transistor is also the reason why we cannot build a transistor by putting two discrete diodes back to back, but that's another story...Similar 'bridge phenomena' can be identified in the workings of FETs and MOSFETs. In all cases it is this 'decoupling' that allows us to model a transistor (in this example for large signals, but a similar reasoning applies to the linearized small signal circuits) with controlled generators - usually in a T topology.Large and small signal equivalent circuits of the transistorWe can turn this three terminal device into a two-port by putting one terminal in common, the base in this particular example.And it is again the 'decoupling' between the input and output currents that allows us to either have amplification (in a CB configuration it is voltage amplification) or match impedances (the current that is paired with a large voltage in the output mesh is seen as a current paired with a small voltage in the input mesh).Practical Application of transistor actionWhen the transistor is in active mode, the input mesh can sustain a maximum voltage, across the transistor BE terminals, of less than 1V (typically, it's 0.7V) and the input current is strictly linked to that voltage. The output mesh, on the other hand, can sustain the much higher voltages of a reverse biased junction, with a current that is essentially independent of that voltage.. Therefore, with an adequate supply, the output port can reach higher voltages than those provided in the input circuit. Since Ie = Ic + Ib, the current in the load will be roughly the same as that flowing in the input circuit. This simultaneous link between input and output current (they are basically the same) and decoupling of the voltages associated with such currents in the input and output meshes, is what makes amplification and impedance matching possible.As a practical application of transistor action we can use an emitter resistor RE to 'program' the current in the input mesh, and then letting the transistor transfer it to the output where it can flow in another resistor, the load resistor RL.How the voltage in the output mesh redistributes itself between CB junction and load resistor depends on how much voltage the current set by the input will develop across the load.Two extreme examples are the following: in the first case we program a current of 3A in the input circuit by choosing RE = 0.78 ohm. The nearly 3A flowing in the load will result in almost all 24V of Vcc to drop across RL, and nearly nothing across the CB junction.circuit 1: A load with big current and big voltageActual values for this circuit areVEB = 0.98 VVCB = - 0.45 V IE = 2.973 A Vload = 23.55 VIn this other circuit, where we program 1mA in the input circuit, only 8 mV will develop across the load resistor, and basically all of VCC will show up across the reverse biased CB junction.circuit2: A load with small current and small voltageActual values for this circuit areVEB = 0.773 VVCB = - 23.992 V IE = 0.972 mA Vload = 0.008 V @ 0.962 mAThe independence of the output current from the collector-base voltage can be easily shown by changing the value of the load resistor, without changing the 'programming' in the input mesh. For example, if in the above circuit we set RL = 10 kohm, we get nearly 10V across the load, 14V across the reverse biased collector junction,VEB = 0.773 VVCB = - 14.378 V IE = 0.972 mA Vload = 9.62 V @ 0.962 mAand (with the idealized BJT model of LTSpice) the current did not even flinch.Footnote:What I am neglecting completely in the typical already simplified analysis of BJT currents that appear on introductory books, is the electron current that from the base goes into the emitterThis current can be made small by strongly doping the emitter and lightly doping the base. It's not exactly negligible but its inclusion would have complicated the exposition. Moreover, the "holes 'exiting' the base" are actually electrons entering the base and recombining with that 1% of the unlucky holes that do not make to the collector. In the charge controlled model of the transistor in a common emitter configuration, they are the reason a bigger current flows from emitter to collector....I said that the purpose of my question was twofold, so I am going to state the second purpose in a separate post that happens to be post # 23.

author: TimFox
author_rank: Super Contributor
timestamp: May 25, 2022, 09:28:01 pm
content: Perhaps the common-base configuration (sketched in the question) is the easiest to explain in the hand-wavy department.You inject current into the emitter from an outside source.Giving the charge carriers free will, once in the base region they see a higher field attracting them to the collector region, and most decide to flow there instead of into the base contact.The ratioIC/IE= "alpha" is slightly less than 1. The usual current gain "beta" =IC/IB= /(1 - ) and is reasonably large.Transistors are actually voltage activated, but when forward-biasing the base-emitter junction by an appropriate voltage (like 0.6 to 0.7 V for silicon), the ratio of collector-lead current to base-lead current is high and roughly constant.Although transistors are often treated as current amplifiers, the physical reality is that they are voltage-operated devices with substantial input (base) current.

author: rfeecs
author_rank: Frequent Contributor
timestamp: May 25, 2022, 09:59:00 pm
content: For a PNP transistor like you have drawn:The first PN junction (emitter to base) is a diode, not a resistor. Ohms law doesn't apply.In a diode, when the P region is higher voltage than the N region, holes flow from P to N and electrons flow from N to P. The transistor is made so there are a high concentration of holes in the P region and a low concentration of electrons in the N region. So there are much more holes flowing from P to N than electrons flowing from N to P. This is why you have current gain.The small current of electrons flow in from the base, resulting in a high current of holes flowing in the emitter. When the holes enter the base, they are swept up into the collector because the collector is negative relative to the base.V1 is the forward bias voltage of the diode and is always around .6V for a silicon diode. V2 is a reverse bias and can be anything from zero volts to a high voltage as long as it doesn't break down the junction.So actually we can't conclude that V2 > V1.

author: TimFox
author_rank: Super Contributor
timestamp: May 25, 2022, 10:21:48 pm
content: However, if we have the same current flowing into a higher (externally-applied) voltage, that is a power gain.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 26, 2022, 12:08:50 am
content: The arguments concerning resistance suggest a too tight reliance on linear theory; which, a transistor most certainly is not .Perhaps it would be illustrative to introduce Thevenin equivalents, and then nonlinear elements (e.g. diodes), and compare average (Rav = V/I) and incremental (Rinc = dV/dI) resistances. This contradicts the prior case, which itself (the contradiction) perhaps needs to be explained away, that earlier teachings are intentionally oversimplified, and can't be taken as complete and whole truths, and few teachers take the opportunity to note this about their material. (Perhaps some other examples might be given, like how you can't take sqrt(-1)... except when you can.)And then to make it clear that, collector current is still irrelevant based on just this; i.e. clearly the collector incremental resistance (h_oe to be technical) is high regardless of Ic (as long as Vce > Vce(sat)) so Ic cannot possibly be due to this either. So we need still more advanced theory to explain this.And this should maybe be taken as par for the course, at this point; because, again, nonlinear systems can be nonlinear in so many interesting ways. Ways that, often, our linear toolkit isn't wholly inapplicable to -- but they must be applied more narrowly and carefully.If, having exhausted all these more conventional explanations, further explanation is still left wanting -- perhaps a minor introduction to semiconductor physics would be in order? For example, the fact that a diode has reverse leakage current is pretty ordinary; whatever mechanistic explanation might be offered to motivate that, is probably fine. Well, we can simply say that, by gluing two diodes together (very closely), it just so happens that the B-E junction causes the B-C junction to leak, indeed with a carefully constructed device, it causes such great leakage that Ic dominates over Ib. And this is a similar effect to the leakage depending on applied light (photodiode / PV) (which however does generate a small amount of power, as seen by nonzero current flowing at zero voltage; we might argue that the B-E voltage drop accounts for this in the BJT, hence we don't see negative Vce(sat) in that case).Tim

author: free_electron
author_rank: Super Contributor
timestamp: May 26, 2022, 12:13:20 am
content: let me answer with another question :flip that emitter and collector around. see what happens .who says emitter and collector are where they are drawn ?if you can answer that , you have the answer you seek.

