transitor: the base pin. - Page 1
Forum section: electronics / beginners

author: PlagueDoctor
author_rank: Contributor
timestamp: December 19, 2014, 08:08:26 am
content: I've always been to embarrassed to ask this question, since I've designed & built several circuits, beambots, etc using transistors, even helped other people out... Any ways.(Assuming NPN type) when applying a current to the base pin to allow the current to flow from the collector to the emmiter(or vice versa) where does the current from the base pin go?I've always assumed either one of two things:The current of the base pin shares the ground with the emmiter pin. (Which doesn't make sense to me)Or the current doesn't actually flow to the base pin, but rather puts it under an electrical tension. (Which makes a lot more sense to me)Or am I way off base? Surprisingly, despite having read quite a few articles about them. Even going as far to learn how they are manufactured. I haven't crossed any text that explains this clearly.Thank you for your time"

author: Simon
author_rank: Unknown
timestamp: December 19, 2014, 08:15:49 am
content: The base current flows from the base and out of the emitter. Why does this not make sense ?The "electrical tension" you refer to is how MOSFET's work where no current flows.Between the base and emitter is basically a diode, the current through it x gain = collector current.

author: LvW
author_rank: Frequent Contributor
timestamp: December 19, 2014, 10:22:39 am
content: It is the the voltage Vbe that causes a current Ie out of resp. into the emiitter - and this current is split into one large current Ic and a much smaller current Ib (which is a more or less fixed percentage of Ic). Therefore, the current Ib of course goes into (or out of) the emitter node according to KCL: Ie=Ib+Ic.

author: Tandy
author_rank: Frequent Contributor
timestamp: December 19, 2014, 11:30:42 am
content: No need to be embarrassed, the best way to learn is ask questions about things you are unsure about. It is better to admit you don't know and learn the answer than pretend you do. There are plenty of people pretend to know and then give out misinformation to others.As others have said the current applied to the base travels to the emitter.When I first started to learn electronics I connected a circuit where I had tied the collector to +v and the emitter to a buzzer. I couldn't understand why the transistor would not switch reliably but eventually realised that the small current couldn't flow between the base and the emitter due to the resistance introduced by the buzzer. This is why you typically use a common emitter circuit connecting the emitter to ground with an NPN transistor and control the amount of current applied to the base using a resistor.A somewhat inaccurate but easy to visualise illustration of a transistor is a big pipe filled with large balls connected to the collector, and a tiny pipe with little balls connected to the base. Each tiny ball that passes through from the base to emitter carries a big ball with it from the collector. The more small balls you pass through the base the more big balls travel from the collector. Both the small and big ball drop out of the emitter.

author: Zero999
author_rank: Super Contributor
timestamp: December 19, 2014, 12:18:52 pm
content: Yes, the current flows into the base and out of the collector.The base-emitter junction forms a diode and this is why, in a common emitter amplifier the base current needs to be limited, normally by a resistor, otherwise an unlimited current will flow, causing it ti overheat.A base resistor is not required in an emitter follower because when the transistor turns on, the load between the emitter and 0V drops a voltage across it which lifts the emitter voltage up, causing the base current to fall, so the circuit balances itself; this is an example of negative feedback and is why the input impedance is high.Normally the base is the input but it's also possible to fix the base at a steady voltage and change the emitter voltage, as is done in a common base amplifier.

author: LvW
author_rank: Frequent Contributor
timestamp: December 19, 2014, 12:23:05 pm
content: A somewhat inaccurate but easy to visualise illustration of a transistor is a big pipe filled with large balls connected to the collector, and a tiny pipe with little balls connected to the base. Each tiny ball that passes through from the base to emitter carries a big ball with it from the collector.OK, this visualization looks rather good but one should mention that it models the BJT as acurrent-controlleddevice. No great problem - such a model works for many applications. However, you always should know what you are doing and, thus, realize that this is a model only. The physical reality is that the BJT is, of course, avoltage-controlleddevice (Ic=f(Vbe)). This is a proven fact.

author: rsjsouza
author_rank: Super Contributor
timestamp: December 19, 2014, 12:23:29 pm
content: As others have said, the base current flows to the emitter.Physically speaking, the base region is very narrow when compared to the emitter and the collector regions, and the flow of current occurs when a positive (NPN) or negative (PNP) voltage (in relationship to the emitter) is applied to it.This causes electrons (NPN) or holes (PNP) to be attracted to it at a great speed and, being narrow, the majority of these charges (electrons or holes) will completely bypass the base region and jump to the collector, which on itself has a positive (NPN) or negative (PNP) voltage applied to it in relationship with the emitter. This voltage on the collector ends up accelerating the charges even more, which causes a large current flow.The higher the base voltage, the larger the current flow between emitter and collector, but only up to a certain limit. This limit is the transition voltage between the base and the emitter, which for silicon is typically 0.6~0.7V.The base current itself is the small percentage of charges that end up staying in the base region (recombination) and flow outside of the device. The relationship between the base current and the large collector current is the transistor gain.(I am almost sure I got all the facts correct, but it is 6AM and I may have missed something)

author: Simon
author_rank: Unknown
timestamp: December 19, 2014, 01:03:08 pm
content: Quote from: LvW on December 19, 2014, 12:23:05 pmreally ?

author: deephaven
author_rank: Frequent Contributor
timestamp: December 19, 2014, 01:14:11 pm
content: Quote from: Simon on December 19, 2014, 01:03:08 pmhttp://www.vanderbilt.edu/AnS/physics/brau/H182/Theory%20of%20Transistors.pdfsection 3.1.3

author: macboy
author_rank: Super Contributor
timestamp: December 19, 2014, 01:30:42 pm
content: Quote from: LvW on December 19, 2014, 12:23:05 pmFor most people here,howit works is far more important and useful thanwhyit works. The voltage developed at the Vbe junction may bewhycurrent flows in the collector, but the function "f" in Ic=f(Vbe) is non-trivial. It is defiantly complex.This is where thehowit works is so important. The Vbe voltage corresponds to a Ib current, and reciprocally, any Ib current gives a Vbe voltage... just like with any forward biased PN junction. But what is great is that in Ic=f(Ib), the function f is much, much, much simpler than in the above case. At its simplest, a first order approximation works very well over a relatively wide operating range. So Ic=G*Ib where G is the current gain or hFE of the transistor. It is a simple linear relationship (understanding of course that this is an approximation). So,howa BJT works is simple, current Ic is current Ib times gain. Can you imagine trying to design a BJT amplifer with only the knowledge that Ic is controlled by Vbe, and with no knowledge of the relationship with Ib? It would be a nightmare. Engineers design BJT circuits with Ib, and account for some Vbe that is implicitly created by that Ib. Physicists know that Vbe controls the flow of electronsthrough both the base and collector, but it stops there. While Vbe controls the current flow, explicitlycontrollingVbe in order to control current flow is not practical. The concept is ridiculous.So, in short, stop preaching about Vbe. This is an engineering forum, not a physics forum, and furthermore, it is the "beginners" section. You are confusing people.

author: dannyf
author_rank: Super Contributor
timestamp: December 19, 2014, 02:02:48 pm
content: QuoteOnly to those who agree with such "fact" of yours.QuoteEmitter. You will often see that emitter current is (1+beta)*Ib, vs. beta*Ib for collector current.QuoteFor BJT, there will always be a base current, however low it is.For FET, there is no "gate current" (in theory and in DC analysis). The tension is the opening / closing of the gate area (channel as it is called).However, in real life, there is gate current, due to leakage and more importantly in AC analysis, as the gate of a FET is essentially a capacitor.

author: LvW
author_rank: Frequent Contributor
timestamp: December 19, 2014, 02:04:06 pm
content: Quote from: macboy on December 19, 2014, 01:30:42 pmMacboy - why so sensitive? Is it really necessary to answer this way?1.)I am not "preaching" - I gave a pure technical comment .2.)In particular, for an engineer it is important to know WHAT he is doing and WHY this is allowed (or not). If you read my reply again you will notice that I did NOT mention any charge carrier behaviour or somethink like that. Hence, why do you mention a "physic forum"?I am sure, everybody who knows the exponential law for the pn junction of a diode will have no problems to accept a similar relation for the transistor.Hence, where do you seee a problem?3.)I am confusing people? Just because I have mentioned the physical reality ? Do you really need some justification derived solely from CIRCUIT behaviour?4.)What people really confuses (and this is my experience in teaching electronics at a university for 25 years):Teaching current-control of BJT`s for beginners and being NOT able to explain all effects to be observed in electronic circuits (without using the voltage-control properties).5.)Did you ever realize WHY the input resistance of a circuit with Re feedback increases? This is theory ofvoltagefeedback! Is this physics or electronics?How do you explain the difference between class-A amplifiers and class-AB or class-B amplifiers?Where do YOU select the desired operational point? Not on the Ic=f(Vbe) characteristics?I hope you now recognize that I am writing as an engineer and not as a physicist.RegardsLvW

author: LvW
author_rank: Frequent Contributor
timestamp: December 19, 2014, 02:06:14 pm
content: Quote from: dannyf on December 19, 2014, 02:02:48 pmIs this a technical comment? Do you need references?I think, it is really a pity that in such a forum like this a pure technical discussion seems to be problematic. Why so polemic?

author: dannyf
author_rank: Super Contributor
timestamp: December 19, 2014, 02:19:30 pm
content: Quotehis/her comments are more wrong from a physics point of view, stemming I guess a complete lack of understanding of what a "current control" vs. "voltage control" mechanism means.

author: c4757p
author_rank: Super Contributor
timestamp: December 19, 2014, 02:24:37 pm
content: Instead of having another tedious current-vs-voltage pissing contest, why not just go read the last twelve we had?

author: IanB
author_rank: Super Contributor
timestamp: December 19, 2014, 02:31:07 pm
content: Quote from: dannyf on December 19, 2014, 02:19:30 pmI am not following your comment. The collector current in a BJT is controlled by the base-emitter voltage. This is how the device works and can be represented with considerable accuracy by device models.The current gain of a transistor is a simplifying abstraction that can be used to construct many aspects of a circuit design, but as a simplifying abstraction it has limitations and is not sufficient to handle all design considerations.

author: zapta
author_rank: Super Contributor
timestamp: December 19, 2014, 02:36:09 pm
content: Quote from: LvW on December 19, 2014, 12:23:05 pmThis looks correct to me. Ic = f(Ib), Ib = g(Vbe) an therefore Ic = f(g(Vbe)). In a similar way, the light emitted by a LED is a function of its voltage. These are not useful models because of their sensitivity and instability but correct nevertheless.

author: Tandy
author_rank: Frequent Contributor
timestamp: December 19, 2014, 02:44:24 pm
content: IMHO this kind of thing is very off putting to a beginner. If I had been the original poster who asked the question I would probably now be thinking bah I don't understand what these guys are talking about I'll just forget it.I appreciate that people are just trying to be helpful in giving a more detailed explanation but too much detailed analysis can be detrimental.For example, when a diode was first explained to me it was using a 1 way water valve example. Such an example does not show things like voltage drop or recovery times or any other aspect of a diode. Using water as an example also quickly breaks down when we need to explain things like EMF. However as a complete beginner this was perfectly adequate to set me on a path of understanding the basic purpose of a diode and an idealised explanation of how it functioned. Later when I had this understanding I was able to progress to a more detailed understanding of how a diode works.I had a recent similar experience not related to electronics. I was considering getting a CNC machine and a vacuum forming machine to make some plastic parts. I joined a CNC forum and asked a few beginners questions trying to establish how to get started only it set off big debates among the knowledgable people over the merits of certain machine types and construction techniques etc that in the end I felt overwhelmed and decided I was unable to understand what they were saying enough to make an assessment of the merit of their opinions. In the end I just gave up.

author: LvW
author_rank: Frequent Contributor
timestamp: December 19, 2014, 02:48:41 pm
content: Quote from: c4757p on December 19, 2014, 02:24:37 pmHad we? Why do you qualify this as a "pissing" contest (instead of giving a technical reply)?Is this not a forum for discussing technical matters?Don`t you agree that students/beginners can learn a lot from such discussions (and also how engineers should NOT interact with each other!)

author: c4757p
author_rank: Super Contributor
timestamp: December 19, 2014, 02:58:04 pm
content: Quote from: LvW on December 19, 2014, 02:48:41 pmExhibit: How I thought about this the last twelve times we had this discussion. This time, I return with a bit more experience actually trying to teach people. It just confuses them. These arguments have been trotted out loads of times before and most are tenuous at best. We're all bashing each other over the heads with our perspectives and it doesn't get anyone anywhere.

author: LvW
author_rank: Frequent Contributor
timestamp: December 19, 2014, 03:00:46 pm
content: Quote from: Tandy on December 19, 2014, 02:44:24 pmTandy - in the following I repeat one sentence from post#1:Or am I way off base? Surprisingly, despite having read quite a few articles about them. Even going as far to learn how they are manufactured. I haven't crossed any text that explains this clearly.Don`t you agree that this question deserves a bit more than only a simple analogy?I agree with you that - in explaining things - we always should try to find a good trade-off regarding the details of techical/physical effects. However, are you afraid that a simple exponential law for the pn junction will overstrain the questioner (who did ask for a "clear" explanation) ?

author: Tandy
author_rank: Frequent Contributor
timestamp: December 19, 2014, 03:18:28 pm
content: Quote from: LvW on December 19, 2014, 03:00:46 pmPerhaps this is down to a different understanding of the meaning of the question. I felt that the OP had read information explaining transistors, some quite technical but none of the explanations answered the (relatively simple) question.Most transistor descriptions I read explain things like the doping of semiconductors to create P & N regions, the way the electrons overcome the barrier, the hFE etc but don't explain in a simple way for a beginner that current is flowing from base to emitter when controlling the transistor. So a simple illustration would provide a simple way of answering the question without complicating it. Only the OP will be able to clarify the meaning of their question and their level of current understanding. I however prefer to start simple in a beginners forum and then they can always turn round and say, the explanation is to basic I want to know more about the detailed workings etc.

author: IanB
author_rank: Super Contributor
timestamp: December 19, 2014, 03:22:00 pm
content: Quote from: PlagueDoctor on December 19, 2014, 08:08:26 amIf we reduce this question to its core, it asks "Where does the base current go?" Answer: it goes out at the emitter.Reading the rest of the question betrays some misconceptions, however. For instance, you do not "apply a current" to anything. You apply a voltage and current flows.Current tends to flow between two different points if there is a voltage difference between them and if there is a path for the current to take.In a normally biased NPN transistor the base voltage is higher than the emitter voltage. Therefore a current will tend to flow from the base to the emitter. An NPN transistor provides a path for this current and so a current does in fact flow into the base and out from the emitter.When the transistor is operating normally, current also flows into the collector and out of the emitter.Since all current must be accounted for, there is an equation that can be written for the transistor:Ic+Ib=IeThis equation says that the sum of the base current and the collector current is the emitter current.

author: Tandy
author_rank: Frequent Contributor
timestamp: December 19, 2014, 03:24:19 pm
content: IanB I like the explanation, very clear.

author: free_electron
author_rank: Super Contributor
timestamp: December 19, 2014, 03:42:07 pm
content: Quote from: LvW on December 19, 2014, 10:22:39 amnonsenseIt is an electrical field that causes electrons to flow into the emitter and out of the base ( for an NPN transistor , using electron model , not conventional model ).in order for electrons to start moving you need to breakdown the recombination zone (often wrongly called the depletion zone. A depletion zone is an area that doesnt have any free electrons. (note : do not confuse this with depletion in depletion mode mosfets. that is classic model , not electron model !)) between the N and P material. The field required for that is in the order of 0.6 to 0.7 volts. once the recombination zone is gone current flows. Vbe is the field across this band. the bipolar transistor is NOT voltage controller. the relation is Ic = Ib x Beta. Pump more current in the base and you let more current flow in the collector.Simple as that. Bipolar transistors are current controlled.i gave a detailed electron model explanation including diagrams already on the forum . search for it.trust me, i've spent 23 years of my life taming electrons to run where i want them to run. i got a pretty good grasp on that stuff.

author: dannyf
author_rank: Super Contributor
timestamp: December 19, 2014, 03:44:08 pm
content: QuoteModels work by simplifying things, to get down to the essence of how the real world works. Unfortunately, what "essence" is varies.Take your diode example. The one directional (check) valve is a good analogy in description the primarily function of a diode -> one direction flow of current.But that's an imperfect description of a diode. Thus the amendment of a forward drop voltage.Unfortunately, that amendment itself is an imperfect description too: that forward drop voltage varies with the forward current, and may require further amendments....The point is that what is right or wrong is a relative concept. Just because a certain simplification doesn't work for a particular case doesn't make it invalid for all cases; and vice versa.

author: LvW
author_rank: Frequent Contributor
timestamp: December 19, 2014, 04:11:33 pm
content: Quote from: free_electron on December 19, 2014, 03:42:07 pmYou must be very sure to declare my position as "nonsense".(By the way: Also the universities of Berkeley and Stanford produce such "nonsense"; did you ever read the doctoral thesis from W. Shockley?)Free-electron, I am not so experienced as you in carrier physics (that is not polemic!).Therefore, as an engineer I prefer to evaluate such questions from measurable properties of a part or a circuit.I kindly ask you to answer one single question:How do you explain therisinginput resistance of a common-emitter stage with Re feedback (if compared with Re=0)?From feedback theory we know that current-controlled VOLTAGE feedbackincreasesthe input resistance and that current-controlled CURRENT feedbackreducesthis resistance. Therefore, is it a voltage or a current acting as a feedback signal ? Does the transistor react upon a change of the voltage Vbe or upon a change of the current Ib (of course, Ib changes also)?As you know, current feedback needs a common node where two currents are superimposed. Where is such a node?

author: free_electron
author_rank: Super Contributor
timestamp: December 19, 2014, 05:02:41 pm
content: perceived resistance you mean...There is a difference between analyzing a circuit and analyzing the fundamental of a transistor.your circuits input impedance may alter but that does not alter anything inside the transistor.Your common emitter resistor gets the sum of two currents ic and ib. this causes a drop over re. You do the math. It involves incoming bias , Vre and the currents.Simply tying a resistor to the emitter does not change the physical properties of the structure inside the transistor. It also does not alter the carbon content of the resistor.