author: aneevuser
author_rank: Frequent Contributor
timestamp: May 26, 2022, 06:31:36 pm
content: I'm not sure exactly what the question is that you're asking. However:1) the student seems to believe that the BJT behaves as if it has a couple of resistors inside it, if I understand what they have written correctly. You need to explain that this isn't the case, else they won't get far.2) I think that the key to understanding BJT operation is:a) that of understanding that the reversed biased base-collector junction has a constant-current property - if the emitter injects I A/s of charge carriers into the base, then (almost) I A/s flows into the collector, and hence flows out of the collector.This is true regardless of the emitter-collector voltage (over some range) - although the rate of drift of charge carriers in the collector varies with the electric field therein, that doesn't change the current that passes through it, which is determined by how much charge the emitter injects (the electric field in the collector will however determine the charge density in the collector, which will be larger when the collector drift speed is smaller - intuitively, when drift speed is lower due to smaller electric field, the charge carriers will be bunched closer together, but we will still have the same number entering and leaving the collector each second)b) to understand that that the current injected into the base by the emitter is determined by the base-emitter voltage V_{BE}, and is exponential in V_{BE} (this is physically due to the fact that the charge carriers on either side of the junction follow (to a good approximation) the Boltzmann distribution - somewhat strangely, given that they're fermions...)c) to understand that the base-current-to-collector-current relationship/beta is more like a correlation than a causation - collector current is "caused" by the V_{BE}, but correlates well with the base current.The extent to which that can be put across to any given student will vary, of course.[prediction: this thread will gradually devolve into the standard heated argument as to whether BJTs are current-controlled or voltage-controlled]

author: rfeecs
author_rank: Frequent Contributor
timestamp: May 26, 2022, 07:03:25 pm
content: Quote from: aneevuser on May 26, 2022, 06:31:36 pmYou cannot show that voltage causes current or that current causes voltage. You have a relationship, not a cause/effect. So better not to bring it up.

author: TimFox
author_rank: Super Contributor
timestamp: May 26, 2022, 07:08:37 pm
content: For devices with a monotonic relationship between voltage and current, it is often useful to label them "voltage-activated" or "current-activated" depending on the curvature.For linear resistors, this label is not useful.For a PN diode, it is easier to consider the voltage to be a function of the current, since the current as a function of voltage is a very steep function, and subject to small shifts in the voltage parameters, especially those to do with temperature.

author: Terry Bites
author_rank: Super Contributor
timestamp: May 26, 2022, 08:19:57 pm
content: The behaviour of the bipolar transistor viewed as semiconductor physics is a little complicated. It takes the magic out of it for me, I can tell you.Picture that the small current from the base bias potential is making the base more transparent to the C to E current flow. You only a little current need to keep the feed enough carriers into depletion regions and push the energy barriers down.Base bias is used to overcome the energy barriers created by the depletion regions. Tons of pictures and animations out there.The depleted regions lack charge carriers and so no current can flow there. Extra charge flowing into the base undoes the depletion, it reduces the width of depletion regions allowing charge carriers in the collector and emitter to be in much closer proximity and do a bit of mingling (and other stuff for exams). This allows current to flow from E to C in an NPN or C to E in a PNP.Theres always more to it but WTF, solder like a boss!

author: TimFox
author_rank: Super Contributor
timestamp: May 26, 2022, 08:25:52 pm
content: In my first electronics class, the teacher started with the basic Shockley equation for a P-N junction current as a function of voltage.Adding the B-C junction on top of the base region, and applying an appropriate voltage on the collector pin attracts most of that current to the collector pin.That governs the collector current as a function ofVBE.From that, one calculates the transconductancedIC/dVBEMaking a less accurate assumption that the fraction of the emitter current that flows in the base lead instead of the collector lead gives the usual parameter (beta), which is not truly constant, and an estimate of the base current.From that, one gets an estimate of the input resistance of the common-emitter circuitdVBE/dIB.We were taught how to reduce the dependence of our circuit on that base current, due to its wide variation.You can get a lot of low-frequency bipolar transistor behavior from those simple assumptions.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 26, 2022, 09:11:47 pm
content: The argument with V2, which is supposed to be the base to collector voltage (in the text it is written as emitter to base like V1) is not really helpfull. The point is, that the base to collector voltage can even change the sign.Using the base configuration and thus alpha is a good way, though one may argue that a NPN transistor may be a little easier as electrons carry most of the current.The understanding of the transistor depends a lot on the picture one has of the diode. The usual picture is balance of the dirft and diffusion currents, which can be quite confusing. Another picture is the depletion zone as a high resistance zone and the bottelneck in forward direction, which is a bit misleading.A 3rd less common picture, but closer to the quantitative modeling is realizing that the recombination process as the one to limits the current flow in forward direction. So in a forward biased PN junction holes from the P side pass through the depletion zone (with essentially no obstacle) and than cause a kind of congestion of holes in the N zone waiting for recombination. The rising concentration of holes in the N zone is what limits the current through the depletion zone, not the low number of carriers in the junction region. The voltage at the PN junction directly translates to the ratio in the hole / electron concentrations on both sides. In forward direction this can increase the minonrity carrier concentration a lot (exponential), but there is not much room to decrease it as it can't get negative and there are only a few to start with. So the generation current (reverse direction) is limited and kind of saturates while the forward current can increase exponentially with the voltage.In a PNP transistor the holes in the base than have the choice of waiting for recombination and flow to the base terminal or passing all the way through the base and going to the collector without a recombination. With a emitter - collector voltage large enough for the holes to be wellcome in the collector (e.g. > some 50 mV or about 2*kT/e) a lot of the holes choose the collector instead of the base. The holes in base still see mainly the other holes and this way the emitter potential and not really the base potential.For the start one can keep the approximation of current control, assuming a constant gain.

author: Sredni
author_rank: Frequent Contributor
timestamp: May 26, 2022, 11:07:18 pm
content: Quote from: Kleinstein on May 26, 2022, 09:11:47 pmWell spotted. Yes, that should be the collector voltage and then there is the question of the sign that should be consistent with the BJT being is active mode. This is a beginner question and that configuration is that used to illustrate the working of a transistor in active mode.Personally I believe a PNP is better to explain a common base configuration because the outside current and the dominant current inside the device (hole current) have the same sign. But this is a matter of tastes.

author: rfeecs
author_rank: Frequent Contributor
timestamp: May 27, 2022, 12:14:44 am
content: Quote from: Sredni on May 25, 2022, 09:19:25 pmI didn't notice that. The level of sophistication of the drawing went way over my head.It might be worth considering the importance of notation. I think most textbooks have some convention of using capital letters for DC values, like bias points, and lowercase letters for AC values. This simple habit helps to keep things straight in your mind.Keeping track of positive and negative is also essential. Holes, electrons, PNP, NPN are easy to get mixed up. Conventional current, electron current, hole current, total current ...

author: bostonman
author_rank: Super Contributor
timestamp: May 27, 2022, 03:21:44 am
content: NPN transistors were always easier for me to understand, but, when it came time for PNP, I always get confused.Something that never quite made sense to me. BJT's are "current" gain transistors and MOSFET's are "voltage" gain.A BJT will not turn on without the base being approximately 0.7v higher than the emitter. Although the beta of the transistor times the base current equals the collector current, if used in a switch configuration, it seems like the math revolves around the voltage.Is a BJT considered a current gain because beta times the base current equals the collector current?In the case of a MOSFET, the drain will have more current than the gate, so that seems like it's also a current gain.

author: Sredni
author_rank: Frequent Contributor
timestamp: May 27, 2022, 03:26:30 am
content: Quote from: rfeecs on May 27, 2022, 12:14:44 amWell, I am on my way to open an art gallery in London. But let's get that Nobel prize, first. :-)Seriously, there is a reason I made that picture as ugly as it is, without copying the original one. I have to say that the original picture had the arrows showing the current directions, which I forgot to add.QuoteI agree, but we must keep in mind that this is the first encounter with a transistor and that at this stage of the study there is only one kind of voltage and one kind of current : the small signal model with its artificial mathematical decomposition of iB = IB + ib (or, as I prefer in ASCII Ib = IBQ + ib) is a few lessons away. There is no bias + signal, just current (or voltage).

author: TimFox
author_rank: Super Contributor
timestamp: May 27, 2022, 03:28:58 am
content: They are both voltage-actuated. The difference is that the DC current in the base is substantial, and the DC current in the gate is not."Beta" is called the current gain, but it can be considered as the reciprocal of the fraction of the emitter current that flows in the base lead.In the linear approximation, both devices have a "transconductance", the derivative of the output current with respect to the input voltage for small variations around a quiescent (bias) point.For a silicon bipolar transistor,gmis approximately the emitter current divided by 26 mV.

author: bdunham7
author_rank: Super Contributor
timestamp: May 27, 2022, 03:33:59 am
content: Quote from: TimFox on May 27, 2022, 03:28:58 amSo a traditional bipolar transistor is still a 'field effect' transistor in way?

author: bostonman
author_rank: Super Contributor
timestamp: May 27, 2022, 03:53:19 am
content: My first education in transistors was given the beta (i.e. 100), and calculating the BJT circuit. When I got my first job, I remember looking at schematics trying to find where the greek symbol beta was located to show the beta value.When I returned to school, I learned more about beta and transistor circuits, however, I've never needed to dedicate myself to the finite calculations. Plus, my education was a technical degree rather than strictly EE.In any case, this was one reason why I asked why BJT's are considered current gain. Also, as for beta, I use to get confused on whether I needed to calculate the base current first, and then estimate where the beta is based on the range so I can calculate the collector current, or get the beta based on the graph.Then I'd get caught in a whirl of basically thermal runaway where the BJT gets warmer, so beta changes, that changes the collector current, etc...Maybe it's my slight lack of understanding BJTs that confuse me about PNP's, but they just don't flow correctly in my head when I see one used in a circuit.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 27, 2022, 05:42:04 am
content: Quote from: bdunham7 on May 27, 2022, 03:33:59 amEven better, there's a SS equivalent to the vacuum tube triode, the Static Induction Transistor. It's made something like a FET but with a grid, which like the triode, allows some electric field through -- rather than the strong shielding effect of a fully-pinched-off channel -- thus greatly increasing g_os (or however you want to label output conductance).The fields in a BJT become clearest when using bandgap diagrams, but that is quite fully into semiconductor physics so probably isn't helpful/illustrative for the OP's purposes; but that is indeed where the voltage dependence comes from.Incidentally, tubes have the same exponential cutoff, for the same reason (Maxwell-Boltzmann statistics of the charge carriers), though it happens to be quite far down the transfer curve (Child-Langmuir law takes over at useful signal currents), and easily masked by leakage currents, so it's not very consequential, and as far as I'm aware, no textbooks bother to discuss it. I've measured it for a 6AL5 for example; themodeled emission factorof 3.15 presumably implies about as much higher an absolute temperature, which... it's not quite that high, as it happens, but that's plausibly part of the explanation. Probably tubes have a "non-ideality factor" to them as well, so that the temperature seems a bit higher than otherwise. (This is the same "n" that commonly appears in the Schockley equation, typically around 2 for Si diodes, which are decided not simmering at 586 K in typical use.)Quote from: bostonman on May 27, 2022, 03:53:19 amFully grasping all the symmetries, once you've already spent so much time learning just the basics, how to use one side of them -- seems confusing, daunting, like relearning all of it; once you realize it's just turning your head sideways, it's quite a lot easier. But it takes that insight to get there, and it can feel really weird / uncertain while you're still getting used to it.Even myself, it was relatively recently I pushed through to fully understand/use the series-parallel transformation. Enjoy a fruit of this labor, the inverted Cockroft-Walton current multiplier (drew this ~2018):This is one transformation that you tend not to use very much, so I suppose it's no accident I didn't learn it sooner (and even then, it was out of curiosity; one can go a professional career without needing it).Tim

author: aneevuser
author_rank: Frequent Contributor
timestamp: May 27, 2022, 07:25:59 am
content: Quote from: bdunham7 on May 27, 2022, 03:33:59 amYes, very much so. The field in question is the electric field across the BE junction. Feynman discusses the behaviour of PN junctions quite nicely in terms of the E field in sections 14.4 and 14.5 here:https://www.feynmanlectures.caltech.edu/III_14.htmlIf you are not familiar with the Boltzmann distribution (which is a idea from statistical mechanics), then you'll have to take the exponential voltage-current relationship on trust though, as he doesn't derive that in this section.

author: aneevuser
author_rank: Frequent Contributor
timestamp: May 27, 2022, 08:01:57 am
content: Quote from: bostonman on May 27, 2022, 03:53:19 amI find that it's easiest to keep the NPN-vs-PNP voltage arrangements clear in my mind by thinking about a BJT push-pull circuit. There, you can imagine a mirror image line though the middle of the circuit, where all the voltages are reflected, NPN voltages above, PNP voltages below.So, for example, if the NPN device has its collector 10 V above its base, then the PNP device will have its collector 10 V below its base.And you can imagine a sine wave, with +ve and -ve half-cycles, driving the base - in the +ve half cycle, the NPN turns on when its base is 0.6 V more +ve than the emitter - mirror imaged, this means that the PNP turns on when its base is 0.6 V more -ve than its emitter, which happens in the -ve half-cycle.

author: Sredni
author_rank: Frequent Contributor
timestamp: May 30, 2022, 02:09:13 am
content: Serving a youthful porpoiseWhat was the second purpose of my question here? To verify that the quoted question was not an unanswerable, not focused or ill posed question. And that it could be easily interpreted (despite an ugly picture and one typo) and answered in a way that could help other beginners with similar doubts.I would go even further to say that the doubt expressed by the student is fairly reasonable for a beginner and provides a couple of hooks that help driving home the concepts of transistor action and how the currents in the input and output meshes are simultaneously linked (IE approx IC) and independent (they show up in two one-ports with a completely different voltage).Some of the gentlem-- gentlepersons who donated their time to answer, might wonder "why did you need to verify that?". The least I can do is to offer an explanation of where I found the original question and why I did bother with all this. Consider the following a general observation on what should constitute good didactics and also a reflection on the quality of another EE-related website, like I have seen many in the chat section (on distributors, customer services, developers, designers). And I would have posted this in the other thread I had opened in the chat section, if it hadn't been deleted already.The original quoted question was asked (not by me) on EE Stack Exchange and nearly immediately closed and downvoted so that it could be deleted after a few days. The stated reason was that it was "not focused". I asked on the corresponding Meta the reasons for closing it (and deleting it) but i received no real explanations for the closure except a tautology ("it was closed because three users with x points decided to close it") and generic bogus reasons ("it was low quality", "you can google it" - yeah like every other question) that were not actually explained. Neither by those who voted to close it, nor by the other high rep users and moderators that follows that Meta.I needed a reality check to see if I was the only one on this planet to see the question as understandable and answerable (if possible even in a way that could address the doubts expressed by the OP, and also being useful to others.)On a more general note (and this is why I called this an EE didactic experiment and wasn't sure whether to post it here or in the chat section), what happened shows how a legitimate (and all in all well formulated, considering it exposed the student's reasoning and gave a few key points to work on) student question can be shut down by a lazy or incompetent teacher... just because.The real problem, though, is when the other 'teachers' band together to defend an opaque decision and cannot even explain why such decision was taken (especially in the case of such a simple and short question). I have seen more talkative witnesses at a mafia trial.One more thing.I left a code at the end of my query on EESE Meta. I did this after modifying it in its longest form that asked for specific reasons why the OP question had been closed - therefore before I got downvoted from 0 to -2 and I received further comments that were not specific and gave generic 'low quality' reasons (someone even suggested I was affiliated with the pic hosting site and could be making money out of their banners - which by the way I never saw)Anyway, it's this code here:3d8098c310eb3583a51c95f338b340c5What is it? Well, it is the md5sum of the following txt file that completes my question and should be read after my last line "I hope it is clear what I am asking":It's about the "we don't know, we cannot know" part. But let's see if my crystal ball is working: I forecast downvotes that will have the effect to hide this Meta question and no real explanation about why the OP question was closed. Except mabye a vague and unsubstantiated "it's low quality" 'reason'. I would not be surprised to see some excuse not to answer the questions posed.And exactly yesterday (well, when I wrote this it was yesterday) my question went from -2 to -3, and still no actual answers.Dang, that gipsy was right. I made a bargain!64c9c56dc1e08ebe882cd07ce33559ec

author: Sredni
author_rank: Frequent Contributor
timestamp: May 30, 2022, 02:18:49 am
content: Quote from: TimFox on May 27, 2022, 03:28:58 amI have to disagree with this. Voltage and current, like electric and magnetic field are concomitant manifestations of a single entity. There is not a cause and an effect (like in an EM wave it's not the change in magnetic field that causes the change in electric field and so on).One can choose to see a BJT as voltage controlled or as current controlled, or even as charge controlled.The 'bias' toward the voltage is due to the fact that in solid state physics we deal with energy and potentials, so everything seems 'caused' by a potential.Then there is the fact that - as you said in another message - it might be easier to see the transistor as current controlled because the voltage would be too sensitive to fiddle with, but fundamentally, one can explain how a BJT works by injecting current - see my answer herehttps://electronics.stackexchange.com/questions/547625/working-of-bipolar-junction-transistor-with-electron-flowthat has a bit of references at the end, including the original paper by Ebers and Moll who had no problem in modeling the BJT as current controlled.