author: LvW
author_rank: Frequent Contributor
timestamp: December 19, 2014, 05:20:11 pm
content: Quote from: free_electron on December 19, 2014, 05:02:41 pmI am afraid, I have expressed myself not clear enough - my fault. Nevertheless, thank you for your answer.Let me explain again:Applying negative feedback to an opamp, of course, also does not change the physical properties of the active unit, but it changes the overall input resistance of the whole amplifier circuit. And that`s what I mean:You can presume I am aware that Re feedback does not change something inside the BJT. But the input resistance at the base node is modified due to feedback.And THAT was my question: Onlyvoltagefeedback increases the input resistance at the base node - and exactlythiscan be observed and measured as a consequence of Re feedback. If currents are the relevant quantities to control the transistor the input resistance at the base must go down!Something wrong?

author: AG6QR
author_rank: Frequent Contributor
timestamp: December 19, 2014, 05:59:16 pm
content: The "voltage control" versus "current control" can be simplified by considering a diode, or even a resistor. Is a resistor "current controlled" or "voltage controlled"?Does voltage across a resistor cause current to flow? Or does current through a resistor cause a voltage to be developed?Ohms law tells us the relationship between voltage and current, but it doesn't tell us which one is cause and which one is effect. Often it's convenient to think of a resistor as voltage controlled (think of a resistance heater, where you apply a fixed voltage from a power supply and you can calculate how much current will flow), but other times, it's convenient to think of a resistor as a current controlled device (consider a shunt resistor, where you apply a current and a proportional voltage will be developed which you can measure). Most of our power sources deliver more-or-less constant voltages, so we usually lean toward thinking that voltage causes current, but in a world dominated by current sources, our thinking might be different. Would that kind of thinking be demonstrably wrong?With a diode, the function relating voltage and current is a bit more complex than a resistor's, and not a straight line, but still, the fact that there's a well-defined function relating voltage versus current doesn't tell us which is cause and which is effect.In my book, it's wrong to insist that the causality can only go one way. There are two perspectives, and both can be useful. Neither is wrong, but neither one, by itself, is the Only One True Right Way of looking at things.For those who would continue to argue it's only one or the other, let me put it another way. Suppose I gave you two diodes: One was voltage controlled, and the other was current controlled. Both had the same I-V curves. What experiment would you do that would unambiguously distinguish between them?Granted, if you insist that diodes are ALWAYS one way or the other, you might have to say that one of those diodes came from an alternate physics universe, but my point is, what measureable characteristic would distinguish between those two universes? Or in other words, how do you know that our universe is the one where the control parameter is the one you say it is?

author: IanB
author_rank: Super Contributor
timestamp: December 19, 2014, 06:21:22 pm
content: The best approach to this question (in all engineering fields, not just electronics) is what happens when you consider models of how a device operates.Any given model has operating variables and model parameters that go into the equations relating those variables. The best models are ones where the model parameters are nearly constant over the whole possible range of model variables. If you have such a model, then with values of those parameters you can make design calculations and operating predictions with considerable confidence.If your model parameters are not constant, then they themselves become variables and your ability to use this model to make design predictions becomes weaker.For instance, "the forward voltage drop of a diode is 0.6 V". In this model the "0.6 V" part is a parameter. It's a good rule of thumb, except the 0.6 V is not at all constant. With higher currents it could be 0.8 V or 1 V or more. If your circuit design has high currents flowing through the diode you are going to get a poor design with this model.There are more precise models of the diode forward voltage that account for varying currents (at the simplest, the graph in the datasheet), and with these models your design can handle large currents.Same with a transistor. The device model Ic = gain x Ib suffers from the fact that the gain is not a constant parameter. You cannot use this model successfully to design circuits in cases where the gain matters unless you introduce a new model for the gain as a function of other things. And those other things may include voltages.

author: dannyf
author_rank: Super Contributor
timestamp: December 19, 2014, 06:38:15 pm
content: QuoteLots.I pointed out this to you in your old thread but apparently you still couldn't understood it. Forget about current- or voltage-controlled bjts, just put in a theoretical model of a three terminal device, either voltage or current controlled, and you can work out the math -> you will see that the input impedance of the said circuit is comparable between the two devices. IE, the observed behavior of the circuit reveals nothing about the physical nature of the device in discussion. Thus, any attempt at answering your question is irrelevant to the discussion.

author: dannyf
author_rank: Super Contributor
timestamp: December 19, 2014, 06:41:41 pm
content: QuoteI am not sure if that comparison is valid for this discussion.Ohm's law certainly doesn't apply to a diode. And in the case of a two terminal device, there isn't anything to control or to be controlled.I think if you look at the physical mechanism for a transistor, you will have to agree that the recombination of carriers (ie current) controlls the collector current in a bjt. ie., it is current controlled. and many models for bjt devices indeed follow that description.

author: free_electron
author_rank: Super Contributor
timestamp: December 19, 2014, 09:20:10 pm
content: Quote from: LvW on December 19, 2014, 05:20:11 pmstill nonsense. you are again describing the behavior of a system. what happens inside the transistor does not change.sure, your system reacts to voltage , the transistor inside still reacts to the flow of the electrons. the physical mechanism does not change. you need to insert electrons in the emitter to remove the recombination zone. the electrons flow into the emitter and out the base. put an ampere meter there and you will measure that current.electrons per second is ampere. ergo : bipolar transistors are current driven. pump up the base emitter current and the collector emittor current goes up . the relation between the two is the beta.it doesn't get more basic than this. stick an ampere meter in the base path. if you get a reading : there is current flowing.do the same with a mosfet. electrons will flow into the gate and once the gate is fully charged the current flow stops. ergo : gates on a mosfet are capacitors. once fully charged that's it. so in transient a gate of a mosfet is current driven , but statically the mosfet is voltage driven.you can sing and dance all you want : that is the nature of the beast.

author: PlagueDoctor
author_rank: Contributor
timestamp: December 19, 2014, 10:01:12 pm
content: Wow! This thread kinda blew up :pThis is not my official reply yet, I just wanted to let you guy's know that I've read all of your comments.Tandy pretty much spot on replied exactly as I would have replied.With the exception of 'flame wars can be confusing'I actually find them quite helpful, even though some of it's is confusing. but that's why we have google/booksunfortunately I've got to go to work. and before I give a real reply, I want to test out on some breadboard what I think I have learned from you guy's.my problem was, up until now I've pretty much used transistors as JFET's, so that's how I though about them. and since all my circuits have been so low power with wide tolerance components, it hasn't been much of a problem. But I'm really glad I asked, since some of the stuff I was thought early on via word of mouth seems to have been way off base.so yeah, I'll get back to you guys when I'm done work, and after I've googled some of the stuff you guys are saying, and experimented on some breadboard. I think I'll also make a little animation to visualize it, just so you guys can check to see if I have it rightThank you guys so much!

author: Simon
author_rank: Unknown
timestamp: December 19, 2014, 10:22:49 pm
content: Quote from: LvW on December 19, 2014, 04:11:33 pmI hate to rain on you but YOU have made this into a more complicated question that need be and we seem to be having a mud slinging match over use of terminology. Granted i have not be practicing for decades and i have no peices of paper to show but i have read enough BJT explanations to become fairly bored and I think we are confusing here the actual functionality of a BJT with how it is MADE to operate in a circuit.The BE junction is a diode and as a diode and LED has a certain voltage to current relationship. I don't know about this particular diode but LED's drop their voltage requirements ("resistance") when they heat up and if a transistor does the same you have a piss poor controllable device. Sure you can feed less than the junction voltage and control it but show me the millions of applications where this is done, I think we are confusing properties of BJT's with properties of circuits containing them.I'm sure the OP is extremely confused now as he/she never asked for such level of physics that take us to splitting the aton if that is really what your on about.

author: Simon
author_rank: Unknown
timestamp: December 19, 2014, 10:29:14 pm
content: Quote from: PlagueDoctor on December 19, 2014, 10:01:12 pmSo are you:1)working precisely with the few mV of range that get this sort of operation (which implies your much more knowledgeable than you make out)2) been blowing BJT's up, surely not ?3)putting a resistor in series and not realizing that your using ohms law to convert your so called voltage input into a current input ?I'll go with 3!

author: PlagueDoctor
author_rank: Contributor
timestamp: December 19, 2014, 11:17:22 pm
content: Quotenone of the above really.1: although I do like to measure to mV I will not claim to posses suck knowledge. but this is the type of operation I've been getting (I work in frequency ranges of ~ 1hz or slower normally, since I don't have an oscilloscope, and tend to use a diode to test the logic of my circuits, and need to be able to see the data with my naked eye.2) I have been using transistors, I check the data sheets every time before I use any component to make sure I can supply the required voltage/current without blowing anything up. and looking at the symbol/name is the first thing I check, to make sure I didn't mix up my parts :p3)no :p when I do that I do it purposefully for dividing voltage and such (I'm quite familiar with ohm's law)I've essentially been using existing circuits to make different logic gates that I didn't have on hand(like those seen herehttp://electronics.stackexchange.com/questions/72187/resistor-values-in-transistor-logic-gates) , or to make H bridges to drive small hobby motors, with some small changes here & there where needed. and assembling them like Lego blocks.so far that has worked for me. though I won't deny I've had my share of problems, and I have no problems admitting that has to do with my lack of understanding, but I've always managed to get my circuits to do what I wanted them to do at the end of the day with enough tinkering and Googleing.anyways, I'm cutting this close, I got to go to work.. like.. now :p

author: jlmoon
author_rank: Unknown
timestamp: December 20, 2014, 01:10:32 am
content: Quote from: free_electron on December 19, 2014, 03:42:07 pmVery good explanation, I was wondering if any one was going to explain what really goes on with that depletion zone ...Opps recombination zone. Great job Free!

author: jlmoon
author_rank: Unknown
timestamp: December 20, 2014, 01:43:36 am
content: Quote from: PlagueDoctor on December 19, 2014, 08:08:26 amPlague,I have a couple of great books I suggest you reference if you have the time as well.1st. Chapters 1 - 4 , Fundamentals of Electronic Devices, David A. Bell; ISBN 0-87909-276-9I don't know the current version in print, but this is one of many basic books that shed some light on the subject.2nd. Highly recommend this one.... Mmodeling The Bipolar Tranisistor, Ian Getreu... ISBN unknownhttp://stores.lulu.com/iangetreuIf you can survive a little complex math, this is an invaluable source of information well worth the reading.Hope this helps,jlmoon

author: amyk
author_rank: Super Contributor
timestamp: December 20, 2014, 02:51:03 am
content: BJTs are so mundane that everyone feels they are qualified enough to talk about how they work.

author: mikerj
author_rank: Super Contributor
timestamp: December 20, 2014, 10:06:58 am
content: Quote from: LvW on December 19, 2014, 04:11:33 pmA bipolar transistor can quite legitimately be viewed as either current controlled or as a transconductance device. Both models have advantages in different situations, and to suggest that one or the other is somehow incorrect is ridiculous and undoubtedly confusing to beginners on the forum who are trying to learn.

author: LvW
author_rank: Frequent Contributor
timestamp: December 20, 2014, 12:23:04 pm
content: Quote from: free_electron on December 19, 2014, 09:20:10 pmThank you. (Shouldn`t we try to treat each other with some respect?).Quote from: free_electron on December 19, 2014, 09:20:10 pmI am afraid you have overlooked that I never didn`t deny the existence of a base current Ib. Voltage control is not equivalent to assuming Ib=0.By the way, are universities (Stanford, Berkeley,..) - in your view - also producing "nonsense" regarding this question? And what about the design-oriented "Art of Electronics"?Perhaps you should expand your horizon a bit (excuse me, but this is my answer to "nonsense")?______________________________________Gentlemen - I am afraid, there is afundamental misunderstandingleading to a discussion style, that - unfortunately - in some parts is unsatisfying and disappointing. Shouldn´t it be possible to exchange technical arguments without being polemic?Let me explain:I think, some of you didn`t discriminate in their answers between the two activities: (a) Designing a BJT stage and (b) Understanding the working principle of the BJT.For designing an amplifier stage (i.e. common emitter) we all use thesame design principlesand thesame equations.There is absolutely no difference - whether we rely oncurrent-controlor onvoltage control.And - of course - for designing the resistive voltage divider for base biasing I take the base current into account (Ib=Ic/beta).But this has nothing to do with theprinciplehow the BJT`s collector current is controlled.And - as far as the "current-control party" is concerned - don`t they use the quantity Vbe=0.65...0.7 volts in their calculation?And, of course, they again are using other Vbe values for clas-A/B or class-B operation. Why? Why do they not start with Ib?Do you know the meaning of the Temp-Co of the base-emitter voltage (-2mV/K) to keep the Ic value constant?Is there any similar figure which describes how beta=Ic/Ib depends on temperature?In short: Everybody is using the same equations - the whole design process is completelyindependenton the question "current or voltage-controlled".However, this does NOT mean that we have two models which explain theworking principleof the BJT. Of course, as for all other electronic parts, there can be only one single explanation.And - in response - to one reply: This is not a chicken-egg (voltage-current) problem. Of course, it is always a potential difference (voltage) which drives a current.No current without voltage.Of course, for designing circuits we make use of the current source concept (it is a concept! only) but we never should forget if we are using a "concept" or a "model", that it not necessarily reflects the physical reality. That`s good engineering practice.I hope, this helps to clarify things.Thank you.LvW

author: dannyf
author_rank: Super Contributor
timestamp: December 20, 2014, 12:33:56 pm
content: QuoteYes, exclusively on your part.QuoteIt has been explained to you then, multiple times, and it has been explained to you here, multiple times, whether a bjt is current or voltage controlled has no bearing how a circuit around such a device is designed.QuoteNone of that has any bearing on this discussion.You may want to remember next time that quoting irrelevant facts or terminologies serve no purpose in a discussion other than to undermine your own credibility, especially when such quotes showed your complete misunderstanding of the quoted items, or your "facts" are erroneous.Hope it helps.

author: wazzokk
author_rank: Contributor
timestamp: December 20, 2014, 12:43:47 pm
content: Nobody knows at the deepest level how physics really works, all we have is various models of how materials behave.What matters is which model is appropriate to our requirements.Dave

author: atferrari
author_rank: Frequent Contributor
timestamp: December 20, 2014, 12:47:54 pm
content: The question in the OP:Quotewas answered in post #2.QuoteBasically, the OP (also known in some forums as the TS - thread starter), got the answer.Done.

author: Zero999
author_rank: Super Contributor
timestamp: December 20, 2014, 01:45:36 pm
content: Quote from: LvW on December 20, 2014, 12:23:04 pmNot true. If the resistance is zero, as in a super conductor, current flows without any potential difference.

author: Simon
author_rank: Unknown
timestamp: December 20, 2014, 01:53:23 pm
content: Quote from: LvW on December 20, 2014, 12:23:04 pmIt clarifies nothing other than your stubbornness! in an ideal world Vbe would be 0V but it's not an ideal world so we have to take into account that there needs to be 0.7V added just to make the thing work. If you want to discuss particle physics go and join a forum that specializes in that!

author: c4757p
author_rank: Super Contributor
timestamp: December 20, 2014, 02:10:19 pm
content: Quote from: Simon on December 20, 2014, 01:53:23 pmwtf...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 20, 2014, 02:38:12 pm
content: My response to the subject:Transistor: The Base PinSpaceballs: The Transistor Base

author: LvW
author_rank: Frequent Contributor
timestamp: December 20, 2014, 03:46:54 pm
content: OK gentlemen - I resign myself.I still think that we are in „A Free & Open Forum For Electronics Enthusiasts & Professionals“ but I don`t like to further respond to „arguments“ like*nonsense.*  Yes, exclusively on your part.* ...quoting irrelevant facts or terminologies serve no purpose in a discussion other than to undermine your own credibility, especially when such quotes showed your complete misunderstanding of the quoted items, or your "facts" are erroneous.* It clarifies nothing other than your stubbornness!I am in the age of 73 - and I still feel able and willing to learn.But I am not sure if this is true for all of you who have answered as quoted above.I came to this conclusion because - up to now -nobodywas able or willing torespond technicallyto the arguments and examples I have mentioned in favour of voltage control (Re-feedback, dVbe/dT=-2mV/K, class-B biasing,...).If I am totally wrong - are you really not in the position to show (with objective technical arguments!) where/why I am wrong instead of starting personal attacks?(But I can imagine what`s behind all this).Final remark (surprising observation):In my life I had, of course, many discussions about technical details - in most cases on a fair and respectful basis.However, as soon as the question is discussed how the BJT is controlled (because of conflicting explanations in different books/articles) very often there is a surprising change:Some people stop exchanging technical arguments and treat this matter as a kind of „religion“ - claiming that they have learned decades ago the current-control property and - hence - this would be the truth. Full stop. Without any justification and without any counter argument against voltage-control.Interesting observation.

author: Simon
author_rank: Unknown
timestamp: December 20, 2014, 04:00:25 pm
content: If you talk amplifier circuits then yes you talk voltage because the feedback system uses voltage. but the transistor is a current controlled device. I have never heard it described otherwise. As soon as you stick a resistor in a BJT circuit you control the circuit (not the BJT) with voltage.

author: kxenos
author_rank: Frequent Contributor
timestamp: December 20, 2014, 04:14:13 pm
content: This thread has become so embarrassing to read. You can't talk to a 73-old university professor like that. It doesn't matter if he's right or wrong, there is a better way to respond. I'm out of this thread

author: dannyf
author_rank: Super Contributor
timestamp: December 20, 2014, 04:29:31 pm
content: QuoteIt is, except that it requires participants willing and able to listen. You have demonstrated neither.QuoteThose are examples that have zero bearing on if a bjt is current or voltage controlled, a theme that has been repeated multiple times for you.Take the tempco question for example, contrary to popular believes, it is NOT a constant. We often use a constant to approximate it but in fact it varies with other factors, including temperature itself.As to emitter / source resistor: it is always "current" feedback, regardless of if a current or voltage controlled device is used. IE., it would be current feedback even if a fet is used.QuoteYou are totally wrong. Others have pointed out where and why. You don't realize that you are totally wrong because you are not willing to listen to the facts presented by the other side, and you insist on asking people to answer questions that have nothing to do with this discussion, which is what frustrated people in this discussion.As to your tendency of calling for authority, that's a show of weakness. Your arguments should stand on their own. The fact that some authorities somewhere else had cited it has no bearing whatsoever in a "technical" discussion.