author: TimFox
author_rank: Super Contributor
timestamp: May 30, 2022, 03:01:55 am
content: As a simple explanation to the novice, it is easier to start with the diode equation, have the base-emitter voltage produce an emitter current, then state that the higher voltage on the collector attracts most of that current.Beta varies all over the map between devices with a given part number, and over a reasonably wide range with respect to emitter current, while the transconductance from the Schockley equation is better defined.For DC, the voltage and current happen together, just like in a resistor, but not linearly related.Otherwise, you need to explain how a low current introduced into the base terminal magically induces a larger current in the collector lead.Of course, then the student should learn the more detailed equations for the DC performance, and then proceed to the dynamic performance.

author: Sredni
author_rank: Frequent Contributor
timestamp: May 30, 2022, 03:12:49 am
content: The explanation of how a small base current can draw a large collect current is given on Streetman, for example.On a less device oriented textbook, Sedra and Smith explain how a diode works by driving it with an ideal current generator and how that current ends up changing Vbe, which in turn...It's a chicken and egg problem.But as an agnostic, I don't want to start a religion war here.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 30, 2022, 04:23:31 am
content: Ah, EESE? Being a community, like any other, it could simply be you'd get different answers at different times of the day or week -- different users online. You're as likely to get flamed here (or just get low quality or off topic comments) as useful comments; the only difference is they have tools available to the users to self-moderate.Being a didactic question -- it could well be better placed onhttps://academia.stackexchange.com/than EE. You run the risk of subject matter experts not being present -- but perhaps the meat of the question is recognizable enough, or the subject can be explained well enough to them, to figure out a solution to the prompt.I guess I'd be surprised if Meta didn't suggest this, but maybe there aren't as many people there, or, that care.Tim

author: magic
author_rank: Super Contributor
timestamp: May 30, 2022, 05:52:45 am
content: Quote from: Sredni on May 30, 2022, 02:18:49 amOne can choose anything, but here's something:You can write a reliable equation for emitter current as a function of base-emitter voltage, or vice-versa, which works for any collector voltage and temperature.There is no straightforward and equally general equation for emitter current as a function of base current. Nor for collector current.Quote from: Sredni on May 30, 2022, 02:18:49 amWell, yes, because the base is a feedback output which gives you approximately 1% of collector current, at least under some conditions, and for some value of "approximately"Quote from: TimFox on May 30, 2022, 03:01:55 amThis doesn't seem to account for saturation.Quote from: TimFox on May 30, 2022, 03:01:55 amYep. The magic involves forward biasing the junction as a result of pulling current from it, and we are back to voltage.BTW, the original question was extremely low quality and a clear sign of somebody being not even close to understanding fundamentals. First, the use of Ohm's law to talk about diodes. Second, the confusion/errors in the definition V2. Third, why should anyone care about Vce being greater than Vbe? What it was that the guy even wanted to know?The only sensible answer to such question is a full, ground-up introduction to BJT theory. You can become the 1000th person to write one if you think there is something novel you can add to the topic, or just tell the OP to RTFtextbook. It appears that the general attitude at SE is to do the latter in such cases.

author: aneevuser
author_rank: Frequent Contributor
timestamp: May 30, 2022, 10:57:35 am
content: Quote from: magic on May 30, 2022, 05:52:45 amYes, indeed. And that is of course because there are well understood physical principles which relate BE voltage to emitter current (Boltzmann distribution), but there aren't any such principles which relate base current to emitter current - that depends on the details of device manufacture.And it seems to me that it's fundamentally silly to promote a BJT as base-current controlled, when circuits generally do not (or never?) "control" base current. Usually, a BJT is controlled by choosing a DC operating *voltage* on the base, and a small signal *voltage* is then applied to the base, and the equations which predict the linearised behaviour of the BJT are written in terms of voltage. It's voltage all the way, no?If we were controlling base current, wouldn't we be seeing programmable current sources attached to the base?

author: TimFox
author_rank: Super Contributor
timestamp: May 30, 2022, 02:04:24 pm
content: The vacuum triode, operated in the normal space-charge-limited regime, is always considered to be voltage actuated (i.e., the cathode current is a function of the grid-cathode and plate-cathode voltages).If you forward-bias the grid-cathode diode, the same space-charge equations give you the cathode current, but now some of the cathode current flows to the positive grid.It is still considered voltage actuated, and the ratio of cathode current to grid current is not considered a control parameter, but rather as a result of the positive grid voltage.

author: rfeecs
author_rank: Frequent Contributor
timestamp: May 30, 2022, 05:23:32 pm
content: It's amusing that Shockley wrote that in both the vacuum tube and the bipolar transistor, "one current is controlled by another":From "BSTJ 28: 3. July 1949: The Theory of p-n Junctions in Semiconductors and p-n Junction Transistors. (Shockley, W.)"https://archive.org/details/bstj28-3-435

author: rfeecs
author_rank: Frequent Contributor
timestamp: May 30, 2022, 05:34:20 pm
content: Quote from: aneevuser on May 30, 2022, 10:57:35 am

author: TimFox
author_rank: Super Contributor
timestamp: May 30, 2022, 05:41:32 pm
content: The Schockley article referencing grid current reminds me of an advanced topic in vacuum-tube electronics: "charge control", as discussed in one of the last great textbooks to include tubes, "Amplifying Devices and Low-Pass Amplifier Design" by E M Cherry and D E Hooper, Wiley 1968. This concept is useful in discussing high-frequency behavior in tubes, but is not normally taught as the first item in freshman vacuum-tube electronics. Instead, one usually started with the Child-Langmuir Law, where the current in a space-charge limited thermionic diode is proportional toV3/2, and then proceeded to add a grid and calculate the field at the cathode due to grid and plate voltage, introducing the parameter  to represent the effect of the grid on the cathode field due to the plate voltage.

author: aneevuser
author_rank: Frequent Contributor
timestamp: May 30, 2022, 06:31:06 pm
content: Quote from: rfeecs on May 30, 2022, 05:34:20 pmYeah, harsh. But fair. Good point well made, and all that kind of stuff. The wording after the "but" is indeed unfortunate - let me reword it to something like "the physical principles which determine it are more easily influenced by device manufacture than those that determine the relationship between V_BE and I_E."Happier?

author: rfeecs
author_rank: Frequent Contributor
timestamp: May 30, 2022, 07:25:15 pm
content: Quote from: aneevuser on May 30, 2022, 06:31:06 pmLooking at a textbook:https://www.chu.berkeley.edu/wp-content/uploads/2020/01/Chenming-Hu_ch8-2.pdfThe equation for collector current (8.2.7) is exactly the same form as the equation for base current (8.3.3). The equation for beta (8.4.5) depends on the same physical parameters as the currents: diffusion lengths, base and emitter widths, doping concentrations and intrinsic carrier concentrations.They depend on exactly the same physical principles.Then again, almost all the textbooks say something like "if the potential across the junction is changed, current flows ... "And most people tend to look at it that way. That doesn't mean it is the only way to look at it.

author: magic
author_rank: Super Contributor
timestamp: May 31, 2022, 08:00:38 am
content: Quote from: rfeecs on May 30, 2022, 07:25:15 pmPerhaps something to do with the "Vbe" appearing in your equations.I wonder where is the "small currents control large currents" explanation of bandgap voltage references or, hell, even current mirrors.

author: rfeecs
author_rank: Frequent Contributor
timestamp: May 31, 2022, 05:00:29 pm
content: Quote from: magic on May 31, 2022, 08:00:38 amYou're right, it's not "current controlled". It's temperature controlled!