author: LvW
author_rank: Frequent Contributor
timestamp: December 20, 2014, 04:35:05 pm
content: Quote from: Simon on December 20, 2014, 04:00:25 pmYes - and this is exactly the problem. Once and again, one should ask themselves if there is something to review.Earlier in this thread I have mentioned specialists from Universities (Berkeley, Stanford,...) as well as the well-know "Art of Electronics".I have offered corresponding references, but nobody was interested.In contrary, it was me who was qualified as "stubborn".

author: c4757p
author_rank: Super Contributor
timestamp: December 20, 2014, 04:37:11 pm
content: Quote from: Simon on December 20, 2014, 04:00:25 pmWell that settles that, Simon hasn't heard it any other way.For those of us whose headsaren'tup our asses - there are multiple levels to the function of a BJT, and you can look at it on any of them. This is why everyone - including some quite knowledgeable people, as well as some quite ignorant ones - is disagreeing about this. Voltage at the B-E junction causes current through the device, most of which overshoots to the collector, blah blah... it depends on how you look at it.Of course, let's not let that get in the way of a good argument.

author: LvW
author_rank: Frequent Contributor
timestamp: December 20, 2014, 04:41:13 pm
content: Quote from: dannyf on December 20, 2014, 04:29:31 pmPerhaps you should refresh your knowledge about feedback and realize in which cases the input resistance at the base/gate increases due to an emitter/source resistance.Or do you even deny that this resistance goes up?By the way: Interesting observation. The S-G path of a FET receives a feedback current.This statement (current feedback for FET`s) shows your qualification for this discussion.

author: Simon
author_rank: Unknown
timestamp: December 20, 2014, 04:46:13 pm
content: Quote from: c4757p on December 20, 2014, 04:37:11 pmAnd we started with a very simple question begging a basic clarification and if I'm not mistaken we have gone to a whole new level a few levels above the normal one.

author: Simon
author_rank: Unknown
timestamp: December 20, 2014, 05:04:10 pm
content: So the evidence presented so far is this wishy washy document :http://www.vanderbilt.edu/AnS/physics/brau/H182/Theory%20of%20Transistors.pdfthat goes into very little detail.If by voltage controlled it is meant that there has to be a voltage difference for anything to happen then that is a poor basis as we can argue the opposite that you can have all of the voltage you like but no current and of course still nothing will work in your model or real life.If I have read right people are saying that the art of electronics supports "the voltage theory" well I've just skimmed through my copy and all formula's that involve voltage also involve resistance values in a circuit!Of course voltage has a part to play but it's not strongly linked to BJT basic behaviour other than the obvious requirement for a voltage in order for current to flow.

author: dannyf
author_rank: Super Contributor
timestamp: December 20, 2014, 05:06:58 pm
content: QuoteIt reminds me of "You want the truth? You can't handle the truth!" - that's a great movie.You keep asking those tangential questions and insisting on others saying things they didn't say.In your case, you want a "technical" discussion but you cannot handle a "technical" discussion. You don't have the ability to comprehend the technical details, nor do you have exhibited any willingness to understand what's being communicated to you.

author: c4757p
author_rank: Super Contributor
timestamp: December 20, 2014, 05:07:16 pm
content: The voltage applied to the BE junction is whatcausesthe current to flow; this is what people mean by voltage-controlled. Some consider the "control" to be what is applied to the pin; some consider it to be what the charge carriers do inside the silicon. Neither of them is wrong, until they start claiming everyone else is.

author: LvW
author_rank: Frequent Contributor
timestamp: December 20, 2014, 05:21:29 pm
content: Quote from: dannyf on December 20, 2014, 05:06:58 pmDear forum memberdannyf., if you are fair you will have noticed that I have - earlier in the thread - in detail explained my example, but I have no problems to repeat it here again:1.)First statement: The input resistance of an amplifier goes up if avoltageis fed back to the controlling inverting node and it goes down if acurrentis fed back. (These are basic rules from feedback theory).2,)Observation: The input resistance of a BJT/FET goes up in case we apply negative feedback using an emitter/source resitance.I kindly ask you to answer two short questions:a) Are both points 1) and 2) above, correct or not?b) Can we derive from the observation in 2) any information about the question if a voltage or a current is fed back to the inverting node?Are these questions "technically" enough?I am awaiting your answer.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 20, 2014, 05:23:56 pm
content: Quote from: Simon on December 20, 2014, 04:00:25 pmSimon please. I know and understand the underlying physics. I can't quote the equations verbatim anymore, but I can assure you, beta drops out as a side-effect of the voltage control.If you still don't believe, try this circuit: logic level input, series diode, series current limiting resistor to base. Emitter to ground, pullup to collector. Measure turn-off time.If it's current controlled, then as soon as base current ceases (without reversing -- the diode proves this), the collector current should cease. But it doesn't, it takes quite a long time to change.Whereas, if you drive it in a voltage mode (such as from a low impedance (but obviously not zero ohm) source, or another diode junction), it goes quite quickly. The usual approach being a simple B-E resistor to discharge base voltage on turn-off. As in the current mirror. Which uses one diode-strapped BJT (note again, base current is tiny, collector current accounts for alpha times the total current) to drive the B-E of another transistor (which unless it's saturated, also draws little base current).Tim

author: IanB
author_rank: Super Contributor
timestamp: December 20, 2014, 05:45:24 pm
content: Further to what T3sl4co1l says, the currents through the device are controlled by charge accumulations in various regions inside it. An accumulation of charge is somewhat akin to a capacitance and the amount of charge varies with voltage. The base current is in a sense a parasitic current that occurs due to the way the device is built. You could wish it wasn't there, and in an FET that wish is granted.

author: Zero999
author_rank: Super Contributor
timestamp: December 20, 2014, 07:58:32 pm
content: Quote from: T3sl4co1l on December 20, 2014, 05:23:56 pmThat's the storage time. You'll only notice a difference, if the transistor is driven into saturation. If the series current limiting to the base is high enough, then the transistor will not go into saturation and will turn off quickly, even with the diode.Another way to ensure it doesn't saturate is to with a Baker clamp. Add a diode with its cathode connected to the collector and the anode connected to the anode of the diode in series with the base. When the transistor starts to saturate, current is diverted away from the base, via the collector to the emitter, bypassing it.

author: Simon
author_rank: Unknown
timestamp: December 20, 2014, 08:18:22 pm
content: So we are still back to driving a voltage through a resistor to create a current. Of course the hfe does change with different currents, as far as I know large collector currents (and therefore large base currents) cause the hfe to go down and I'm sure there will be other effects, that is the whole point of the feedback circuit and then people start claiming that the properties of the whole circuit are those of the BJT on it's own.I rather get the impression that the illusive points and evidence that has not been properly explained amount to splitting hairs in the practical realm of electronics and we are not really here to look at physics and the original question was certainly not asking for it.

author: c4757p
author_rank: Super Contributor
timestamp: December 20, 2014, 08:19:26 pm
content: Quote from: Simon on December 20, 2014, 08:18:22 pmI get the distinct feeling that you have no idea what the rest of us are talking about.

author: Simon
author_rank: Unknown
timestamp: December 20, 2014, 08:37:00 pm
content: Well maybe I'll find out one day

author: Tandy
author_rank: Frequent Contributor
timestamp: December 20, 2014, 08:56:07 pm
content: To be fair I have always understood thermionic valves as being voltage controlled and BJT to be current controlled subject to the required voltage to cross the junction. Clearly this received wisdom is not universally accepted and is causing a somewhat heated debate.In all situations we are dealing with models that best represent the component operation and these models can never fully represent the physical properties of the component. Certainly during my education the current controlled model was taught so that is what I have used in my own work. It seems to have served me well over the years but I am always open to expanding my understanding where possible.You will see that I have stayed quiet in this debate hoping to learn something from the debate but I feel it is rather adversarial and confusing. It would be interesting to know how these opposing opinions come about, is it due to limitations in the models used or a misunderstanding somewhere? I would appreciate more detailed explanations and links to references where possible because I am always happy to learn something new, even if that proves wrong something I have understood as being a fundamental principle all this time.

author: Simon
author_rank: Unknown
timestamp: December 20, 2014, 08:59:26 pm
content: Quote from: Tandy on December 20, 2014, 08:56:07 pmAs I said i have never heard of this alternative explanation since I saw this thread.

author: IanB
author_rank: Super Contributor
timestamp: December 20, 2014, 09:08:51 pm
content: Quote from: Tandy on December 20, 2014, 08:56:07 pmThere has been very little actual theoretical content in this thread (such content would have to contain an examination of physics, device models and equations). Here is a reference that goes over the theory in some detail:http://ecee.colorado.edu/~bart/book/book/chapter5/ch5_1.htmIn physics influences are usually conveyed and expressed in terms of fields. In the electrical/electronic domain we have electric fields and magnetic fields. Electric fields are produced by voltages and magnetic fields are produced by currents. Inside a transistor the various effects are mediated and regulated by electric fields produced by accumulations of charge in parts of the device, and thus by voltages. For a device to be controlled by current, we would have to find magnetic fields mediating the actions of that device. Magnetic fields are not significant in the operation of a transistor and therefore current in a transistor is not a controlling cause, but rather is a derived effect, or an outcome.This is not to say that simplified models that assume base current to be a controlling cause are not useful, but such models are just that: models that make simplifying assumptions to suit certain purposes.

author: dannyf
author_rank: Super Contributor
timestamp: December 20, 2014, 10:31:26 pm
content: QuoteModels don't create the underlying mechanism - which is the topic of the discussion. Instead, the underlying mechanism determines the models used.If you look at the bjts, it is all about carriers (electronics and holes = negative charts + positive charges if you will) -> ie, it is a current - driven mechanism.For the fets, it is about the electronic field that makes the majority carriers difficult to go through -> it is a voltage-driven mechanism.All the discussion about tempco or diode equations or circuits are nonsensical and irrelevant for this discussion.

author: free_electron
author_rank: Super Contributor
timestamp: December 21, 2014, 04:13:20 am
content: Quote from: wazzokk on December 20, 2014, 12:43:47 pmnot true. There exists equipment that can actually track electrons flowing in material (e-beam probing ) . IBM even has a machine that can arrange individual atoms to form the IBM logo... they have made a twenty atom transistor with it. and yes, it does work. they can track the electrons flowing. not simulated. measured.e-beams are frequently used to sniff out leakage in integrated circuit design as you can count individual electrons.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 21, 2014, 04:14:14 am
content: Quote from: Hero999 on December 20, 2014, 07:58:32 pmYes, storage time is a huge part of it, in saturated mode. The collector can sit there drifting fortens of microseconds, even though the base terminal current is absolutely zero!QuoteThis helps, but helps the most as a combination approach.Just now I set up a circuit which I expect will meet your approval: the schematic as shown above, with 1k base resistor, 0/5V square wave input, a pair of 1N914, a 2N3904, and a 220 ohm collector load (to a +12V supply).The waveform: well darn, my camera battery is low, so I'll have to describe it. Anyway,On the input falling edge, collector voltage does begin to rise immediately (the Baker clamp is indeed doing its job). Starting from "saturation" (about 0.8V), for the first 270ns, it rises approximately linearly, to 4V. Then the decay is approximately exponential, with a time constant of about 300ns. The 10-90% rise is 680ns.In the same time frame, base voltage barely falls at all, starting around 800mV and only falling to 600mV after a whopping 2.5us. Given the Ebers-Moll equation says 60mV/decade, that means the collector current has only decreased by 2000x in that time, or reaching 25uA. While that's quite a bit more "off" than it started, it still has a seriously long way to go before being truly cut off (~nA!).Whereas if I simply add a B-E resistor of 1k, the input-falling propagation delay is 110ns, collector rise time is 85ns and base storage time is 160ns.Still not convinced it's voltage not current? Come on, you people are hopeless...Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 21, 2014, 04:21:38 am
content: Quote from: free_electron on December 21, 2014, 04:13:20 amNot only is electron science in general quite well understood, semiconductor physics itself has been quite well understood for much of the last century, with the advent of quantum mechanics.In fact, the field-effect transistor was the first to be theoretically evaluated (back in the '30s I think), but it had to wait for materials science and process engineering to catch up (because of somewhat less well-understood theoretical aspects, which were nonetheless figured out later, which eventually factored into the development of practical commercial devices).The BJT is fairly complex, theoretically speaking, but it was predicted and engineered in a quite reasonable manner. It took until Bell Lab's famous model for a practical design to arrive, though triode type semiconducting devices were known before then (for instance, if you form two point-contact diodes very close together, you get majority charge carrier diffusion between the two junctions, and it acts as much more than simply two diodes on the same substrate!).Tim

author: free_electron
author_rank: Super Contributor
timestamp: December 21, 2014, 05:23:08 am
content: Quote from: Tandy on December 20, 2014, 08:56:07 pmexcellent reference. he really throws in the kitchen sink ( including things like 'early effects' and all the other oddities most people have never heard about)now to come back to the debate.for the sake of clarity can we define a few things ?- an electron has a charge- the core (nucleus) of an atom also has a charge- for a condustor ( and most atoms): the charge of the electrons cancels out the charge of the nucleus so the net charge is zero.- a moving charge is a current .- the number of charge carriers per second is a quantity we call ampere.right ?now, before we delve into transistors let's take a look at something more simple.: an electric conductor , simplified, a wire. we will start with an ideal wire first ( no resistance)A wire is made from a material with movable electrons. a wire' in rest ' has enough electrons to fill the outer shell of all the atoms in the wire.a wire in rest has no charge across it. There are charges moving internally as the electrons spin around their nuclei but, since the sum of the electron charge is equal to the sum of the nuclei charges the net effect is zero. it all cancels each other out.now, let's push in an electron on side of the wire shall we ?what happens : we unbalance the system. there is now an extra, unbound electron, in this stub of wire. so something has to give... none of the existing atoms want anything to do with this electron. so , as soon as there is a pathway for it to be shoved out of the wire it will be.close the loop and the wire will spit out an electron for every electron shoved in the other end.as electrons are charge carriers and moving charge carriers is called 'current '.... you got the idea right ?of course in order to get an electron to move we need to apply a field.now. step two. let's take a resistor. ( or , a not so ideal wire if you would like )the principle still stands : one electron in is one electron out. electrons are particles , matter. laws of physics state you cannot create or destroy matter. ( unless in nuclear reactions). if i send 20 electrons in that resistor , 20 will come out. none extra are created and none are destroyed in this process.Electrons are accelerated by the applied electric field. strenghten the field ( up the voltage) and you up the acceleration. Electrons only move at the speed of light in a vacuum. in material the run slower.so an electron entering the resistor is accelerated by the applied field. so it gains kinetic energy. as it starts traveling into the wire it will sooner or later hit an atom and lose some of its kinetic energy. enrgy also cannot be created or destroyed , only converted. The electron is still propelled by the initial push it got from the external field so it keeps going towards the exit , all the while slamming into several atoms and losing some of its kinetic energy to the atom. this causes the atoms of the wire to jiggle faster thus producing ... heat !but, eventually all twenty electrons sent in , will exit as none of the atoms int he resistor want them ! their charges are in balance.In short we have been dealing with materials called 'conductors' : materials where the charges are in balance. Every proton has an electron so the net charge is zero. shoe in an electron and it needs to come out.Now it gets complex :Semiconductor.a semiconductor is a material where the charges, for a single atom, are NOT in balance. there are not enough electrons to satisfy the number of protons (sometimes called holes) in the core.Now, atoms are pretty clever little buggers and, in an INTRINSIC semiconductor, they will arrange themselves in a crystal like lattice so they can 'share' some of the electrons on their outer shell. As the electrons are spinning they no longer run around a single core but actually run in figure eight patterns around two cores. so now the atom cores 'see' enough electrons.if you take a such a material and shove in an electron on side , one will roll out the other side. in an intrinsic semiconductor material the charge is not balanced absolutely ( if you were to count the number of electrons and number of protons there is a mismatch ) , but on a per-atom level they are balanced. as each atom has enough electrons for its number of protons. just not all the time as the electrons keep circling mulitple cores. note : this is heavily simplified. in reality it is more a brownian motion , 'electron soup' if you will.a lump of intrinsic semiconductor material has no net charge and behaves like a wire. shove an electron in and one will come out. there is no need for the extra electron. the atom pairs sharing have no use for it. Sure, one of the two atoms sharing a single electron could absorb it and be content but then the other atoms is left with an electron missing so it is going to seek another sharing partner. Resulting in there being now one electron too many in the newly formed sharing couple. So that electron is spit back out.Now we are going to alter this intrinsic material by shoving in impurities. called 'doping'. we can shove in a material that has spare electrons or a material that doesn't have enough electrons.doping the intrinsic semiconductor starts an electron exchange process.if you feed extra electrons, one of the paired atoms will absorb it temporarily causing its bonded twin to be unhappy now and seek another partner. three is company .. , also in electron land and The newly partnered atoms now have an excess electron being spat out . This process goes on and on. These 'traveling electrons' hop from bonded atom pair to bonded atom pair and are called free electrons. this is N doped materialif you feed a material with shortage of electrons ( holes ) the same happens, only now does the 'vacant spot' ( the hole) travel around. the lacking material snatches an electron from a bonded atom pair , one atom in the bond is now unhappy and goes in search for an electron it can borrow from an adjacent pair. causing a marital rift in that pair , divorce and another atom going in search for an electron ... so the 'vacant spot' also travels.now. if we take such a doped piece of material and we shove in electrons things happen differently. i forgot the metallic - oped semiconductor process operation. it is the shottky effect ( a shottky diode is essentially a diode made from one doped material ( p material) and a simple conductor not relevant for this discussion. i;d have to read up on that again.but , if we take a lump of P material and a lump of N material and we slam them together something happens. in the contact surface the free electrons from the N material will 'fall' in the holes of the P material making atomic bonds (the recombination zone) . since in that region there are now no mobile charge carriers, this is effectively an isolator !now, if we apply a field to such a diode. , in the right direction , we will be able to send electrons into it , but none will come out the other side ! you will need to keep increasing the external field until you hit the point where the recombination zone is gone. essentially this zone travels in the material. add some electrons in the N material by applying a field and they will push the recombination zone to the exit ( other end) once you hit enough potential the recombination zone is gone and electrons flow freely. this potential is called the forward voltage. ( 0.6 volts in typicla silicon diode) and is depending on the levels in the valence bands of the used material.now. if we make a bipolar transistor we get two such recombination zones. by applying pressure to the base-emittor (biassing) we get that thing to go in conduction. now we have established an electron flow there. ( a current). that electron flow now pushes out the second recombination zone and allows collector current to flow. ( i have described that process already once in detail here on the forum, i'm not going to repeat it here )change the intensity of the emittor-base current and the intensity of emittor-collector current changes. ( electron model currents )so , to put the dot's on the 'i': for a properly biased bipolar transistor ( bipolar as it used both electrons and holes as charge carriers) the collector current has a relation to the base current. once base current flows, both recombination zones are gone and collector current flows. send in more electrons to the base and more will flow toward collector. depending on the strength of the doping there is an amplification factor.ergo : such transistors are current controlled.internal material resistance cause losses, so you get all kinds of side effects such as temperature dependency of certain factors . there is also charge accumulation due to uneven material properties ( microcracks in the lattice , uneven distribution of the impurieties etc ) so you get all other kinds off effects. if you flood in enough electrons to completely collapse the recombination zone you are in 'saturation' and then other effects kick in as well. these are well understood and described in various models and equations. but none of that erases the base principle :make base current flow to get collector current. increase base current to get more collector current. so : current controlled current ampliefier.now do you get it ?

author: c4757p
author_rank: Super Contributor
timestamp: December 21, 2014, 05:57:37 am
content: Quote from: free_electron on December 21, 2014, 05:23:08 amThis goescompletelyagainst every explanation I've ever seen of PN junctions, where the depletion region is constricted inward from both sides, not pushed out one end... how do you plan to "send electrons in" without removing electrons from the other side at the same time?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 21, 2014, 07:18:58 am
content: Right, if nothing's coming out the other side, you have a capacitor. Electrons pile up until the voltage is equalized against the source. Which is an approximate model of a diode at V < Vf (including negative, i.e. reverse bias).Although the depletion region does shrink under forward bias, the usual story given is that the applied voltage skews the electronic band gap diagram; in particular, when Vf > E_g / q_e, the diode is in forward bias. This causes holes to leak into the electrons region, and vice versa; electrons and holes are freed purely by thermal means (minority carriers), so there is a statistical population of charge carriers (number and energy), which is where the exponential characteristic arises from in semiconductors.Which is why, although a depletion region can be approximated as a capacitor, it's actually a very leaky capacitor, if at all. DC current flows across the junction, due to random carriers generated in the gap (reverse bias), or in bulk (forward bias; charge carriers move into the depletion region, helped along by the flow of current).If depletion region thickness were the only factor, one would expect quantum tunneling to be an important part: in fact, it is not, and the depletion region is much thicker than this characteristic length in most diodes. Diodes made with particularly strong doping (and therefore very low breakdown voltages and very thin depletion regions) do exhibit tunneling, and are called tunnel diodes.There's also something quantum about the Zener effect, which is a low voltage breakdown phenomenon, but I don't remember offhand how it works.Tim

author: LvW
author_rank: Frequent Contributor
timestamp: December 21, 2014, 10:34:30 am
content: Quote from: free_electron on December 21, 2014, 05:23:08 amIt`s just a claim and, more than that, not logical.Nobody has denied that there is a "relation" between both currents.We simply have two events (currents) caused by the applied voltage. Such things happen in our world (electronic, natural) from time to time.I have another question:How do you explain theEarly effectwith your current-control view without using the electric field (voltage) causing the Ic increase?(By the way: Did you forget to answer my technical question in post#61?)

author: wazzokk
author_rank: Contributor
timestamp: December 21, 2014, 12:28:37 pm
content: Quote from: free_electron on December 21, 2014, 04:13:20 amApologies for my sloppy description. I was trying to say that even electrons, protons, quarks etc. are still, I believe, a model for the behaviour of something, string theory maybe, we have yet to understand. And so we use models apropriate to our needs.Respectfully Dave

author: LvW
author_rank: Frequent Contributor
timestamp: December 21, 2014, 12:53:40 pm
content: Quote from: Simon on December 20, 2014, 05:04:10 pmHi Simon,excuse me but I totally have forgotten to answer your request regarding Horowitz/Hill.Here is an excerpt from chapter 2.09 "Transconductance":Clearly our transistor model [LvW: current-control] is incomplete and needs to be modified in order to handle this circuit situation, as well as others we will talk about shortly. Our fixed-up model, which we will call the transconductance model, will be accurate enough for the remainder of the book.And at the beginning of chapter 2.10:But to understand diff. amplifiers, log. converters, temp. compensation, and other important applicati0ns you must think of the transistor as a transconductance device - collector current is determined by base-to-emitter voltage.(This justifies my "stubborness" you have attributed to me.)

author: free_electron
author_rank: Super Contributor
timestamp: December 21, 2014, 05:55:10 pm
content: Quote from: c4757p on December 21, 2014, 05:57:37 amDon't get all upset about what i will state now, and i will go out on a limb here... Everything you know is wrong... Not your fault, nobody ever explained it to you the right way. They sketch a pciture but it is not complete. Even i don't have the full picture as i never really had to delve that deep. The people that make transistors do. I got what i know from them, some of that stuff i don't grasp but other stuff i do.So let me try to fill in a few gaps.A diode that is not in conduction (whether forward or reverse) is indeed a capacitor.Remember the varactor diode ? You bias that thing in reverse . Apply more voltage and the recombination zone grows thicker , thus capacitance goes down.A diode in forward also has capacitance. In this situation you apply a field , causing electrons to move to collapse the recombination zone. A field travels.. As charged particles . Charge carriers are electrons. Traveling charge is a current.To get a diode into conduction you need to send some electrons in. You can actually measure that.Take a voltage source of a few hundred millivolt, well below sending the diode in conduction. Take a very sensitive coulomb meter and you will see a current for a very short time enough to charge the barrier. It takes time for a dide to turn off as the internal charge needs to come out to allow the recombination zone to grow. Diodes have turn on and turn off times due to the capacitance !Other things like the miller effect are similar things.Look up 'capacitance of forward biased diode'http://web.ewu.edu/groups/technology/Claudio/ee430/Lectures/pn_mos_2.pdfSo back to the transistor : as long as you do not get a base current flowing no collector current flows.See what happens if you try to apply 10 volts to the base emittor of a transistor. It thermally self destructs. The 9.4 unnneeded volts cause such high pressure in the resistance causeing so many electrons to bang around that you destroy the pathway.Bipolars are current controlled. I stand by that statement.

author: c4757p
author_rank: Super Contributor
timestamp: December 21, 2014, 06:05:11 pm
content: But nothing you just said contradicted what I said!Quote from: free_electron on December 21, 2014, 05:55:10 pmRight, obviously.QuoteRight - now put that coulomb meter on theotherside. I'd bet serious amounts of money (if I had them...) that you'll see equal amounts of charge going out the other end, whether or not the diode is in conduction - hence the depletion region will be constricted inward by the symmetric change in charge, not pushed towards one end like you said.Quote from: free_electron on December 21, 2014, 05:55:10 pmIt's a bikeshed. Most of us aren't arguing about how BJTs work, we mostly seem to understand that. Whether we realize it or not, most of us seem to be bickering about which aspect of the transistor's behavior is thecontrol- something about which the charge carriers don't give a damn!

author: free_electron
author_rank: Super Contributor
timestamp: December 21, 2014, 06:43:23 pm
content: Ah, if you want to put it that way. What is the control ?Collapse all the equations and what do you get for a single transistor (not a system built around one) ?Ic = ib x beta.Now tell me what the symbol I represents.Connect the dots.And that's all i will say about that. This is an endless discussion leading nowhere. It's righ there in that equation.

author: c4757p
author_rank: Super Contributor
timestamp: December 21, 2014, 06:50:09 pm
content: Right, and people are arguing because there areotherways to collapse those equations!

author: mtdoc
author_rank: Super Contributor
timestamp: December 21, 2014, 07:00:14 pm
content: Interesting discussion and fun for an electronics newbie like me to follow. Lots to learn from the more educated and experienced posters when they disagree.The following seems apropos:

author: Zero999
author_rank: Super Contributor
timestamp: December 21, 2014, 07:03:43 pm
content: After doing some more research, I've realised no one here is right. BJTs are neither current nor voltage controlled devices. The truth is as BJT is a temperature controlled device!

author: c4757p
author_rank: Super Contributor
timestamp: December 21, 2014, 07:19:17 pm
content: Nuh-uh, they're obviously light-controlled. What, are you people leaving themin the box?

author: free_electron
author_rank: Super Contributor
timestamp: December 21, 2014, 07:27:30 pm
content: Quote from: Hero999 on December 21, 2014, 07:03:43 pmnonsense. they are smoke controlled. once the smoke escapes it's game over. that is why we have clean rooms. we take all the smoke from the room and deposit in the transistor package. that is why that room is so clean. oops. now the dirty secret is out ... shouldn't have said that ... the semiconductor police is now out to come and get me . they will take me a way to the funny farm where i will have tea with unicorns until eternity. damn !

author: free_electron
author_rank: Super Contributor
timestamp: December 21, 2014, 07:45:53 pm
content: Quote from: c4757p on December 21, 2014, 06:05:11 pmyou will see a charge displacement but the current in will not be equal to current out. some electrons stay behind to bind with the atoms to erase the holes. once the thing goes into conduction electrons in will equal electrons out. turn off the diode and current flows a bit longer than the point where it stops conducting. those are the electrons now being released to reform the recombination zone.Now, as for the 'symmetrical' that is not right . it depends where the electrons are on the shell to knock em off. or seed them in. i can't remember it exactly but the layer shrinks assymetrically. i depends on the dopant used and how strong the bonds are. where they sit in the periodic table has an impact. boron, antimony, phosphorus and arsenic are the traditional dopants for silicon or germanium processes. you need a materials chemist to tell you how that stuff happens.and then that is only theoretical. there are flaws in the crystalline structure of the semiconductor. during implantation we shoot the cristalline structure to shreds. part of that is repaired during the drive-in process step. after implanting the wafers go in an oven. this does two things : it drives the dopant deeper in the wafer vertically and it heals the cristalline structure.the deposition in vertical is not uniform. there are more dopants per cubic micron at the surface than 2 micron deep. so the deep regions actually kick in first as you need more charge at the surface to break down the zone there.area wise (seen from the top ) the doping is pretty uniform as the wafer spins through the beam thousands of times when being implanted. vertically it is not.so that barrier does not dissolve symmetrically.on a transistor the barrier does shift. keep in mind that we are dealing with layers that are a few thousand atoms thick. i think 10.000 atoms is required for a micron. that base layer isn't a micron, it's less. and the whole recombination zone isn't the full width of the base either.the distance between the b-e barrier and the c-b barrier is very small. the zones are built-off asymetrically. the be zone collapses first. it needs to as it hinders the c-e connection. once the b-e zone is gone electrons shoot from the emittor to the collector ( electron model) and start collapsing the recombination zone there.so once again : it is the electrons flowing from emitter to base region that make electrons from emittor shoot into the collector.think of the damn thing as a venturi. stop the electrons fro coming out of the base and none will come out of the collector.so again : current controlled. no traveling electrons , no amplification. sure you can blabber that you need a field first to make them move in the first place , but you can put a million volt9provdied the part doesn't flash over) at the emittor , if the base does not emit a single electron you have no current there. , nothing out of the base is nothing out of the collector.a mosfet is different : there you do build a static field in the part , that attracts charge carriers on the other side of the barrier ( the gate isolation ) and then you can send electrons through that pile of atteacted carriers . so mosfets are voltage controlled ( they have current flowing in and out changing state from field to no field and vice versa. ( on and of )bipolars are current controlled.

author: IanB
author_rank: Super Contributor
timestamp: December 21, 2014, 08:09:11 pm
content: Quote from: free_electron on December 21, 2014, 07:45:53 pmIt is one of the most fundamental laws of electricity that current in equals current out. The is Kirchhoff's current law. I really do not think that semiconductor devices violate this law. If you measure the currents at all the terminals of an electronic device the sum is going to be zero.

author: c4757p
author_rank: Super Contributor
timestamp: December 21, 2014, 08:12:12 pm
content: To be fair, that fundamental law is more properly that current in equals current out + current stored - but I'm still not convinced. This one'll take some consideration...I'll give him that the depletion region doesn't shrinktrulysymmetrically, this will obviously be subject to all sorts of variations. But it sure as hell doesn't get pushed out one end.

author: free_electron
author_rank: Super Contributor
timestamp: December 21, 2014, 08:31:05 pm
content: Quote from: c4757p on December 21, 2014, 08:12:12 pmfundamentally it is electrons out = (electrons in - electrons stored). moving electrons are current. stored (non-moving) electrons are charge. potential energy <-> kinetic energy.when i say ' pushed out ' this is in reference to the transistor. in a diode that region sits at the junction of the n and p material. thats why we call it the junction.in a transitor the two regions sit very close together , they are really only separated by the width of the base.so when you start deconstructing the b-e junction that barrier becomes thinner. as it shrinks asymetrically it has the perceived effect of shifting towards the base and away from the emittor. once gone, the electrons flow freely into the base region. as the base is doped more weakly than the collector , and the potential at the the collector is (typically) higher ( unless you run in inversion mode) some electrons will happily run into the base, others , which are mechanically further away from the attraction of the base potential , will shoot into the collector.in simple words : the collector is a wider pipe which is easier to traverse than the base pipe. and the 'pull' from the potential at the collector is higher than that of the pull from the base.so if you want to do the 'voltage controlled' thingy it would actually be the collector that has control... is it has a stronger pull.but it all comes back to the fact that no base current is no collector current, irrespective of what voltages you put there. if you dont succeed in collapsing the b-e barrier and get a current to run there, then no current runs in the e-c path.no current flow in b-e is no current flow in c-e. ergo : current control.

author: Syntax_Error
author_rank: Regular Contributor
timestamp: December 21, 2014, 08:43:06 pm
content: This is a very interesting discussion.free_electron, if I understand your posts correctly, the following seems a logical extension of what you are saying:As the E field pushes charges into the diode substrate, the diode should technically become charged during forward biasing operation, i.e. not strictly neutral. Can this be measured with either an electroscope or some sort of E field meter?

author: free_electron
author_rank: Super Contributor
timestamp: December 21, 2014, 09:08:12 pm
content: Quote from: Syntax_Error on December 21, 2014, 08:43:06 pmyes. this can be measured (i'll show how in a moment) . the diode is charged BEFORE it goes into conduction. it behaves like a capacitor until conduction. the perceived effect of this is the turn on time of a diode. you apply a field , current starts flowing in , but nothing comes out until the barrier is charged enough so it becomes conductive. then stuff comes out. that delay between current in and current out is directly proportional with the physical size of the diode. make a big bulky diode of 10 square meters and it will be slow as molasses. make it a square nanometer and it will be lightning fast. why ? capacitance ... (what we call capacitance is charge storage. we defer electron flow to a later point in time )same goes for turning off the diode. stop cramming electrons into a diode that is in conduction. the current flows a bit longer at the output. where do those guys come from ? they are released from the junction . so the recombination happens , electrons are spat out and the barrier is recreated.you don''t need special equipment to measure this.two current probes , a scope and a pulse generator. look at the phase shift of the currents. if you dont have current probes : a diode followed by a resistor. measure vin and voltage across resistor. you will see the shift. That is the turn on and off time of the diode. cause by charging and discharging the barrier.How much is that charge you ask ? measure the drop across the diode, look a the turn on time you can find out how many coulombs... does the coin drop now ?there are of course some non linear effects due to material physics and resistive losses and temperature making stuff jiggle faster or slower.No laws of physics are broken as , integrated over infinite time, electrons-in = electron-out of this system. there is just a temporary hold on some of them to destroy/reconstruct the barrier.that 'electron storage' is seen as the phase shift in current. kirchoffs laws apply only to effects integrated over time. that is why you can use them for ac signals with impednce like capacitors and inductors. look at them frozen in time and they make no sense as they do not account for stored charge. you need maxwell for that crap ...Kirchoff only works on conducting diodes. not to look at transient behavior. ( actually it does but you need to track every damn electron as it travels through material at infinitesimal time fragments.)i'm not pulling this stuff out of my ass you know. this is how these things work. It's all really very simply if you think about it. problem is we are always being told half-complete stories. i know we have to start somewhere , but i feel too much is left out in classical literature to explain it in enough detail so it makes sense. later on you can slap on the -very complex- mathematics.

author: Syntax_Error
author_rank: Regular Contributor
timestamp: December 21, 2014, 09:50:26 pm
content: I don't think anyone's accusing anybody is pulling stuff out their ass. I think the various players here all genuinely think their understanding of things is the correct version, and I think to various degrees, they may all be. This sounds quite politically correct, but I'm not trying to make it that way on purpose. I think theres a bit of cognitive dissonance going on, and it is being handled in the most commonly human way: instant rejection of the new information that drives the dissonant feeling.free_electron, your information regarding fields and charge accumulation make perfect sense to me. I always wondered about this, since for the life of me I have never been able to understand how voltage and charge could ever be separate, unrelated quantities, based on my understanding of their definitions. In order for a node to have a different potential than another node, it's charge *had* to be different, or it's physical location within an external electric field would have to account for all of it's voltage. It makes so much more sense to me that the voltage (in terms of P/Q) of a node is due to the presence of charge, however small. I relate the quantity of charge to the voltage related to it as the capacitance. A diode with very small capacitance would accumulate very little charge for any given voltage. A diode with large capacitance would accumulate more charge for that same voltage, and conversely less voltage for the same charge.Edit: Not to imply that the forward voltage drop is due to accumulated charge. Rereading my post made it seem like this is what I was implying.