author: Sredni
author_rank: Frequent Contributor
timestamp: June 01, 2022, 01:24:57 am
content: Quote from: T3sl4co1l on May 30, 2022, 04:23:31 amThis is not the first time I have witnessed the closing of potentially very interesting question out of the ignorance of those who voted to close them, and asking why a question was closed is like being appointed defense attorney in a witch trial :-) . The verdict can't be wrong, mods/curators side with each other, and the reasons given (if any) are opaque to say the least.Arbitrariness and unaccountability. I don't think even giving them a bodycam would change anything. :-)The youthful porpoise of EESE should not be that to boost one's ego and/or punishing seemingly lazy users. You will do no good to the site by deleting their valid questions because they did not format the text to your liking, or they did not adopt your notation. Nor will you succedd in educating them when they do not immediately reply to comments (because, well, the Earth is not flat), or even when they are just plain lazy. For the simple reason that even when they are in the wrong either they just don't learn or they are never the same student: they keep changing every time. Therefore, if the purpose is to provide answers to the community, fix the godd***n question, not for the lazy ba**ard, but for all other users that can take advantage from the ensuing answers (this is how the site generates traffic, after all).But of course, one needs to understand the question or put their ego apart to type a space here and there, instead of voting to close it. This is especially damaging to the site when users who do not care about 'punishing' people for their bad formatting or for speaking English as a second language, might have started working on the question, to only find it closed when they log in to answer.From time to time I test the site to see if something has changed and would make it worth it to go back, but they seem unable to evolve. They lost me a long ago as a contributor (I make exceptions to maintain the answers I gave on the Lewin ring and for some rare occasion when I stumble on something particularly irking) but I really wished they did not drag the level down too much, tho.So many potentially interesting questions wasted to show off their 'power to close'. It's kind of sad.(On the bright side, if you need to select a resistor to light an LED...)QuoteYeah, this is the impression I received: nobody really cares. They seem to be there for brownie points.

author: Sredni
author_rank: Frequent Contributor
timestamp: June 01, 2022, 01:27:54 am
content: Quote from: magic on May 30, 2022, 05:52:45 amSee? This is what I mean when I say that they close potentially interesting questions out of their own ignorance.There is this vicious rumor that a diode in not a resistor. But it is. It is a resistor in that its state is described by the instantaneous values of voltage and current and not their derivatives (or integrals). Being nonlinear, the device resistance is a function of the current - or the voltage (I am still an agnostic). And I am not talking of the differential resistance, but to full blown resistance computed as the ratio of the voltage across the diode and the current flowing through it. It all depends on how this student has been exposed to the concept in class: not all the world is alike.And this brings us to another interesting point that could have been pointed out in an answer to that question: the fact that a transistor does not only amplify 'differential signals' about the biasing point (as was suggested by one of the comments). In one of the examples I gave, an high current but low voltage supply can be used to drive a high voltage load. This is using the transistor as an amplifier: 3.3V input, 10.some V output. A common collector can be used to amplify current (not the differential signal, the whole shebang) and a common emitter can do a little bit of both.And no, you do not need to do a full course on transistors to answer the question. I will edit my answer by formatting the unnecessary details in a small font (just that) - you do not need that, but I felt it would have made the answer more general.All this student was asking was: if Rpn<Rnp and IE>IC how can I say Rpn*IE < Rnp*IC? And the answer is: because IE is basically the same as IC. Yet, the real explanation is that of how IC can be essentialy equal to IE no matter how I choose VCC in the battery-only circuit or the load resistor in the resistor-laden circuit.And that can be explained by the fact that in a reverse biased diode the characteristic is flat (the current is essentially identical to the reverse saturation current, no matter what the reverse voltage). Personally, this is what made me really understand how BJT works.(I cannot help but notice that nobody pointed out that in one of my examples with added resistors I get |VCB| < VEB :-), the reason this happens in the second type of circuit can be shown graphically by introducing the load lines.)

author: Sredni
author_rank: Frequent Contributor
timestamp: June 01, 2022, 01:37:44 am
content: Quote from: aneevuser on May 30, 2022, 10:57:35 amThe body of relationships relating transistor parameters to voltage is the consequence of the framework into which solid state theory has evolved. Everything is reconduced to energy levels, so everything electronic seem to be caused by electric potential and voltage.But you cannot have a voltage without displacing charge - first in your battery/generator and then at your device's terminals. This is at first a matter of surface charge, but then it's the charge flowing inside the conductors that allows you to mantain the surface charge in a configuration that will sustain the voltage. It's a chicken and egg situation. And even in the device, if you want to change the thickness of the depletion region, and hence the voltage that opposes the flow of charge, you need to supply charge or your voltage won't change (see the elementary treatment of the diode in Sedra & Smith, for example). Heck, the diode's built-in voltage itself the result of charge movement initiated by difference in concentration.Even the MOSFET requires a lot of current during transients, because if you want to change the voltage (in what is fundamentally a capacitor) you cannot do that without moving charge.If anything, the model that better describes electronic devices is the charge transfer model: you move charge from here to there (there is a current) and at the same time the potential in space changes. The potential is just a manifestation of the position of charges in a given configuration. Yes, if you succeed into fixing the charges in that position, from then on you can have a potential field without current (electrostatics) but to reach that configuration (in an empty space with charges at infinity) you need to move charges first. And the potential in all space will change at the same time as your configuration of charges change (actually, the information on the new charge's position will propagate at the speed of light.)Then, once you have reached your configuration (of charge and potential), if you want to change the potential in space, you need to add, remove and in any case move charges. Again the potential will change at the same time the charges change configuration.QuoteHow do you think I set the collector current in my two examples? By choosing VEB? No - I programmed IE by choosing a resistor of value (VEE-0.7V)/IE (this is by modeling the input diode as a vertical line at 0.7V).

author: bdunham7
author_rank: Super Contributor
timestamp: June 01, 2022, 01:43:57 am
content: Quote from: Sredni on June 01, 2022, 01:27:54 amYeah, sometimes when people refer to resistance or impedance (zener impedance for example) they mean dV/dI, not V/I. So there's some ambiguity there. But to be called aresistorand have the term mean anything at all, I think it is implied that V/I = dV/dI or at least is very close. Another way of saying it is calling it 'ohmic'. Calling any device a 'resistor' just because there is a V/I number at any given point reduces the term to meaninglessness. IMO, of course.

author: Sredni
author_rank: Frequent Contributor
timestamp: June 01, 2022, 02:30:31 am
content: Well, you won't find carbon powder or thin metallic film inside, but as far as the VI characteristic go, that is a nonlinear resistor from the point of view of circuit theory.Of course when you consider second order effects, you will have all kind of parasitics, starting from the two capacitances. But if we work at DC, as specified from the original circuit...

author: bostonman
author_rank: Super Contributor
timestamp: June 01, 2022, 02:48:30 am
content: QuoteWhich book is that from? I tried working backwards from the hyperlink, but couldn't get to the source.As for where this topic has gone and the disagreements, as with many topics, people understand things differently. As an example, maybe it's my basic knowledge of BJTs and MOSFETs that keeps my questions and interpretations basic.With both a BJT and MOSFET, I look at it as the base/gate starting the whole process. Neither one will turn on unless voltage and enough current is on the base/gate. At that point it's just ohms law. If the BJT is turned on fully, the DC voltage on the collector is 10v, then (with a perfect transistor) Ic would be 10v / Rc. If Ic = Ie, then it's Ic/RE for the voltage.In the case of a MOSFET, it's basically (in my head) similar steps. Hence, this is why I never grasped why a BJT is a current amplifier and a MOSFET is a voltage amplifier.Due to seldom needing to calculate transistor circuits, sometimes I get confused when a resistor is on the emitter.With a BJT and the emitter tied to ground, the base is (with a perfect transistor) 0.7v. If a resistor is on the emitter to ground, then that changes the base voltage which changes the base current, which changes the collector current because beta changes, etc... Now I'm confused.On a side note, at my old job, we use to give basic electronics tests to applicants. My boss would give a quick overview of the test and I'd over hear him say that the resistor on the emitter and the resistor going from base to ground are in parallel and the circuit should be easy to calculate. I use to laugh because he moved up the corporate ladder by making upper management believe he was some great engineer who ran large groups at previous jobs.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 01, 2022, 03:48:56 am
content: Quote from: bdunham7 on June 01, 2022, 01:43:57 amRight -- the whole point of highlighting resistance, it being this whole big thing, big enough to call it a "law" -- is that it's this, perhaps unexpectedly, near perfectly linear (very flat, over a very wide range of scales) characteristic happens to show up, and so we celebrate that simplicity, for it makes our jobs so very much easier.And so anything that is significantly nonlinear, is most definitely not a resistor, at least not any good of one. So It leads me to wonder where the student even got this notion in the first place, did they come up with this themselves? What is OP's relationship here, are they tutoring? Is this entirely hypothetical (fabricated)? If so, is it even a realistic scenario -- is it based on some other real experience?In short, it strikes me that, it was likely someone's duty to inform the student that, no, in fact transistors are NOT resistors, they cannot be treated that way, and it seems like OP is stretching this idea much farther than it can go.So I guess I'd be inclined towards the self-learning angle, that seems the most reasonable origin. But OP has no duty to entertain this attempted application of theory; I'm all for, not disregarding ones' ideas outright, but acknowledging it should be enough, and then immediately moving on to why it doesn't work, and how ones' perspective can be shifted to a more practical direction.(Again, usual caveats apply: I'm very far from an expert on teaching theory; no offense intended, just being my usual frank self, trying to get at the core of things here.)Tim