author: IanB
author_rank: Super Contributor
timestamp: December 21, 2014, 09:53:31 pm
content: Quote from: c4757p on December 21, 2014, 08:12:12 pmYes, I understand that, and in a monopolar device like a Van de Graaf generator there is a transient accumulation of charge. But the generator has a big metal dome and a current imbalance of a few microamps produces a voltage rise of thousands of volts.A P-N junction is a bipolar device with two terminals, like a capacitor, and it is on a totally different microscopic scale compared to a VdG. So sure, on some scale there may be a transient charge accumulation term, but we must be talking of scales like femtoamps and picoseconds. Being a bipolar device the minutest charge imbalance will produce a current on the other terminal.Is there a supporting reference somewhere for this?(To give something for comparison, consider an actual, big capacitor. Feed it with an AC signal and measure the current at each terminal. Can you measure a phase difference between these currents?)

author: free_electron
author_rank: Super Contributor
timestamp: December 21, 2014, 10:32:14 pm
content: Quote from: Syntax_Error on December 21, 2014, 09:50:26 pmkeep in mind that ,as the charge increases, the capacitance increases. in a standard capacitor this doesnt happen.a higher voltage, but below conduction decreases the thickness of the barrier. thinner barrier = larger capacitance for the same surface area ( capacitance is inverse of distance between the plates )so you may start of with a small capacitor but it actually increases as the barrier starts to collapse.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 22, 2014, 12:30:27 am
content: It's worth noting that Kirchoff's law (it's not a rule, despite being labeled as such in most references) is absolute. But if you want to niggle every little detail, you must understand the scope of that law.- Kirchoff's is absolute at DC.- At AC, it is instantaneous and pointwise. Currents can only be truly said to sum at a node if that node is infinitesimal. Alternately, if the speed of light is infinite.- Kirchoff's is local. If you consider a current flowing in a wire, but don't consider the transmission line nature of that wire, then you might lead yourself into a trap, where the two ends of that physical conductor apparently violate KCL. This is incorrect. When the wire is correctly modeled as a two port transmission line, then it becomes obvious that, at each end, you are measuring the current between the active wire and ground (or whatever it's physically closest to). The physical mechanism is, the current is being transported as displacement current in the electromagnetic field. Where the transmission line itself leads, is completely irrelevant -- which is one of the fantastic features of transmission lines, that they offer a path for signals to flow, independent of the environment nearby. (The real world isn't quite so fairy-tale, but transmission line transformers can be constructed which do an excellent job of this over a wide frequency range.)The proceeding conversations about the nature of charge and depletion regions is superfluous (you're arguing the wrong thing), because for any frequencies where we can reasonably discuss the nature of semiconductor devices (i.e., DC to ~GHz), the pointwise / infinite-speed-of-light model is sufficient. You push current into one terminal, the same total amount necessarily must flow out the others, period, no ifs, ands or buts.Tim

author: free_electron
author_rank: Super Contributor
timestamp: December 22, 2014, 12:56:46 am
content: i am not arguing with kirchoff. kirchoff is right , but only for static systems. you cannot use kirchoff to solve transient effects.take three capacitors , connect them in star. two go through a resistor to ground. apply a between the third input of this network and ground. good luck with kirchoff..accoording to kirchoff no current flows as capacitors block dc. yet electrons have moved !the diode does not violate this . a portion of the electrons is temporarily stored and released time-shifted. the total sum of electrons in is equal to the total sum of electrons out. kirchoff was not violated. the storage is happening as electron-proton bonds in the semiconductor material when the recombination zone is deconstructed. that zone discharges them eventually when it is reformed as the diode turns off.nothing wrong there

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 22, 2014, 01:16:46 am
content: Quote from: free_electron on December 22, 2014, 12:56:46 amUh, I just said you can.QuoteThe current flowing through a capacitance is calleddisplacement current.Tim

author: c4757p
author_rank: Super Contributor
timestamp: December 22, 2014, 01:23:58 am
content: You can'tseriouslybe claiming that Kirchoff can't solvethis...You're screwing with us, right?

author: photon
author_rank: Frequent Contributor
timestamp: December 22, 2014, 05:31:10 am
content: Quote from: T3sl4co1l on December 22, 2014, 12:30:27 amBut I thought we were talking about pushing electrons? Displacement current does not consist in moving electrons and only plays a role in EM waves, not in transistor theory, as far as I know.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 22, 2014, 05:54:48 am
content: What do you think makes electrons "push"?As one of the four of Maxwell's equations (in the standard Heaviside format), displacement current is an inseparable aspect of E&M theory on any level. Whether it's pushing electrons in a quantum mechanical solid, or waves propagating in space (displacement current exists whether or not there is matter present to be polarized by the field).Ed: slightly disingenuous because electrons also obey the Pauli Exclusion Principle. That's more of a statistics thing than a pushing thing, but I guess they're equivalent since energy is energy.Tim

author: photon
author_rank: Frequent Contributor
timestamp: December 22, 2014, 07:28:41 am
content: Quote from: T3sl4co1l on December 22, 2014, 05:54:48 amThe E field. The magnetic field has no effect on an electron.

author: LvW
author_rank: Frequent Contributor
timestamp: December 22, 2014, 10:25:49 am
content: Wow - up to now more than100 contributionstriggered by my claim (replies#2 and#5) regarding the control mechanism of BJTs.I didn`t expect that - on the other hand it reflects the (surprising, funny, unbelievable) situation that many decades after the BJT was invented there still exist (in books, lecture notes, articles, papers) two different explanations why/how the BJT works.I must admit being not a physicist I cannot contribute too much to the discussions/explanations on charged carrier level.I rather feel as an engineer who is able only toobserve and to interpret measurements.More than that, I am trying to bring these results in relation to commonly accepted rules and principles.And there are some phenomena which can be explained - as far as I know - based on thevoltage control principle only(Shockley`s exponential relation Ic=f(Vbe)).Some of them:* Voltage feedback principle (emitter resistance) with increasing input resistance at the base;* Early effect - caused by (electrical) field increase;* Transfer characteristic for diff. amplifier (tanh function);* Working principle of a voltage source based on the „Vbe multiplier“ concept.* Principle of translinear circuits (introduced by Barry Gilbert).Up to now, nobody was in the position to show how these effects/principles can be explained using the current-control approach.More than that, I didn`t hear about one single circuit principle/effect which can be explained on the current-control principle only.(To avoid misunderstandings: Of course, I do not deny the fact that there is a base current Ib=Ic/hfe, which must be taken intoaccount during the design process; however, we are talking here about the control mechanism only).Thus, I see no reason to treat the BJT as a device having a collector current that iscontrolledby Ib.In contrary, tests/measurements have convinced me that both currents (Ic and Ib) are controlled and determined by Vbe only (Shockley`s equation).

author: free_electron
author_rank: Super Contributor
timestamp: December 22, 2014, 01:19:22 pm
content: Quote from: c4757p on December 22, 2014, 01:23:58 amThat schematic is right , but , i did not say apply a pulse. I said apply a step.Meaning flick the switch on that power supply so its out put goes from zero to let's say 5 volt and stays at 5 volts forever.All your resistors are 1k. caps are 1 microfarad , 2 microfarad and 3 microfarad. You can assume ideal parts, no leakage etc.Please, humor me. Apply kirchoff. All your currents will be zero , yet that ampere needle of that power supply did deviate the moment i threw the switch.I agree you can do kirchoff for an ac signal, as you can calculate the impdeance of the capacitor at that frequency. But not for a transient.That is what i am trying to explain when italk about whathappens when a junction goes from non co duction to co duction and back (including the voltage across the junction going to zero).Anyway, i've had enough of this topic. We are talking two different things here : what happens in the box and how you use the box.I just saw a little light go off when LvW wrote 'schockleys equation' ic = f(be).He is talking the input characteristic of the transistor. And indeed that curve is used to make amplifiers and figure out a circuit. You can make a diagram with 4 curves , this one sitting top right.So yes, for all means and purposes : when designing with the box : that is function you need.When designing the box itself that function is a result of doping and the physical structure.I've been inside the box too long...But, as a closing thought, using that curve : draw a line that shows ic versus ib and look where they intersect.

author: dannyf
author_rank: Super Contributor
timestamp: December 22, 2014, 01:37:19 pm
content: QuoteWow!Q: how many eevblog experts does it take to understand Kirchoff?A: as many electrons as in the Dead Sea.

author: free_electron
author_rank: Super Contributor
timestamp: December 22, 2014, 01:54:55 pm
content: Quote from: dannyf on December 22, 2014, 01:37:19 pmWell i would like to see an answer, because i don't know how to do it.I can do kirchoff for an ac signal. But not for a step. Kirchoff will tell you that the currents in that thing are zero. Yet there was clearly an inrush current. You may be able to figure out how high that inrush current was (pretty easy actually as the caps are essentially a short at that point). And you may even be able to figure out what the current is if the caps are at 1/5 th of their max charge (although i threw a spanner in that one by making them different so they will reach their 1/5 th point at a different point in time )The math gets very complex, very quickly. Same reason simulators have tremendous problems with capacitors and cant solve the matrix. Put two caps in series in a sim and see what willl happen. It'll throw a fit.Anyway all that stuff is moot. We're arguiing the wrong thing. There is nothing wrong with kirchoff , no laws of physics are harmed i. The i ternal function of transistors, there is only temporary charge storage.If you want the raw maths : i posted a link to that document that shows how a diode goes into conduction and out. Have a ball.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 22, 2014, 04:28:09 pm
content: Quote from: free_electron on December 22, 2014, 01:19:22 pmDoes this meet with your request and expectations?http://seventransistorlabs.com/Images/Kirchoff_for_FE.pngNote: the blue trace is colored to show that it is drawn completely over the red trace. Of course, since they match up exactly, I can't show both the red and blue traces, in the same place, at the same time...QuoteWhat?Then how do you think transient simulation, or the real world itself, ever functions?In the pure time domain, capacitors do not have reactance, they have capacitance. The fundamental capacitance equation is:I = C * dV/dtYou perform an integration over time to solve the circuit for its transient response.In the SPICE case, it's done in small, approximate steps, a necessary solution given the messy things SPICE normally has to deal with. This linear circuit can be solved symbolically though, for any arbitrary input which satisfies the rules of electrical signals (i.e., excluding pathological functions from the darker side of analytical calculus).QuoteHow does a conducting junction become non-conducting? That's the definition of a junction, that it is always conductive, and therefore connects the component pins which share that junction. Just because no average current is flowing through it in some instantaneous moment, doesn't make it not a conductor. Again, if you'd like to bring that all the way down to the quantum level, it remains true. (In fact, quantum mechanics does an excellent job proving this, since, classical measurement requires that you have current flow to measure the resistance: if R = V/I has exactly no current, R is undefined (but certainly not non-conducting). QM fluctuations are always present, and depend on the medium; the fluctuations in a conductor are characteristic, as are the fluctuations in 'empty' space -- the vacuum state.)Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 22, 2014, 04:33:31 pm
content: Quote from: photon on December 22, 2014, 07:28:41 amI expect a vast number of experiments would contradict that second part... But, at least within present technology semiconductors, it is the case that magnetic field and spin have no impact on the current flow.Well... no, I can't even allow for that. Because Hall effect sensors are very much a real, useful application of magnetic fields upon electrons within a semiconductor.It would be more accurate to say, general purpose semiconducting devices (those not intended for interacting with magnetic fields) are generally insensitive to magnetic fields. Your average BJT probably doesn't do anything different until a tesla or a few, at which point, weird things start happening (I would guess, reduced hFE, nonuniform current density leading to early failure of power transistors... um... probably not much else though?).As for the E-field, precisely: change in the E-field does not occur for free, because the field stores energy. And a change in that energy is driven by -- guess what -- displacement current. (Regarding free EM waves: that current, in turn, necessitates a magnetic field, which arises from the changing electric field -- the four states, from the four equations, for the four quadrature phases of a complete circle (relating directly to the fact that d^4/dt^4(sin t) = sin t), and hence the propagation into infinity of an electromagnetic wave.)Tim

author: Dave
author_rank: Super Contributor
timestamp: December 22, 2014, 04:35:31 pm
content: Quote from: free_electron on December 22, 2014, 01:19:22 pmSure, no problem.Quote from: free_electron on December 22, 2014, 01:19:22 pmIn the steady state, yes, all currents will be zero. However the magic of the laws is that theyalwaysapply. You can quite easily analyse the behavior of the circuit when a voltage step is applied to it.Here is the circuit with everything marked, to avoid potential confusion:KVL1: Vg= VC1+ VC2+ VR1KVL2: Vg= VC1+ VC3+ VR2KCL: i = iC2+ iC3

author: photon
author_rank: Frequent Contributor
timestamp: December 23, 2014, 07:53:47 am
content: Quote from: T3sl4co1l on December 22, 2014, 04:33:31 pmI was referring to the force law F = q(E + v x B) in the Maxwell model. F gives the force on an electron in terms of the E and B field. If the electron is moving, i.e. a current, then both the E and B field contribute to the force on the electron. If the electron is not moving, i.e. a charge, then only the E field contributes. I was attempting to add some carefulness in the use of the words "electron" and "current" in this discussion.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 23, 2014, 11:16:50 am
content: Oh, as in a static electron.Of course, electrons are constantly milling about thermally -- but we can refine that even more pedantically by saying, not just an electron, but a population of them whose net velocity is zero, so it still works in that case.Tim

author: dannyf
author_rank: Super Contributor
timestamp: December 23, 2014, 12:22:26 pm
content: Wow, this has really turned into a stump-the-champ exhibition for our experts.I will provide a different way of thinking, using Kirchoff,.The two legs are identical. So they can be viewed as one branch, with 2C and 1/2R in serial, which itself is in serial with a C on the upper body.You can then combine the 2C from the lower leg with the C from the upper body, yielding a 2/3C.Ie., this circuit, viewed from the source, is identical to a RC circuit with 2/3C and 1/2R.The peak current is then V/(1/2R), declining from there exponentially to zero.You can further conclude that the current in the lower two legs are 1/2 of the current going through the upper body.Yes, Kirchoff holds.

author: Dave
author_rank: Super Contributor
timestamp: December 23, 2014, 06:27:43 pm
content: Quote from: free_electron on December 22, 2014, 01:19:22 pmQuote from: dannyf on December 23, 2014, 12:22:26 pmNope.Quote from: dannyf on December 23, 2014, 12:22:26 pmIt certainly does.

author: dannyf
author_rank: Super Contributor
timestamp: December 23, 2014, 09:28:53 pm
content: QuoteThe same process follows in the analysis.

author: Dave
author_rank: Super Contributor
timestamp: December 25, 2014, 05:44:43 am
content: Quote from: dannyf on December 23, 2014, 09:28:53 pmSorry, nope.As soon as you have two different capacitors in the bottom branches, voltage and current waveforms no longer follow a simple exponential rise or fall. You need a different approach to analyzing the behavior of the circuit.One of the ways to do it would be to write out some differential equations:vg= vC1+ vC2+ R1*C2*dvC2/dtvg= vC1+ vC3+ R2*C3*dvC3/dtC1*dvC1/dt = C2*dvC2/dt + C3*dvC3/dtThese are actually the same three equations I wrote in my post above, just written with differential equations.Quote from: Dave on December 22, 2014, 04:35:31 pmYou would then have to mash the three equations together and solve for vC2, for example. But solving for a Heaviside (step) input voltage function would be quite difficult in the time-domain, so a better solution would be to switch into the s-domain (Laplace transform).The same three equations then become:vg= vC1+ vC2+ R1*vC2*C2*svg= vC1+ vC3+ R2*vC3*C3*svC1*C1*s = vC2*C2*s + vC3*C3*sThe Heaviside function is quite simple to write in the s-domain:vg= 1/sSolve for desired quantity and perform an inverse Laplace transform, so you get the output function in the time domain.This is how one would solve the circuit analytically. The process is rather tricky and time consuming, but (contrary to what some in this thread claim) itcanbe done. Kirchoff's laws apply. Always.It is a good idea to help yourself solve problems like these with programs like Matlab.______________________________________________We could simplify our life and approach the problem by slapping everything into SPICE and running the simulation. This is what you get, if you choose C1=2µF, C2=3µF and C3=1µF:You can see something interesting happening, which may seem somewhat counterintuitive at first glance. The current through C3goesnegativeat one point (V(n3) is voltage across R2, therefore directly proportional to current in that branch).Just after switching, the capacitors are all at 0V and the currents are only dictated by the resistors. Both resistor values are equal and therefore the currents are equal. The function of voltage on a capacitor is 1/C * integral(iCdt). Because C3is only a third of C2, the voltage on the former starts increasing three times as fast as on the latter. However, at the end of the transient, they both need to settle at the same voltage (as they are connected in parallel), so the current in C3has to go negative in order for the voltage to drop and match the other capacitor. Interesting, huh?