author: Sredni
author_rank: Frequent Contributor
timestamp: June 01, 2022, 03:57:54 am
content: Quote from: T3sl4co1l on June 01, 2022, 03:48:56 amLeon O. Chua, Charles A. Desoer, Ernest S. KuhLinear and Nonlinear Circuits1987, McGraw HillI would call this THE bible of circuit theory.p. 50 : section 1.2: The nonlinear resistorthe first example is the ideal diode.For BostonMan:Chenming C. HuModern Semiconductor Devices for Integrated Circuits2009, Pearson384 pp.(pdf version on author's websitehttps://people.eecs.berkeley.edu/~hu/)Modern Semiconductor Devices for Integrated Circuits, First Edition introduces readers to the world of modern semiconductor devices with an emphasis on integrated circuit applications.Written by an experienced teacher, researcher, and expert in industry practices, this succinct and forward-looking text is appropriate for anyone interested in semiconductor devices for integrated curcuits, and serves as a suitable reference text for practicing engineers.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 01, 2022, 04:37:13 am
content: Quote from: Sredni on June 01, 2022, 03:57:54 amI don't have this book handy, sorry. Are you saying that the motivation for analyzing it as a "resistor" comes from this reference then?Tim

author: Zero999
author_rank: Super Contributor
timestamp: June 01, 2022, 11:19:52 am
content: Quote from: rfeecs on May 26, 2022, 07:03:25 pmhttps://www.eevblog.com/forum/beginners/transitor-the-base-pin/https://www.eevblog.com/forum/projects/_n-channel_-vs-_npn_-name-confusion/

author: TimFox
author_rank: Super Contributor
timestamp: June 01, 2022, 01:42:31 pm
content: At best, Ohm's Law is an accurate approximation to the current vs. voltage in a two-lead passive component that we call a resistor.Even resistors found at Mouser, such as high-megohm thick-film short surface-mount parts (e.g., 50 megohm 0603 package) have measurable non-linearity.But, if we are to confuse new students by calling any two-terminal passive component a "resistor", do we stop before we hit the NE-2 neon bulb?There is a well-worked-out theory of non-linear components, including resistors and capacitors, but I don't believe it treats charge storage in a PN diode.

author: magic
author_rank: Super Contributor
timestamp: June 01, 2022, 01:59:04 pm
content: I would like to propose the radical notion that you are wrong and it actually is resistors which are a subset of diodes, namely those diodes with variable saturation current which satisfy:Is = e(ln(Vf)-Vf/Vt+X)where Vf is the forward voltage, Vt is the thermal voltage and X an arbitrary constant which characterizes the particular specimen ofresistordiode.In other words, you can always fit any model to a single operating point. But the result is not guaranteed to be helpful in any way or worth caring about.Talking about static "resistance" of a collector is even more meaningless, because Vce and Ic are two almost completely independent variables and can be anything without affecting the other much. Their static ratio truly tells you nothing at all about anything.

author: magic
author_rank: Super Contributor
timestamp: June 01, 2022, 02:36:40 pm
content: Quote from: bostonman on June 01, 2022, 02:48:30 amIf I understand correctly, you refer to the case of a series connection of: a voltage source, base resistor, NPN, emitter resistor, ground.That's one of the legitimate uses of β, which is to say, the result will vary between transistor types.If Ib is the base current,Vsupply = Ib·Rb + Vbe + Ie·Re = Ib·Rb + Vbe + Ib·(β+1)·ReKnowing the values of resistors, Vbe and β+1 it's easy to get Ib and all the rest from there.To be strict, Vbe is the logarithm of Ie times some constant, but you can take 0.7V and it's likely close enough.

author: Sredni
author_rank: Frequent Contributor
timestamp: June 01, 2022, 08:12:20 pm
content: Quote from: T3sl4co1l on June 01, 2022, 04:37:13 amYou can find the first 15 pages of the second chapter on the website of Berkeley University.https://inst.eecs.berkeley.edu/~ee100/fa08/lectures/EE100supplementary_notes_3.pdf(if your google-fu assists you, there is even the whole second chapter online - at another uni. I link this from Berkeley because that's where Chua, Desoer and Kuh taught.)And no, it's not only that book, even if it is one of the most authoritative textbooks in circuit theory. You can find examples of diodes as nonlinear resistors in Smith & Alley's "Electrical Circuits" book, and here and there in lectures and recitations. One example:https://6002.catsoop.org/_static/S20/s20notes/Recitation11.pdfwhere it says explicitly that a "iode is another common example of a two-terminal nonlinear resistor."It's everywhere. Well, maybe not on Forrest Mim's notebooks (I haven't checked) but I believe any university level textbook of circuit theory should carry this simple notion.Unless... unless all new books have been Upgrayedd already (I am pretty sure the process is under way).You can find the notion - in a way or the other - also in commercial EE sites. Here's Cadence:https://resources.pcb.cadence.com/blog/2019-what-is-linear-and-nonlinear-resistanceit shows a diode as an example of nonlinear resistance. Gives the correct definition, even though it successively focus on the differential resistance (which is linear, by the way!).

author: TimFox
author_rank: Super Contributor
timestamp: June 01, 2022, 08:15:52 pm
content: Your original post involved confusion in a novice student.I submit that he will be further confused if you call the diode a resistor: he already mis-applied Ohm's law in his sketch.A diode, by definition, is a two-terminal device.

author: rfeecs
author_rank: Frequent Contributor
timestamp: June 01, 2022, 08:45:20 pm
content: Quote from: Sredni on June 01, 2022, 03:57:54 amMy "bible" was it's predecessor, so I guess it's the "old testament":Charles A. Desoer, Ernest S. KuhBasic Circuit Theory1969, McGraw HillI bought a used copy from the local university bookstore when I was in high school. I decided one summer to read it from cover to cover. I got maybe half way through.The section on the nonlinear resistor is very similar. Some sentences are identical. There is also a definition of "voltage controlled" (current is a single valued function of voltage) and "current controlled" (voltage is a single valued function of current) in that section.

author: TimFox
author_rank: Super Contributor
timestamp: June 01, 2022, 08:47:08 pm
content: That distinction is important, since some device functions are not monotonic, and the inverse function is multi-valued.

author: magic
author_rank: Super Contributor
timestamp: June 01, 2022, 09:04:23 pm
content: Their definition of a nonlinear resistor seems to admit devices capable of conducting any current that can be expressed with a rational number when biased with irrational voltage and arbitrary irrational current when biased with a rational voltage, so it is a fairly general notion indeed.However, they state very explicitly right at the beginning of 1.1 that Ohm's law is the linear correlation of current and voltage and that it applies to what they call linear resistors, i.e. ordinary resistors that respect Ohm's law. None of that justifies your student's talk about "Ohm's law" with regard to PN junctions.TL;DR, but I very much doubt that there is a single place in the whole chapter where they actually bother to divide the instantaneous voltage by the instantaneous current of a nonlinear resistor. Because it makes no sense to do so.Even if they do, they certainly don't call the resulting ratio "resistance" - that word only appears in reference to the constant R in Ohm's law for linear resistors.Long story short, they use the term "nonlinear resistor" to refer to things that aren't real resistors, but they don't mean it to imply that all those devices suddenly have typical resistor properties and can be analyzed as resistors. It's just a silly name that they came up with.

author: bdunham7
author_rank: Super Contributor
timestamp: June 01, 2022, 09:50:54 pm
content: Quote from: magic on June 01, 2022, 09:04:23 pmWell, tunnel diodes are said to have 'negative resistance', but the meaning of that is clearly dV/dI, not that V/I is negative at any point. I suppose a photo diode could actually have a negative V/I, but...i think the term 'dynamic resistance' as used with zeners would be a reasonable term to distinguish dV/dI from ohmic resistance.