author: LvW
author_rank: Frequent Contributor
timestamp: December 26, 2014, 02:23:40 pm
content: Gentlemen - I am afraid, some of you don`t like to continue the discussion (disput?) about the control mechanism of the BJT (major subject of this thread).As we all know, we didn`t arrive at a - more or less - common conclusion.Nevertheless, let me try a kind of summary:Roughly, I have counted five contributors supporting the voltage-control approach and also five forum members heavily defending the current-control alternative (sorry to say - but in many cases with polemic attacks rather than technical arguments).So I am still waiting for answers to two simple technical questions (contained in my reply#61) as well as some comments to examples I have listed in my reply#105.Is there really no member of the „current-control“ group who is able or willing to answer?I have to correct myself a little. I got two answers. Here are they:(1)"any attempt at answering your question is irrelevant to the discussion. It clarifies nothing other than your stubbornness "(2)"you insist on asking people to answer questions that have nothing to do with this discussion,"Lack of technical arguments? I`ve got the impression that - for some of you - the question under discussion seems to be a „religious“ one (just a matter of faith).Here is another comment I got:"The fact that some authorities somewhere else had cited it has no bearing whatsoever in a "technical" discussion."In principle, I agree to this.However, I think - from case to case - it might be interesting to know the name of the „authority“.As an example, here is the answer to a corresponding question from one of the leading developers in the world of electronics:Barrie Gilbert(Analogue Devices):The old current-in, current-out seems view simple at first, but that's about as far as it goes.We clearly agree that the BJT should be seen in the same way as an MOS device, explaining that the DC base current of the BJT is actually due to a defect (of sorts) and only a nuisance.At Analog Devices we have made BJTs (under special conditions) having a DC beta of over 25,000.RegardsLvW

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 26, 2014, 04:31:55 pm
content: Well if you insist;Quote from: LvW on December 20, 2014, 05:21:29 pmThis is ambiguous, because both voltage and current can be fed back in series or parallel.Indeed, if they are fed back in combination, the amplifier impedances converge (with rising Av) to a constant ratio V/I = resistance.There is no necessity that the input or output ports have the same or different impedances, because the feedback can be wired to account for either purpose.QuoteThis also makes the question much more complicated than necessary, because you are talking about the impedance of a good capacitor: the FET case. The impedance is already infinite (nearly) at DC, but capacitive at high frequencies, neither of which is very easy to measure. What frequency should it be measured at? How are we to determine whether the feedback has an increasing or decreasing effect on it?Tim

author: dannyf
author_rank: Super Contributor
timestamp: December 26, 2014, 05:27:54 pm
content: QuoteYou don't seem to be able to understand it on a device level, nor a circuit level - both of which we have tried on you.Here is another attempt - which I indicated quite a while for you.The following is a simulation of two ideal devices, G1 on the left is a voltage controlled current devices and F1 on the right is a current controlled current devices. Both are configured into a gain stage.The graph shows a dc sweep where Vin goes from 0v to 100mv, and Vout1 / Vout2 goes from 5v to 4v, as expected - Vout2 was shifted upwards by 0.1 so it is not right on top of Vout1.Now, F1 and G1 are completely different devices in that one is voltage-controlled and the other is current controlled.Question for you, which of them is a bjt and which is a mosfet?:0

author: G0HZU
author_rank: Super Contributor
timestamp: December 26, 2014, 07:31:10 pm
content: In all my 25 years as an RF designer I've never ever heard any of my colleagues argue/bitch over the labelling of how a BJT is 'controlled' in terms of voltage or current. However, they do often argue over which 'model technology' is most appropriate for a particular design or study task but this is usually an argument for/against using small signal s parameters instead of trying to build a decent SPICE model.So I don't really have a horse in this (pointless?) labelling race but I do have a suggestion. Try analysing the BJT in its old school configuration. i.e. in common base. This is how I was taught to understand the basic physics of an NPN BJT at college many years ago. It makes the device much easier for a beginner to understand because you can see that the electrons flowing in at the emitter (input) are all meant to transfer across to the collector (output) in an ideal BJT. i.e. current gain = 1.In reality some of them fail to transfer and these are seen as wasted electrons that uselessly flow to the base. But the idea is that they nearly all transfer to the collector load resistor. There's lots of interpretations of what TRANSISTOR means but I was taught that for the simple common base configuration:transistor = 'transfer input current >> collector resistor'So you can get power gain by choosing a suitable collector resistance.This very simple 'transfer' configuration shows that the common base BJT will follow the Shockley equation for device current in that the device current will be a function of Vbe. So it's a bit like a basic diode except that the electrons transfer to the collector rather than flow to the base. It's difficult to argue that the BJT performance is being 'controlled' here by how many failed* electrons arrive at the base.* Failed as in failed to transfer to the collector.Some may argue that current out is a function of current in for the above case or other people may prefer to simply argue for a current control label because Ic = Ib * beta when in the common emitter configuration but then I'm not really bothered about attaching a voltage or current control label to the device. I don't think it adds much value.So I'm not trying to pick a winner here.

author: jaxbird
author_rank: Frequent Contributor
timestamp: December 26, 2014, 08:41:14 pm
content: Oh, I am intimately familiar with the base, she always claim she needs noting, but we all know that current is what she craves, just a little bit she says, but we all know it ends up being much more than that, often way beyond our budget, but what can we do, she needs the current now to perform what we expect of her.

author: dannyf
author_rank: Super Contributor
timestamp: December 26, 2014, 09:56:34 pm
content: Nicely done, jaxbird.

author: jaxbird
author_rank: Frequent Contributor
timestamp: December 26, 2014, 10:19:14 pm
content: Quote from: dannyf on December 26, 2014, 09:56:34 pmThanks, I find a different perspective can help general understanding

author: dannyf
author_rank: Super Contributor
timestamp: December 26, 2014, 10:33:44 pm
content: However, you didn't explain how the 2nd base or the 3rd base works. I think it may help some people here,.If you could work in the concepts of holes vs. electrons, exchange of charges, emission of electrons, collection of majority carriers, etc., it would impress people more, potentially.

author: jaxbird
author_rank: Frequent Contributor
timestamp: December 26, 2014, 10:45:47 pm
content: Quote from: dannyf on December 26, 2014, 10:33:44 pmNot premium service, some original research required :pMight not be fully related to the inner workings of common semiconductors

author: G0HZU
author_rank: Super Contributor
timestamp: December 27, 2014, 12:42:01 am
content: Base region or base pin? Just to clarify, in my simple analysis in #121 I'm referring to base as the base 'pin' or component leg as per the thread title. I'm just trying to post up a simplistic model of the common base amplifier for a beginner. I'm posting up how I was initially taught about the BJT when I was a spotty student and the common base is the easiest to understand.In a simple common base analysis a beginner can crudely model the device as current out = current in (if alpha is very close to 1) or they can use the Shockley equation to try and calculate Ic based on Vbe etc.But please note that I'm not taking sides. I don't have a horse in this race. In my opinion many of the deeply technical posts in this thread are just going to confuse and alienate a beginner.

author: miguelvp
author_rank: Super Contributor
timestamp: December 27, 2014, 01:28:06 am
content: Quote from: jaxbird on December 26, 2014, 10:45:47 pmIt requires a capacitor that can raise it's capacitance overtime but being careful that the capacitor doesn't have enough to feed other bases and moves on to a newer circuit altogether.

author: LvW
author_rank: Frequent Contributor
timestamp: December 27, 2014, 09:51:07 am
content: Quote from: T3sl4co1l on December 26, 2014, 04:31:55 pmI do not "insist" - but I think that a sequence of questions and answers is the best method to clarify things (and opposite opinions).Quote from: T3sl4co1l on December 26, 2014, 04:31:55 pmVoltage always is fed back in series (and currents in parallel).But that`s no answer.It was a simple yes/no question (Is the rising of the input impedance due to Re feedback an indication for voltage control; yes or no?).Quote from: T3sl4co1l on December 26, 2014, 04:31:55 pmQuote from: T3sl4co1l on December 26, 2014, 04:31:55 pmI didn`t mention FET`s case at all. Sorry - but this has nothing to do with the question under discussion.Final remark: In my various posts I have listed some effects (on BJT unit level) as well as corresponding circuit examples which can be explained only because the current Ic is determined by thevoltageVbe (rather than the current Ib, which is existent because it cannot be avoided). In addition there are many experts with high reputation (Barrie Gilbert) who are supporting the voltage-control mechanism.On the other hand - I didn`t hear or read aboutone single examplewhich would justify or proof the claim of current control.Are you really not able to reconsider your opinion?LvW

author: LvW
author_rank: Frequent Contributor
timestamp: December 27, 2014, 09:57:14 am
content: Quote from: dannyf on December 26, 2014, 05:27:54 pmYes - I know. I am not able to understand.Perhaps I would inprove my understanding if you could give an answer to my technical questions. So - I would have the chance to learn something.

author: LvW
author_rank: Frequent Contributor
timestamp: December 27, 2014, 10:14:58 am
content: Quote from: G0HZU on December 26, 2014, 07:31:10 pmThank you for this sentence. This underlines my claim thatall of us(if we follow the voltage- or the current-control approach) follow the same steps for designing an amplifier stage. That means: For designing BJT circuit it it not important at all on which "model" we a re relying.But the funny and surprising situation is as follows (as I have explained in a former reply) : During such a design process all of us treat the current Ib as a kind of "unwanted" (parasitic) current that must be, of course, taken into account. Nevertheless, some people still believe that Ib would determine the value of Ic - and they don`t realize the contradiction.As I have mentioned - no problem for designing; however, if we start toexplainthe working principle of some circuits (I have given examples) the contradiction becomes obvious (example: Re feedback, Early effect) .Quote from: G0HZU on December 26, 2014, 07:31:10 pmYes - exactly this is the situation (wastedelectrons)! Thank you for being the first who is mentioning the common-base configuration.LvW

author: Zero999
author_rank: Super Contributor
timestamp: December 27, 2014, 02:55:27 pm
content: Quote from: dannyf on December 26, 2014, 05:27:54 pmThanks for posting that.Indeed the response is the same, irrespective of whether it's current or voltage controlled.Now look at the input impedance of the current controlled amplifier and you find it's pretty similar to what you'd expect from a BJT amplifier.What simulation package do you use?I tried LTSpice but it doesn't have a current controlled current source, so I used the voltage controlled current source and connected a 1R resistor across the input terminals and it works quite well.

author: LvW
author_rank: Frequent Contributor
timestamp: December 27, 2014, 04:19:10 pm
content: Quote from: Hero999 on December 27, 2014, 02:55:27 pmHi Hero999 - up to now, I didn`t comment this example from dannyf because it does not meet the point (for my opinion).However, because of your answer - I have changed my mind.This example does not meet the point (that means: The subject of our discussion) because it describes CIRCUITS with external elements.But this was not the problem to be discussed in this thread. Instead, we were discussing the control mechanism INSIDE the BJT.Let me take anothersimple example: It is commonly accepted to treat the classical opamp as an voltage amplifier (does this apply also for the "current-control group" in case of BJT input stage?). However, with external elements the whole CIRCUIT can be used, for example, as a current amplifier or as a current-to-voltage converter. Buth this operation does not say anything about the transfer characteristics of the opamp unit alone. Do you know what I mean?More than that, several times I have emphasized the fact that during DESIGN of an amplifier circuit it does not matter at all how/why the collector current of a BJT changes its value.There is no doubt, that I can send a current into the base and watch if and how the current Ic varies. But this cannot tell us anything about the control mechanism because the same is possible with a voltage.But the situation changes as soon as we have toexplainto somebody else WHY the BJT (or a circuit with a BJT) shows a certain behaviour.I have presented many examples, which cannot be explained with current-control - on the other hand, nobody was in the position to proofe the opposite.(I don`t know if I have mentioned already that, for example, the EARLY effect can only be explained using voltage control. It`s that simple.Perhaps it is even better and more clear to say: The existence of the Early effect, in fact, prooves that the BJT is a voltage controlled part.)

author: IanB
author_rank: Super Contributor
timestamp: December 27, 2014, 05:10:39 pm
content: Quote from: LvW on December 27, 2014, 04:19:10 pmHere I think lies the root of all the problems. You see, in physics and in science generally there is no answer to the question of why something happens. Science can only answer the question of what happens, and what will happen.We do experiments and gather data, and then we try to fit a mathematical model to the data. If we have a good model it will predict the results of other experiments we haven't done yet. If we have a bad model it will not, and we discard that model. The best models, the most convenient models to use, have fewer parameters that are easier to determine. We try to settle on the best models if we can, because they suggest we have somehow got "closer" to what is really going on.So trying to ask what "controls" something else, if we mean anything other than what is the simplest model, is futile.Try a question from another field, aerodynamics. Why do planes fly? One person will say that the air flowing over the wing flows faster over the top causing lower pressure above the wing than below it. This pressure differential multiplied by the wing area produces a lift force that holds the plane up.Another person will argue that this is totally wrong and cannot possibly explain why a wing produces lift. The real explanation is that the wing when moving through the air deflects air downwards, like the draft you feel when standing underneath a helicopter. By Newton's law of motion, that every action produces an equal and opposite reaction, the air deflected downwards lifts the plane upwards.Thisis why planes fly.So it seems we need to argue about this instead. Is the lift on a wing produced by pressure or by air flow?

author: idpromnut
author_rank: Unknown
timestamp: December 27, 2014, 07:17:40 pm
content: Quote from: IanB on December 27, 2014, 05:10:39 pmPressure. If it were angle of the wing, a wing with the bottom surface parallel to the ground (or perpendicular to the force of gravity) would not generate lift.Also, I am not sure if you intended this, but the air-flow / pressure are one and the same explanation.

author: IanB
author_rank: Super Contributor
timestamp: December 27, 2014, 11:43:45 pm
content: Quote from: idpromnut on December 27, 2014, 07:17:40 pmAnd see, this is the reason such discussions often founder. Can you point to one place in my post where I mentioned angle of the wing? I said "A" or "B" are alternative explanations. And you chime in with 'oh, but it can't be "C" because...' Well, fine, it can't be "C", but who suggested it was?QuoteQuite right. If air flow wasn't directed downwards there would be no lift force exerted on the wing. I'm glad we agree about that.

author: dannyf
author_rank: Super Contributor
timestamp: December 28, 2014, 12:56:02 am
content: QuoteYou are so not understanding what the other poster was saying. Lift doesn't have to be generate by pushing down air. And rotorcraft doesn't work the way you think it does.Read an Aerodynamic 101 textbook.

author: free_electron
author_rank: Super Contributor
timestamp: December 28, 2014, 01:22:40 am
content: i've been chatting with a few ex collegues on this.It is really very simpleA bipolar transistor as amplifier has two defining parameters :the common emitter current gain is called Beta ( or hFE)the common base current gain is called Alphaboth are .... ratios. <- fill in the blank ....

author: c4757p
author_rank: Super Contributor
timestamp: December 28, 2014, 01:30:49 am
content: Quote from: free_electron on December 28, 2014, 01:22:40 amYou talk as if these are the only defining properties of the transistor and everything else necessarily falls out of them. I could say the same thing about the parameters of the Ebers-Moll equation, and I'd have a more accurate model as well (as yours doesn't account for anything translinear).Also, these are only one parameter, each is directly a function of the other.

author: c4757p
author_rank: Super Contributor
timestamp: December 28, 2014, 01:38:38 am
content: Of course - we all know what'sreallycontrolling the things:This dude, and a bit of magic smoke to keep him happy

author: timb
author_rank: Super Contributor
timestamp: December 28, 2014, 04:19:50 am
content: Quote from: c4757p on December 28, 2014, 01:38:38 amWhere does his poop and pee go?!?!Sent from my Tablet

author: timb
author_rank: Super Contributor
timestamp: December 28, 2014, 04:21:31 am
content: Quote from: jaxbird on December 26, 2014, 08:41:14 pmWith enough current she might just open her holes for you.Sent from my Tablet

author: miguelvp
author_rank: Super Contributor
timestamp: December 28, 2014, 04:27:54 am
content: Quote from: timb on December 28, 2014, 04:19:50 amCommon sense will indicate it will go to the collector, but it goes into the emitter.

author: c4757p
author_rank: Super Contributor
timestamp: December 28, 2014, 04:33:05 am
content: Quote from: miguelvp on December 28, 2014, 04:27:54 amThat depends on whether it's NPN or PNP...

author: miguelvp
author_rank: Super Contributor
timestamp: December 28, 2014, 04:36:13 am
content: Quote from: c4757p on December 28, 2014, 04:33:05 amdefinitely a Poop 'N Pee so collector it is (I think)

author: timb
author_rank: Super Contributor
timestamp: December 28, 2014, 05:48:24 am
content: Quote from: miguelvp on December 28, 2014, 04:36:13 amBrilliant!Now we're finally discussing what this thread truly is: Shit.Sent from my Smartphone

author: free_electron
author_rank: Super Contributor
timestamp: December 28, 2014, 06:59:34 am
content: Quote from: c4757p on December 28, 2014, 01:30:49 amDoesnt matter. A bipolar transistor is a current driven current amplifier.The gain is defined as ratio of two currents.And i'm done with this topic. Do whatever you please. I don't care anymore. It can be approached from both perspectives. Fact remains : no base current is no collector current.

author: LvW
author_rank: Frequent Contributor
timestamp: December 28, 2014, 10:17:13 am
content: Quote from: IanB on December 27, 2014, 05:10:39 pmReally "futile"?IanB - here I am not with you. I know what you probably mean; and I agree with you - as far as "natural laws" are concerend.However, based on these fundamental laws and rules, we have a lot of phenomena which can and must be explained - starting with our definitions for "voltage" and "current". Both terms are used for finding other rules and formulas which, of course, can be explained (voltage divider, superposition,...).And the same applies - for my opinion - to the BJTs control mechanism.More than that, I think it is really necessary to explain this mechanism - at least when students have problems to solve some obvious contradictions.Imagine the following situation: Introducing the BJT I have stated that Ic is determined by Ib using the simple relation Ic=B*Ib.Now - after some additional lessons - I explain thetemperature dependenceof the curent Ic and tell them that - in order to keep Ic constant - I must reduce the voltage Vbe by-2mVper degree temp. change.Of course, some students will ask:Huhhh? We thougt it is the current Ib that controls Ic. Suddenly it is the voltage Vbe?(That`s what I have experienced often.)And why this value of -2mV/K ?And I must answer and explain that this value is not only measured but that this value wascalculatedbased on charged carrier physics in the pn region of the transistor.Do you understand the dilemma? It is really not sufficient to say "doesn`t matter , current or voltage" .I have to make a decision - and (as I have demonstrated with various examples): A current-control mechanmism cannot explain how BJTcircuitsreally work.

author: IanB
author_rank: Super Contributor
timestamp: December 28, 2014, 12:47:14 pm
content: Quote from: LvW on December 28, 2014, 10:17:13 amThis is really a matter of teaching and order of presentation. You have first presented a simple model of the operation of a transistor, of the form f(Ib, Ic, B) = 0, where B is a parameter and Ib, Ic are model variables. Given a value of B, then for any value of Ib you can fix Ic, or for any value of Ic you can fix Ib. But it is important to explain that this is a simplified model that abstracts away many details, and is only valid in a narrow region of operation.Later, it will become necessary to introduce more elaborate models of operation, models that include variables Vbe, Vce, Ib, Ic and perhaps the operating temperature, T. With such a model in place, it becomes possible to talk about the influence of T on the other variables. One can also talk about when it is appropriate to use the simple model and when the more detailed model is needed.But even with the more detailed model, it is not really possible to say that one variable is a cause and another variable is an effect. This is choice is a matter of perspective and intent. The model variables are in relationship to each other. If you fix some variables of your choice, then according to the degrees of freedom of the model other variables will be determined. For instance you could choose a desired value of Ic and determine appropriate values of other variables to achieve this.

author: Zero999
author_rank: Super Contributor
timestamp: December 28, 2014, 01:35:52 pm
content: Quote from: LvW on December 28, 2014, 10:17:13 amOf course VBEdecreases by 2mV per degree temperature change because it is a diode junction. You could take an ideal current controlled current source and connect a diode in series with the input and its input will behave like a BJT's base emitter junction.This debate is silly. I've been talked both models in college. Most of the time I used the current controlled approach but it isn't perfect as it doesn't take everything into account but neither does the voltage control model. If it did, then I wouldn't have to worry about the base current or blowing a BJT up when driving it without a resistor.Let me guess it's most likely the older people here who argue one way or the other?With age brings wisdom but it also reduces ones ability to accept new ideas.

author: LvW
author_rank: Frequent Contributor
timestamp: December 28, 2014, 02:22:07 pm
content: Quote from: IanB on December 28, 2014, 12:47:14 pmQuestion 1: Why should I first present a model which is false and which is not able to explain/verify various effects? I would agree if the "more elaborate" model (to be presented later) simply would be anextensionof the first model. But that is not the case.Questioin 2:Can you, please, show meone singleapplication which is "valid in a narrow region of operation" and where it is advantageous to use the current-control approach?Question 3: I am sure that in each electronic course the pn diode is explained prior to the BJT. That means: The students are familiar with Shockley`s equation.Is there any reason to believe that students would not accept that the same equation applies for the pn junction within the BJT? Is there any good reason to deviate from this knowledge and - suddenly - teach something which is simplyfalse? I do not understand such an attitude.

author: LvW
author_rank: Frequent Contributor
timestamp: December 28, 2014, 02:55:24 pm
content: Quote from: Hero999 on December 28, 2014, 01:35:52 pmHero999, there is a misunderstanding - perhaps I am responsible, but I am not sure."but neither does the voltage control model. If it did, then I wouldn't have to worry about the base current or blowing a BJT up when driving it without a resistor."I think and I hope that I have mentioned several times that - of course - there is base current I have to "worry about" during design of a BJT stage.But this simply is a mathematical matter - nothing else. In practice, nobody connects a battery with 0.65 volts across the B-E path.Of course, I take into account that the top resistor of a base voltage divider carries a current (I1+Ib) and the lower resistor only the current I1.But - what has this calculation to do with the "model-question" or the "control-question" ?Please, can you answer this last question? I am really open to learn.I really have problems to understand the way of thinking of all persons who vote for the "current-control" option.Where is the advantage? Show me one single application where this approach has - at least - advantages.Because you have used the term "silly": The classical BJT gain stage with a low-resistive voltage divider and Re-feedback is designed (after fixing Ic, Rc and Re) starting with VOLTAGES (at the emitter and the base node). Then, as a next step, we calculate the base resistors - of course taking Ib into account.All of us follow this sequence - am I wrong? - nevertheless, some people think that they have established a current-control mechanism.In my view:This is silly.If there would not exist some external requirements (power consumption, input resistance) we could make the divider so low-resistive (I1=50*Ib or more) that we even could NEGLECT the base current.I don`t know if you were (or still are) engaged in teaching electronics. But my position and my arguments result from corresponding experiences.Do you think that students are satisfied with two different explanations ? Hence, to me this is by far not a "silly debate".Why do you think I am so engaged in this discussion (which sometimes merges into a religious matter)?

author: LvW
author_rank: Frequent Contributor
timestamp: December 28, 2014, 02:58:14 pm
content: Quote from: AcHmed99 on December 28, 2014, 02:55:01 pmI suppose you are not engaged in teaching electronics, correct? So you can say "I don`t care about it". Be lucky.

author: dannyf
author_rank: Super Contributor
timestamp: December 28, 2014, 02:58:31 pm
content: QuoteWow, I knew that your understanding of device physics is superficial but I never thought it was this bad,.The tempco is no repudiation of the bjt's current controlled mechanism. In this context, it is purely the side effect of your driving the b-e junction with a (constant) voltage source.Ic = B * Ib still holds, except that if you hold the b-e junction's forward voltage constant, Ib will vary by temperature.Alternatively, you can drive the b-e junction via a (constant) current source, and the tempco's impact on Ic is completely eliminated.Quite frankly, you can do yourself a huge favor by trying more to understand the basics, and trying less to use terminologies that you have little understanding of - it doesn't help you much (ie. at all).QuoteAll models are false / wrong, and that's why models are useful - they help you isolate the important factors so you can focus on what makes sense for your particular study / applications.Different models can be proposed to explain (different) aspects of the same phenomenon. That doesn't mean that one is correct and others wrong.If someone has to break that to you, you are in deep @#$@, I mean, trouble.

author: amyk
author_rank: Super Contributor
timestamp: December 28, 2014, 03:05:04 pm
content: Quote from: dannyf on December 28, 2014, 02:58:31 pm...just like wave-particle duality. Fortunately, the majority of physicists have accepted that there can be different ways of looking at the same thing.

author: IanB
author_rank: Super Contributor
timestamp: December 28, 2014, 04:04:21 pm
content: Quote from: dannyf on December 28, 2014, 12:56:02 amWell it does, actually. Heavier than air craft need an upward force to keep them in the air. This upward force is generated by accelerating some of the surrounding air downwards to generate a reaction force (F = ma). Where airplanes differ from rockets in this regard is that airplanes use the surrounding air for their reaction force while rockets generate their own exhaust gases.

author: LvW
author_rank: Frequent Contributor
timestamp: December 28, 2014, 04:52:17 pm
content: Quote from: dannyf on December 28, 2014, 02:58:31 pmDannyf, I know that you have some problems to handle some new - and for you perhaps "uncommon" - information. This may explain your polemic attitude.Nevertheless, as long as you do not answer my technical questions (my former posts) it is perhaps better to avoid such personal attacks.Quote from: dannyf on December 28, 2014, 02:58:31 pmBack to the value of -2mV/K: Your reply is correct, but the information content is zero.Nobody has denied that in the ideal and unrealistic case (constant current Ib) the current Ic would remain roughly constant. In this case, the temperature dependence of Ic from Vbe has no effect on Ic. That`s logical, is it not? It is a truism - nothing else. In this case, Vbe is floating and is adapting itself corresponding to the tempco.However, may I remind you (as I have stated in my recent post) that this value ofd(Vbe)/d(T)=-2mV/Kwascalculatedbased on the carrier distribution in the pn region? And the result of this calculation is that a VOLTAGE change of 2mV/K can compensate this temperature caused change in Ic.Do you understand the meaning of this figure? A reduction in Vbe brings the current Ic back to its original value!That is the main point which - among others -proovesthat Vbe is the quantity which influences Ic. In this calculation, Ib does not appear at all .

author: miguelvp
author_rank: Super Contributor
timestamp: December 28, 2014, 05:15:48 pm
content: Quote from: IanB on December 28, 2014, 04:04:21 pmAlthough some air pushes down due to the angle of attack, most of the lift happens when air particles are accelerated on the top of the air foil causing a difference in pressure from the bottom foil.At least that is my understanding, but I haven't done any airmodels for a long time since I was a kid but I did get my parents to buy me subscription to a airmodel course by post. So I'm rusty since that was a long time ago.

author: c4757p
author_rank: Super Contributor
timestamp: December 28, 2014, 05:23:39 pm
content: Almost everybody in this thread needs to be slapped on the head with the wordscorrelation does not imply causation.

author: dannyf
author_rank: Super Contributor
timestamp: December 28, 2014, 06:01:13 pm
content: QuotePitch can create lift, but it also increases drag thus slows down the airplane which in turn decreases lift.If anything, the up side of a wing's leading edge does push up the air a little.QuoteYes, you can have zero pitch (or even slightly negative pitch) and generate lift - that pretty much says it all about where lift comes from.The same with two boats in parallel motion get sucked together.

author: IanB
author_rank: Super Contributor
timestamp: December 28, 2014, 06:09:17 pm
content: Quote from: miguelvp on December 28, 2014, 05:15:48 pmAgain, the angle of attack. What is this fixation with the angle of attack?

author: miguelvp
author_rank: Super Contributor
timestamp: December 28, 2014, 06:31:53 pm
content: Quote from: IanB on December 28, 2014, 06:09:17 pmIt affects the lift coefficient:source:http://en.wikipedia.org/wiki/Lift_coefficient

author: IanB
author_rank: Super Contributor
timestamp: December 28, 2014, 07:00:21 pm
content: Quote from: miguelvp on December 28, 2014, 06:31:53 pmWell, doh!But why does it keep getting introduced into this thread?No wonder so many have been having trouble with transistors. There is an almost complete inability to marshal facts, to separate the relevant from irrelevant, and to sort out the relationships between the facts that matter.Is it not obvious that there is a parallel between current and voltage in transistors, and between pressure differentials and downward air deflection in wings?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 28, 2014, 09:56:02 pm
content: The fundamental problem is that people follow blindly rather than reasoning it out themselves. Practical experiments can be designed to positively or negatively prove different aspects of working hypotheses. This isn't sociology here, it can be tested and repeated with complete certainty.I presentedexactlysuch an experiment earlier in the thread! Amazingly enough, it was summarily ignored by all!The other fundamental problem is, the people who teach those who follow blindly, also follow blindly. School curricula NEVER say "oh by the way there's more, but we're just teaching you THE SIMPLIFICATION NOW because we don't want to overwhelm you". They just say, "oh, there's no such thing as sqrt(-1), if you get that, you've done it wrong." "OH BY THE WAY THERE'S COMPLEX NUMBERS"Shit like this and it's no wonder everyone seems to unanimously have problems with math, science, any technical subject!Tim

author: Electro Fan
author_rank: Super Contributor
timestamp: December 28, 2014, 09:56:25 pm
content: Thanks for the informative and entertaining thread.I happened to see the article below.The article might just be presenting semantics or it might offer some meaningful distinctions or it might be wrong - I really don't know - in my mind voltage is needed to enable current (so even after reading this thread and the article I don't get how something can be current driven without also being voltage driven) - but I can't claim to have seen any electrons, holes, or other such phenomenon. I just try to observe results, learn from others, study, and hopefully understand - especially when the proverbial light bulb goes on, or off(And after reading this thread and the article I'd say my light bulb is flickering)Seriously, I don't want to heat up the debate but rather just contribute info that might possibly help explain something or at least help a friendly discussion.Happy New Year EEVers! EF---Difference Between BJT and MOSFEThttp://www.differencebetween.net/technology/difference-between-bjt-and-mosfet/The transistors BJT and MOSFET are both useful for amplification and switching applications. Yet, they have significantly different characteristics.BJT, as in Bipolar Junction Transistor, is a semiconductor device that replaced the vacuum tubes of the old days. The contraption is a current-controlled device where the collector or emitter output is a function of the current in the base. Basically,the mode of operation of a BJT transistor is driven by the current at the base. The three terminals of a BJT transistor are called the Emitter, Collector and Base.A BJT is actually a piece of silicon with three regions. There are two junctions in them where each region is named differently ‘“ the P and N. There two type of BJTs, the NPN transistor and the PNP transistor. The types differ in their charge carriers, wherein, NPN has holes as its primary carrier, while PNP has electrons.The operation principles of the two BJT transistors, PNP and NPN, are practically identical; the only difference is in biasing, and the polarity of the power supply for each type. Many prefer BJTs for low current applications, like for switching purposes for instance, simply because they’re cheaper.Metal Oxide Semiconductor Field-Effect Transistor, or simplyMOSFET, and sometimes MOS transistor,is a voltage-controlled device. Unlike the BJT, there is no base current present. However, there’s a field produced by a voltage on the gate. This allows a flow of current between the source and the drain. This current flow may be pinched-off, or opened, by the voltage on the gate.In this transistor, a voltage on an oxide-insulated gate electrode can generate a channel for conduction between the other contacts ‘“ the source and drain. What’s great about MOSFETs is that they handle power more efficiently. MOSFETs, nowadays, are the most common transistor used in digital and analog circuits, replacing the then very popular BJTs.Summary:1. BJT is a Bipolar Junction Transistor, while MOSFET is a Metal Oxide Semiconductor Field-Effect Transistor.2. A BJT has an emitter, collector and base, while a MOSFET has a gate, source and drain.3. BJTs are preferred for low current applications, while MOSFETs are for high power functions.4. In digital and analog circuits, MOSFETs are considered to be more commonly used than BJTs these days.5.The operation of MOSFET depends on the voltageat the oxide-insulated gate electrode,while the operation of BJT is dependent on the currentat the base.

author: LvW
author_rank: Frequent Contributor
timestamp: December 29, 2014, 09:27:17 am
content: Instead of commenting the last reply I like to enclose a pdf file showing a slide created by one of the world leading semiconductor developers: Barrie Gilbert.

author: Zero999
author_rank: Super Contributor
timestamp: December 29, 2014, 01:23:32 pm
content: Quote from: LvW on December 29, 2014, 09:27:17 amThat may be very helpful for designing an analogue multiplier but it's no good for calculating the base resistor value when one wants to use a transistor as a saturated switch to turn on an LED; in that case it's much better to consider the transistor as a current controlled switch. Trying to use the voltage controlled model for a saturated BJT switch will cause big fuck ups "3.3V should be a high enough voltage to turn on the BJT enough to allow 200mA to flow through that motor but fuck the BJT and MCU are now dead and there's lots of smoke!"

author: LvW
author_rank: Frequent Contributor
timestamp: December 29, 2014, 04:21:42 pm
content: Hero999 - thank you for this remark.This gives me the opportunity to state that all of my previous contributions, statements and remarks are related to small-signal operation of the BJT (e.g. as an amplifier) only.

author: G0HZU
author_rank: Super Contributor
timestamp: December 29, 2014, 04:41:04 pm
content: Quote from: LvW on December 29, 2014, 09:27:17 amHi LvMI still don't have a horse in this race. I think 'current control' and voltage control' are both risky labels to attach to a BJT if the winning label comes at the total demise of anything else.At work I'm happy to design using behavioural models like the Ebers Moll or Gummel Poon (or even small signal s2p models) right up to many GHz and for simple stuff I'll just use a calculator.If I had to comment on your Barrie Gilbert slide I'd say that the Vbe equation suggests that the BJT can be modelled as a voltage controlled current source based on numbers bashed into the supplied equation. So the equation suggests that Vbe is the daddy.But under the hood I think there is a lot more to it and the meaning of the word 'control' is open to (mis)interpretation. Maybe a room full of semiconductor physicists could debate this one.But I doubt their answers will make me any better at designing RF amplifiers or other circuits using BJTs. Although it would be a fascinating debate for an onlooker, I'm not sure I 'need' to know what they might say. Even if I understood it all it won't affect the manufacturer supplied models I use and it probably won't change how I design amplifiers or simple switching/control circuits using BJTs. That's why I don't see much value in it all.Voltage 'control' or current 'control'? I'm not aiming to pick a winner on this one. I'd be happier betting onbothhorses to be flawed or limited losers

author: free_electron
author_rank: Super Contributor
timestamp: December 29, 2014, 05:24:39 pm
content: Quote from: LvW on December 29, 2014, 09:27:17 amDo you have the whole document ? the fact that this was presented at a conference on bandgaps leads me to believe this is part of a simplified explanation how to design band gap circuits. Bandgaps work because they send a constant (temperature compensated) current through a forward junction. for a diode at constant current the forward drop is constant (in this case vbe). Thie voltage across the first junction junction is then used to control a second junction. a delta amplifier cancels out the error and in doing so rejects the temperature creating an ultra stable voltage equal to the bandgap of silicon ( around 1.2 volts)in that case the base current is indeed considered a parasitic. you don't care as it is driven from an opamp and set by the emitter resistor in the bandgap circuit. you are after the Vbe in this system. currents in bandgaps are parasites.

author: LvW
author_rank: Frequent Contributor
timestamp: December 29, 2014, 05:30:44 pm
content: Quote from: G0HZU on December 29, 2014, 04:41:04 pmHi GOHZU,I agree with you - nearly up to 100% - as far as you are speaking to me as a designer.And - that`s what I have mentioned several times in this thread (see my posts#148 adn #153):Independent on our "religious attitude" (voltage or current control) - we all are using the same formulas and even the same design steps for designing amplifiers. Hence, it is complete nonsense to say "I am using the current-control or the voltage-control approach". There is only one single method to design a common emitter stage with Re-feedback and a resistive voltage divider.However - the situation is completrely different for somebody who has the task to EXPLAIN the working principle of the BJT to students.Students are not satisfied to hear: "Doesn`t matter - why and how it works, apply a set of formulas - and be happy if the circuit works..".I gave some examples (tempco -2mV/K, Early-effect,...) which can be explained with the voltage control principle only (I intentionally avoid the term "model" which may give rise to misunderstandings). Therefore, I see no reason why I should start with the current-control approach (some people believe it is easier - it is NOT).But at he end of your contribution you have mentioned a very important aspect: ..."the meaning of the word 'control' is open to (mis)interpretation"Yes - that´s true.Let´s take a basic example: An opamp wired as a current-to-voltage converter (current source at the inv. input which has a feedback resistor).My description: We use a voltage controlled opamp to realize a circuit that produces an ouput voltage which is controlled by an input current.In short: The circuit is current controlled, but the active device (of course!) is still voltage controlled.And that is the terminology I have used in my previos contributions: The controlling quantity for the BJT (alone!) is that quantity which directly acts upon the electrical field within the depletion area of the BJT, which in turn determines the current Ic.