author: TimFox
author_rank: Super Contributor
timestamp: June 01, 2022, 09:56:41 pm
content: Yes, the tunnel diode is a good example of "negative resistance", where the slope of the I-V curve is negative (opposite from that of a resistor) over a limited region.It is also an example of a non-monotonic curve of current (dependent variable) vs. voltage (independent variable), which has other implications.These "incremental" or "dynamic" resistances (derivatives) are useful in small-signal analysis, for example using the negative conductance of the tunnel diode to cancel the positive conductance (loss) of a resonant circuit to make an oscillator. Negative resistances (including tunnel diodes, op-amp circuits, and vacuum-tube transitron circuits) all need external power to function properly.

author: aneevuser
author_rank: Frequent Contributor
timestamp: June 02, 2022, 07:33:27 am
content: Quote from: Sredni on June 01, 2022, 01:24:57 amWell, I'm not surprised by this: StackExchange is fundamentally broken-by-design. It seems to work on the principle that a question has an "answer", and those involved seem ignorant of the fact that most of the value provided by asking a question is the back-and-forth discussion that a good question generates - often when I ask a question, I'm aware that I don't really know what I don't know well enough to formulate a good question, but I expect the following discussion and explanation to clarify it. SE doesn't allow that - discussion is deleted, answers remain.I suspect that the model just about works for its original purpose ("what setting do I use to make the compiler optimize for X architecture?" or similar software questions); it fails miserably in a more general context. All IMHO, of course.

author: aneevuser
author_rank: Frequent Contributor
timestamp: June 02, 2022, 07:38:51 am
content: Quote from: Sredni on June 01, 2022, 01:27:54 amIf I understand this correctly, you would consider all two-terminal devices (or indeed anything physical that you could put a voltage across - a piece of wood?) as a resistor?If so, that seems to generalise the word "resistor" to meaningless.

author: aneevuser
author_rank: Frequent Contributor
timestamp: June 02, 2022, 07:49:08 am
content: Quote from: Sredni on June 01, 2022, 01:37:44 amI don't see how this relates to my question about controlling a BJT by setting the base current.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 02, 2022, 10:44:45 am
content: Quote from: bdunham7 on June 01, 2022, 09:50:54 pmI use "average resistance" to mean instantaneous V/I. I think this is a fairly common use. And yeah, a power source is often defined as negative current (sourcing), while loads are sinking (whose definition? SPICE for example: take a positive voltage source wired to a resistor, V1 will be negative, R1 will be positive), so an illuminated photodiode is an example of a diode with negative average resistance, over at least some portion of its curve.Also, another instance: "marginal" in economics is synonymous with "incremental".Tim

author: Sredni
author_rank: Frequent Contributor
timestamp: June 02, 2022, 05:41:47 pm
content: Quote from: aneevuser on June 02, 2022, 07:49:08 amWe were talking about current control vs voltage control. The common base configuration is probably the best in showing how you can consider the transistor as a current controlled device: you pump a big current IE in, you get an essentially identical big current IC out, with a little leakage as base current.And what you set, when you design the circuit, is the current.As to how a small base current can induce a much bigger current in the emitter and collector, you can find an explanation in Streetman or even in Hu. I believe I posted a link to a SE answer where I give several references.But we can explain it with the Stormtrooper analogy:Let's say we put a single electron in base, and this electron is in the form of a rebel alliance soldier with a target on his back. The emitter is filled with stormtroopers who see the lone rebel and start shooting at him - the 'light bullets' are the holes. Stormtroopers being stormtroopers, they will mostly miss their target - let's say they have an accuracy of (1-alpha): out of 100 shots only one will be on target - all the others will hit the collector's wall.One rebel in base will result, on average, in 99 holes in the collector wall.A small incursion of rebel soldiers in base (let's say 0.1 mA) will result in a large flow of light bullets coming from the emitter - almost all missing their targets - and a giant number of holes in the collector's wall (let's say beta times? 10 mA).

author: Sredni
author_rank: Frequent Contributor
timestamp: June 02, 2022, 09:12:35 pm
content: Quote from: aneevuser on June 02, 2022, 07:38:51 amWell, if you burn it and turn the charcoal into a fine powder, add a little bit of binder (for example clay) and put a terminal here and there - that's a carbon resistor.QuoteHas anyone here ever heard of the 'tetrahedron of state'? Of effort and flow variables? Of constitutive relations?In circuit theory we have two main quantities: electric charge q and magnetic flux phi. Their time derivatives are current and voltage. These are two fundamental relations that come out of Maxwell's equationsi = dq/dt v = dphi/dt(If you prefer seeing it the other way around, the time integrals of current and voltage are electric charge and magnetic flux.)How many relationships can we find between these four variables (q, i, phi, v) ? Apart from the derivative and integral ones, we have four possibilityv vs i ---> resistancev vs q ---> capacitancephi vs i ---> inductancephi vs q ---> memristanceA resistor (or, if you prefer, a generalized resistor) is a two terminal device whose constitutive relation is between the values of voltage and current.An inductor's constitutive relation is between magnetic flux and current (or, if you prefer between voltage and the derivative of current)A capacitor's constitutive relation is between voltage and charge (or, if you prefer, between voltage derivative and current)A memristor's constitutive relation is between magnetic flux and electric charge (or, if you prefer, between the integral of voltage and the integral of current)I hope you won't ask me now: "but a capacitor has voltage and current too, so why is not a capacitor a resistor as well"? Or, "what about a SQUID"?The relationship between the variables is in general a nonlinear and time-variant one. In the case of the resistor it is also called 'generalized Ohm's law'. Some shorten it to just Ohm's law. In high school they only know about the 'ungeneralized' Ohm's law, the linear one, and they think that resistors can only be linear.

author: TimFox
author_rank: Super Contributor
timestamp: June 02, 2022, 09:34:42 pm
content: "Ungeneralized Ohm's Law" seems to be just that the current is an arbitrary function of the voltage, and vice-versa.In physics, for linear conductors, we use the vector relationJ=sE, where the current densityJand the electric field (voltage gradient)Eare vectors. For an isotropic linear conductor,sis a scalar, but for anisotropic conductors it is a tensor.(I have no problem with the well-worked-out theory of non-linear components, but I thought this thread was about teaching transistors to novices.)

author: aneevuser
author_rank: Frequent Contributor
timestamp: June 03, 2022, 10:52:31 am
content: Quote from: Sredni on June 02, 2022, 05:41:47 pmShort of time to engage fully, but..1) "considering" the transistor as a current controlled device doesn't mean that it is indeed such a one - I prefer to consider "current control" as a mere observation of the correlation of two currents, which is then promoted to a (non-existent, IMHO) causal relationship. Nothing in this thread so far has changed my mind (but being the most open-minded of open-minded individuals, I'm happy to be persuaded otherwise, by sensible arguments)2) in your calculation of base current, the base current was computed using V_{BE} i.e. the base current was the dependent variable, and V_{BE} was one of the independent variables. Surely when we talk about controlling a quantity, we want to consider it as one of the independent variables in our experiment. Unless I'm confused, your own argument is promoting voltage control, no?QuoteI have Streetman, but not Hu. I don't recall Streetman promoting current control much, but I will take a look - haven't looked at it properly for a long time.QuoteI'm rather more interested in the physical principles of current control than analogies, TBH.And one point has occurred to me: isn't it the case that recombination in the base (which leads to base current) is something that transistor designers seek to minimise via base thickness and doping? - i.e. they seek to eliminate the very mechanism by which you claim the transistor is controlled? That looks like something of a logical problem in the argument of those who promote current control.Anyway, au revoir. I have a Queen to celebrate!

author: aneevuser
author_rank: Frequent Contributor
timestamp: June 03, 2022, 10:56:22 am
content: Quote from: rfeecs on May 30, 2022, 05:23:32 pmI took a brief look at that paper - the current control comment seems to appear only at the end, when he is making an analogy with tube behaviour. He doesn't go on to discuss it more fully, so it looks like something of a throwaway comment. I'll look at it again when I've got more time though.

author: Kleinstein
author_rank: Super Contributor
timestamp: June 03, 2022, 12:29:40 pm
content: For the DC case one can use both ways (voltage control and current control) to understand the transistor. They both work about equally well, maybe the current control a little more for the beginners and the voltage control if you need to include details like the early effect. Still both ways work OK.It gets a litte different in the AC / time dependent case, especially turning off a transistor in emitter configuration from saturation. One can directly turn off the base current externally or even reverse the direction, but the collector current does not instantly follow the base current. It is only when the base votlage goes down to near zero that the collector current stops. To understand, model the transient or AC behavior the base - emitter votlage becomes the more important parameter. The base current sees complications from internal capacitance and storred minority carriers.

author: rfeecs
author_rank: Frequent Contributor
timestamp: June 03, 2022, 04:56:17 pm
content: Quote from: aneevuser on June 03, 2022, 10:52:31 amFor a modern transistor (since the 1960s), recombination in the base is not a major contributor to base current. If it was totally eliminated, you wouldstillhave base current.The base current is (almost entirely) the current flowing from the base into the emitter.I'm not sure anyone ispromotingcurrent control. I, for one, am simply saying that you can't prove it is one way or the other, and that it doesn't really matter.

author: aneevuser
author_rank: Frequent Contributor
timestamp: June 03, 2022, 05:43:03 pm
content: Quote from: rfeecs on June 03, 2022, 04:56:17 pmYou are of course correct here: I was mentally lumping the two causes of emitter current into one, when I made my comment. However, it is still the case, is it not, that in, say, a PNP device, the electron component of emitter current (I think they call it injection current? - need to check) is deliberately made as small as possible by doping, so I think my point here isn't entirely undermined - we don't want base current, if possible.But I wasn't aware of your second comment about recombination current being so insignificant, so I have learnt something useful anyway. I'll have to go and look up the figures.QuoteWell, maybe...QuoteMost of the time, it doesn't. But people do argue about it a lot, don't they?Anyway, since Mr Sredni has said that Streetman talks about the current control POV, I am recusing myself from this discussion to go and read it. I may return as a turncoat and try to convince you all that BJTs are indeed current controlled.

author: Zero999
author_rank: Super Contributor
timestamp: June 03, 2022, 05:45:43 pm
content: I thought that nearly all of the base current flows through the emitter, until the transistor approaches the saturation region, when a significant amount starts to flow into the collector, as the base-collector junction diode forms a Baker clamp.

author: Sredni
author_rank: Frequent Contributor
timestamp: June 03, 2022, 06:20:12 pm
content: Quote from: aneevuser on June 03, 2022, 10:52:31 amI am on the same page as rfeecs, here. I do not claim that BJT are solely current controlled. I claim that you can either see them as current controlled or voltage controlled. Or charge controlled, for that matter. The idea that they are 'fundamentally' voltage controlled is the result of a double... bias: first the way solid state physics is all about energy levels, and therefore potentials; and then the fact the 99% of the sources we use in our daily life are best approximated as voltage sources.I do not see cause-effect in physical relations. Just relations. If you can bring in the time, in a way or the other, we could discuss cause and effect.QuoteActually I considered VBE as a constant. I used the approximation of the input diode as a vertical line at 0.7V. The current does not depend, by definition, on VBE because 0.7V is associated with ALL possible values of (positive) current.This of course does not 'prove' current control (I am an agnostic: to me both voltage and current control are equally valid), but it shows that in this application it's easier to consider the transistor as current controlled.QuoteIn the stormtrooper analogy, you don't want too many rebels in base. They would either shield the collector walls with their bodies or even go get them 'troopers in the emitter if they are too many.But you still need a few to attract the stormtroopers' fire, otherwise you won't get any holes in the opposite wall.No rebels in base, no flood of laser-fire.QuoteWhatever it takes to get those pints down, right?

author: Kleinstein
author_rank: Super Contributor
timestamp: June 03, 2022, 06:36:31 pm
content: How the current is divided between IB1 and IB2 depends on the purity of the matrials. In both cases there is recombination: either in the base or in the emitter (including the contact). Due to the doping levels only rather few holes are send to the emitter. However the high doping level usually also comes with a shorter life-time and the emitter is usually large and thus more volume ( though planar transistors have rather small emitters).

author: TimFox
author_rank: Super Contributor
timestamp: June 03, 2022, 06:36:40 pm
content: "Actually I considered VBE as a constant. I used the approximation of the input diode as a vertical line at 0.7V. The current does not depend, by definition, on VBE because 0.7V is associated with ALL possible values of (positive) current.This of course does not 'prove' current control (I am an agnostic: to me both voltage and current control are equally valid), but it shows that in this application it's easier to consider the transistor as current controlled."With this assumption, of course the current is independent ofVBE, since you don't allow it to vary.I do not agree with this logic to disprove that the current depends onVBE.

author: Sredni
author_rank: Frequent Contributor
timestamp: June 03, 2022, 10:23:14 pm
content: It was never my intention to disprove that the current depends on VBE.My point is that voltage and current are concomitant, so neither is the cause of the other.I was replying to an objection that in order to design a transistor circuit one need to set VBE to a certain value. Nope, if it fits me better, i can set the current to a certain value (and that is what I did).Given the exponential dependence of I from VBE, there is too much sensitivity on the value of VBE to be able to set it to a desired value in practice. I seem to recall you said something similar in a post of yours in the first page. I said something similar in this answer on SE a few years earlier (https://electronics.stackexchange.com/questions/252791/why-is-a-bjt-considered-current-controlled).I am not setting VBE: I am setting IE and let VBE to be whatever value happens to be on the exponential curve for a value of VBE close enough to the 0.7V I assumed.

author: RJSV
author_rank: Super Contributor
timestamp: June 05, 2022, 06:09:37 pm
content: From student's perspective (they are the 'client' being served, by any teaching effort), that initial glimpse is way way too complex...I would start with a student by maybe showing (those) complex graphs, while introducing speaking as follows:"The transistor models can get complex, but for a common emitter circuit, you've got an established input current, to the base, that is multiplied in the collector (output) current."Then, you could start introducing (those) complex graphs. Sorry, but semiconductors ARE complex, depending on how far in a person wants to scrutinize.That first diagram, applied voltages without (current limiting) resistors made me cringe. Some students might choose the path of least resistance, and tune out the semiconductor physics / and math.I wouldn't do well, in academic tests, there, but do fine in any bench / circuitry understanding. 'Second order poles...' and things like that, go way over my, shrunken head (lol).

author: TimFox
author_rank: Super Contributor
timestamp: June 05, 2022, 06:46:44 pm
content: In my view, as expressed above in several posts, when teaching bipolar transistors to those completely new to the concept, one should treat the emitter current as a function of the base-emitter voltage, and the base current as a small fraction of that emitter current. However, that is a description of the device itself.To make a useful simple circuit, such as a common-emitter AC-coupled voltage amplifier, the next step is to show how the biasing circuit results in an acceptable range of quiescent current (over device substitution and temperature), and how the resulting circuit amplifies small-amplitude AC voltages applied to the base through the appropriate capacitors to larger-amplitude voltage swings on the collector.

author: Sredni
author_rank: Frequent Contributor
timestamp: June 10, 2022, 03:51:48 am
content: Quote from: RJHayward on June 05, 2022, 06:09:37 pmYes, it boils down to what the student has seen in class and what is their background. But the focal point is that when IE enters on one side, IC = alpha IE exits on the other. Then the student can decide to either accept religiously the fact that alpha is nearly 1, or if they are curious enough, investigate why alpha is nearly one and IC does not depend (withing limits) from either the collector voltage or the load resistance.QuoteAnd this is one of the reasons I reputed the question interesting. It exposes the misconception that the circuit with the transistor hooked up to two batteries lacks something. It does not.It does not require resistors, and it shows amplification at DC and not necessarily small signal amplification.The circuit with batteries alone is shown in all major electronics textbooks (Millman Halkias, Sedra Smith, Bell, Jaeger Blalock...always in the first pages of their treatment of BJTs) and it's perfectly fine. You just need not to exceed the maximum allowed voltages.The circuit I posted is an actual simulation and with regulated power supplies you can avoid resistors and even resistive cavbles and nothing will happen to the transistor.As a matter of fact, the absence of resistors is essential to explain why (output voltage V2 = VBC) > (input voltage V1 = VEB) is consistent with the assumption Rnp > Rpn and can be illustrated in a single pictureTransistor with naked batteries. No problemo.where the load lines of the naked batteries are just vertical.If you add resistors and push the operating point too close to the axis, even if the output battery is much bigger than the input battery, you could end up with VBC < VEBResistance in not futile. It adds degrees of freedom.exactly because the load resistance gives you the freedom to move the operating point of the output characteristic.(Sorry for the late posting, but I had to scan the pictures and I had them wrong - I used the CE chars)