author: LvW
author_rank: Frequent Contributor
timestamp: December 29, 2014, 05:43:54 pm
content: Quote from: free_electron on December 29, 2014, 05:24:39 pmNo sorry.But (as given already in my reply#118) here is another statement from B. Gilbert :The old current-in, current-out seems view simple at first, but that's about as far as it goes.We clearly agree that the BJT should be seen in the same way as an MOS device, explaining that the DC base current of the BJTis actually due to a defect (of sorts) and only a nuisance.At Analog Devices we have made BJTs (under special conditions) having a DC beta of over 25,000.

author: Zero999
author_rank: Super Contributor
timestamp: December 29, 2014, 05:51:24 pm
content: VBEcan be a parasitic and just as much of a nuisance when designing an emitter follower. In that case, it would be nice if the BJT had no VBEand was completely current controlled. That way there wouldn't be that nagging VBEto limit the maximum output voltage It would make designing a low drop-out regulator much easier, rather than having to have a common emitter on the output which can be unstable, there could be a nice emitter follower which could saturate close to the positive rail, rather than VBEbelow it.

author: miguelvp
author_rank: Super Contributor
timestamp: December 29, 2014, 06:16:30 pm
content: Re: Barry Gilbert if you notice the slide ruler you will realize that he is using this for analog computing so he only cares about adding logarithmic relations off two BJT transistors.Interesting one hour long talk:Check @9:10 for a glimpse of multiplication using bipolar transistors like a slide ruler will.But with that model you loose linearity but handy for his purpose.

author: LvW
author_rank: Frequent Contributor
timestamp: December 29, 2014, 06:50:24 pm
content: Miguelvp - thank you for providing the link to Barries speech.By the way: The slide under discussion (The magical Vbe) is at the time27:26.(To free_electron: No relation to bandgap circuits.)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 29, 2014, 08:45:04 pm
content: Note that the simplified Eber-Molls equation accounts for the linear range only. The full model accounts for saturation due to the B-C diode (effectively, its own Baker clamp), but at a cursory look, it does not account for charge storage effects, the reduction of hFE in saturation, or saturation resistance effects (Vce(sat) is typically much more than single or double digit mV!).Tim

author: Ian Getreu
author_rank: Newbie
timestamp: January 02, 2015, 05:33:16 pm
content: (This is very similar, in parts, to a post I made in another blog).To answer some of the questions that have been rolling around, the BJT is best thought of as a voltage-driven device. In the common-emitter configuration for example, the Vbe causes the B-E junction to be forward biased. This generates a diode-type current of electrons into the base. The "efficiency" of the BJT is determined by how many of those electrons are swept across the base into the reverse-biased C-B junction and are collected in the collector region (hence the names "emitter" and "collector"). The electrons lost in the base come out the base terminal. The measure of the efficiency is the ratio of collector current to base current (ie the beta). This is why the base must be very thin in order to have a BJT - just putting 2 dioides back to back does not create a BJT."Controlling the BJT with the base current" is not what happens. Even though it appears to be fed by a current, the appropriate Vbe is developed that will support that base current.The point about "don't confuse me with physics, I'm an engineer" (not the exact words but true to the sentiment) is not worth discussing. For any engineer to do a good job, he/she must understand what is being used and how it works.Show me a good designer and I'll show you a good modeler; show me a good modeler and I'll show you a person who understands how it works.This is totally basic to engineering.(This paragraph may be considered as self-serving - if so, I apologize in advance for that).I have written a book many years ago that clearly explains the operation and theory behind the bipolar transistor - without going unnecessarily into mathematics. It covers most of the questions and topics in this discussion - as far as the BJT model is concerned. The book is called "Modeling the Bipolar Transistor" and is available at lulu.com/iangetreu. It is oriented towards the simulation programs like SPICE, so the first half covers all the models from the simple Ebers-Moll to the Gummel-Poon model - concentrating on obtaining an understanding of how the BJT works and how to understand the models that are used with a minimal reliance on equations (though I must admit that the Gummel-Poon model does use quite a few equations). The second half covers how to measure the parameters that are needed for a model. For comments on the book, seehttp://www.electronicspoint.com/threads/bipolar-transistor-spice-modeling-book.222326/I hope I have helped.

author: dannyf
author_rank: Super Contributor
timestamp: January 02, 2015, 07:37:21 pm
content: QuoteI thought you did a great job articulating why it is best thought to be a current-controlled device.

author: dannyf
author_rank: Super Contributor
timestamp: January 02, 2015, 07:38:34 pm
content: QuoteWhat if you drive the base with a current source? Would that cause the underline model to be "current controlled"? If not, why?

author: IanB
author_rank: Super Contributor
timestamp: January 02, 2015, 07:53:48 pm
content: Quote from: dannyf on January 02, 2015, 07:37:21 pmQuote from: dannyf on January 02, 2015, 07:38:34 pmI think your trolling skills are very weak. Must try harder.

author: miguelvp
author_rank: Super Contributor
timestamp: January 02, 2015, 08:15:57 pm
content: And I was so proud of myself for what it seemed the impossible task to stop this thread, but LvW had to PM someone's 4 and a half year old post to revive this.To no fault of Ian Getreu, btw welcome to the forum.

author: Zero999
author_rank: Super Contributor
timestamp: January 02, 2015, 09:59:33 pm
content: Actually I thought danny had a point there.In reality the current controlled vs voltage control debate is all moot. If it's power switching then the BJT should be treated as a current controlled switch but for signal the voltage controlled current source often makes more sense.If you think we should always use the voltage controlled model when what voltage is safe to apply to the base to saturating a transistor with no current limiting resistor? You can't just connect the base to the output of an MCU, even at 3V and not expect anything bad to happen.Oh and Ian Getreu,Why did you sign up here? To sell more copies of your book?

author: miguelvp
author_rank: Super Contributor
timestamp: January 02, 2015, 10:13:13 pm
content: Quote from: Hero999 on January 02, 2015, 09:59:33 pmIf you follow the link he gave, you'll realize that LvW sent him a PM just today, so I guess he dragged him here.http://www.electronicspoint.com/threads/bipolar-transistor-spice-modeling-book.222326/

author: Zero999
author_rank: Super Contributor
timestamp: January 02, 2015, 10:50:03 pm
content: Quote from: miguelvp on January 02, 2015, 10:13:13 pmThat's sad. He felt like he had to bring one of his friends over to support him and keep this shitty thread gong.

author: dannyf
author_rank: Super Contributor
timestamp: January 02, 2015, 11:10:55 pm
content: QuoteAppeal to authority is the weakest of all arguments one can possibly make.

author: Ian Getreu
author_rank: Newbie
timestamp: January 03, 2015, 01:05:53 am
content: Looks like I was invited into quicksand.My apologies - I will not be contributing any more.

author: wiss
author_rank: Frequent Contributor
timestamp: January 03, 2015, 01:11:47 am
content: I'm still popping popcorn. ..

author: miguelvp
author_rank: Super Contributor
timestamp: January 03, 2015, 01:27:38 am
content: Quote from: Ian Getreu on January 03, 2015, 01:05:53 amNo apologies needed, We love for you to stick around regardless of the reasons someone invited you over.We don't get many people of your caliber in here.

author: dannyf
author_rank: Super Contributor
timestamp: January 03, 2015, 01:32:08 am
content: QuoteNo reason to run just because someone disagrees with you.Take the challenge, and let your facts speak for you.

author: c4757p
author_rank: Super Contributor
timestamp: January 03, 2015, 01:36:48 am
content: Quote from: dannyf on January 03, 2015, 01:32:08 amThis godforsaken thread is as good a demonstration as any that no matter what facts you present, no matter which side you're on and how well you represent it, any time you try to make an argument in favor of one perspective on something over another, you're soon to be piled upon by a hundred combative numpties with free time and an axe to grind about nothing in particular.Please, do, stick around, but don't feel the need to contribute to this endless back-and-forth. We've had this argument countless times before and it's not going anywhere.And dannyf, do stop trying to provoke people, please. It's pathetically transparent.

author: free_electron
author_rank: Super Contributor
timestamp: January 03, 2015, 03:07:33 am
content: I just found the root of the confusion this topic.Some people talk about transitors , others about transistors ...

author: LvW
author_rank: Frequent Contributor
timestamp: January 03, 2015, 09:28:19 am
content: Quote from: Hero999 on January 02, 2015, 10:50:03 pmDo you really think that I need more support - in addition to Berkeley, Stanford, Horowitz-Hill, Barrie Gilbert, and finally: W. Shockley (doctoral thesis)?

author: dannyf
author_rank: Super Contributor
timestamp: January 03, 2015, 01:08:13 pm
content: QuoteWhat others think doesn't matter. Do you think you need more support?

author: LvW
author_rank: Frequent Contributor
timestamp: January 03, 2015, 01:31:53 pm
content: Quote from: dannyf on January 03, 2015, 01:08:13 pmNo - I am still waiting for a technical verification for the "super-contributor`s" view.

author: dannyf
author_rank: Super Contributor
timestamp: January 03, 2015, 01:57:43 pm
content: QuoteThen don't fuss about it.

author: LvW
author_rank: Frequent Contributor
timestamp: January 03, 2015, 04:05:25 pm
content: Hello to all,maybe some members of this forum couldn`t care less - but this is my last contribution in this thread. More than that, I have decided not to further continue answering questions in this forum, which is entitled as „the leading place to discuss ...advanced technical questions on any aspect of electronics“.Wow - good title.I have entered the forum in April 2014 (up to now: 150 posts) with the imagination that it could be interesting and valuable to seriously discuss techical issues with competent persons - on a fair and objective basis and with some respect against other opinions. However, I must admit that I was too optimistic. But - fortunately - the internet provides some alternatives.Here is the core of the problem of this thread(post#1):„When applying a current to the base pin to allow the current to flow from the collector to the emmiter(or vice versa) where does the current from the base pin go?...I haven't crossed any text that explains this clearly.“And this was my answer:However, you always should know what you are doing and, thus, realize that this is a model only. The physical reality is that the BJT is, of course, a voltage-controlled device (Ic=f(Vbe)). This is a proven fact.Because I knew that some textbooks and other contributions just state the opposite (the value of Ic would be determined by Ib) I was not surprised that some people didn`t agree with me. However, I was surprised about the way how these persons have participated in the „discussion“: Not with technical arguments/explanations but with polemic and personal attacks.I think, roughly 5 or 6 contributors - more or less - were supporting my view (which was supplemented with many technical examples, effects and explanations) and another 5 forum members maintained on the contrary.In principle, no problem - a good starting point for exchanging arguments.But, interestingly, all „defenders“ of the current control principle were not able to justify their opinion.Their claims can be summarized as follows (my words):The current Ib does exist - hence, because of the nice equation Ic=B*Ib, it is the current Ib that controls Ic.More than that, they even refused to provide technical comments on circuit examples and prooves I have mentioned (at least, in my view). Instead, I could read sentences like:„Any attempt at answering your question is irrelevant to the discussion; you don't seem to be able to understand it on a device level, nor a circuit level; you don't have the ability to comprehend the technical details; this debate is silly; pissing contest“.Other nice comments were:„nonsense; complete lack of understanding of what a "current control" vs. "voltage control" mechanism means; splitting the atom if that is really what your on about; such quotes showed your complete misunderstanding of the quoted items; it clarifies nothing other than your stubbornness“.Apart from such personal attacks, the biggest surprise is that some forum members (engineers?) do not notice their own contradictory behaviour. They are designing a classical amplifier stage (RE-feedback, voltage divider base biasing), without able to realize what they are really doing:* providingvoltagefeedback (although one of the specialists claimed that it was current feedback),* providing a stiff base biasing with avoltage,* designing a stage with voltage gain that isindependenton the BJT`s beta value (but only on the transconductance gm); this even holds without any feedback stabilization. Did they ever notice this effect?They remember the Ic=f(Vce) set of curves (parameter Ib) and believe, that these curves can proove that Ib determines Ic.And they do not understand that theslopeof these constant Ib curves (Early effect) is one of the clearest proof for voltage control.Instead, they still think (blindly) that they follow the current-control model without knowing what they really are doing.Surprising, funny and crazy.Why? Because they seem not to be able to critically review their own views.Perhaps they are good technicians/designers - but a good engineer knows what he is doing - and why! - and he is always able to explain the various effects he is observing.OK - that was my final word (Longer than intended, sorry for that).Happy New Year to all of you.LvW

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 03, 2015, 04:10:50 pm
content: Quote from: LvW on January 03, 2015, 04:05:25 pmGeez... and this ain't even Usenet...You're really letting the trolls win, today.Have a beer (or whatever you prefer to relax with), forget this thread exists, nevermind that some people are assholes (a possible theorem is that all are, even ones' self, which may have illuminating philosophical value), and go on tinkering with your projects.Cheers,Tim

author: Zero999
author_rank: Super Contributor
timestamp: January 03, 2015, 06:08:09 pm
content: Quote from: LvW on January 03, 2015, 04:05:25 pmThen there are the rest, like me who think this is a bullshit debate and that both the current and voltage controlled models apply, depending on how a BJT is used.You've even said it yourself that the voltage control model is only useful at signal level.Quote from: LvW on December 29, 2014, 04:21:42 pmWhat I don't understand is if you don't like personal attacks, then why do you keep posing in a thread which results in more personal attacks? This thread started to die out a couple of days ago and you could've left it but you decided to invite a friend and revive it, then only to complain about more polemic attacks?Why torture yourself?

author: free_electron
author_rank: Super Contributor
timestamp: January 03, 2015, 09:41:12 pm
content: Quote from: LvW on January 03, 2015, 04:05:25 pm??!?? what ? Ie = Ib + Ic . Any textbook on transistors will tell you that. where the hell did that question come from...ah, i see original question.. man this thread has been taking some twist and turns and we are now waaaay off the original question.QuoteThe collector current is plotted in function of the base-emitter voltage , which is essentially a diode curve ( If vs Vf ). Once you have left the 'nik' in the curve ( to non linear section where the thing goes into conduction , you a have a relatively linear curve.So you could say by modulating the Vf ( Vbe in case of transistor) you control the If ( Ic in case of a transistor )Let's go back to the electron model ( forget the equations for now . this is elemental reasoning)take an NPN Transistor in rest .I apply an electron source between emitter and collector. Base is tied to emitter. I can try to accelerate the electrons by increasing the field strength ( the 'voltage'). In reality there may be a few stray electrons ( leakage ) and eventually the field per distance will get so large it will destroy the barriers and flash over with a fried lump of silicon as end result.so, practically speaking no electrons flow ( electrons are charge carriers. flowing electrons are moving charge. moving charge is called current )the structure between collector and emitter is NPN. two back to back diodes. we all know that diodes conduct current only in one direction. ( doesn't matter if we are conventional or electron model. ) put two back to back and you cant get anything through. no matter what polarity you apply one will decrease its recombination zone , the other will grow it.nothing contradictory in my statements.now. i will unclip the base from the emitter and connect to a second electron pump. i will slowly increase the field strength in an attempt to get electrons flowing. when i hit a certain field strength i will succeed in sending an electron across the recombination zone between emitter and base. if i keep increasing the field strength i will send more and more electrons into the base , removing the recombination zone there. This 'p' region now becomes flooded with electrons ( keep in mind this p region is very small in relation to the n regions of the collector and emittor. you cant just take one cubic meter of p material , a cubic meter of p and another cubic meter of n material and get a transistor. that doesn't work. even if you scale it down 3 decades to 1 millimeter of n , 1 millimeter of p and one millimeter of n again , it still wont work as a transistor. it only starts working at the micron level .so , the base region is now flooded full of charge carriers running from emittor to base. this includes the part of the base region that touches the lump of collector material we called collector. the emitter-base field accelerates the electrons and some will start skipping the base ( which for them is a right angle turn so to speak ) and flow into the collector.the conductivity of the base material is less ( the doping) than the collector.This creates a relation between the number of electrons that go into base and the ones that go into collector. The emitter delivers a certain amount of electrons per second. assuming this is constant there will be a ratio between the ones that go into the base and the ones going into the collector. It is a self regulating mechanism . If too many go into the collector, the ones flowing out of the base decreases , starting to rebuild the recombination zone , so now less can enter the collector and more come back to the base. ( remember that our electron source delivers a constant amount of electrons per time unit. ) That is why flow of electrons in the base controls the flow of electrons in the collector.Now you can try to slap names and labels and equations on that and you can write things in terms of applied field strengths or in terms of the ration of electrons. You can call these 'lost electrons' ejected through the base a parasitic current and you can do all kinds of other mathematical trickery. All that stuff works fine. some like it in terms of field strength , some like it in terms of coulombs per second.One fact remains: that if no charge flows from emittor to base FIRST , that recombination zone between emitter and base does not disappear and electrons don't even get a chance to go into the collector. There will be no charge flowing in that pathway. the regulation mechanism is determined by the doping in the material .So my statement remains : no base current is no chance of collector current .Sure you can play with the field strength and write your equation in terms of the base emitter field. But the fact remains that it is this field that accelerates the charge carriers from emitter into base. if that charge does not flow from emitter to base , no charge will flow from emitter to current. moving charge is called current. negative charge carriers are electrons.If you want to write stuff in terms of voltage control you will see that there is an 'offset' in your equations. That 'offset' voltage is required to get the current flowing.Ic = f(Vbe) provided vbe > the field required to get it in conductionyou can eliminate the offset in calculations by switching to current model. if ib is zero then ic is zero.you pick your poison.

author: c4757p
author_rank: Super Contributor
timestamp: January 03, 2015, 09:49:12 pm
content: Quote from: free_electron on January 03, 2015, 09:41:12 pmRepeat after me: "correlation does not imply causation".Correlation does not imply causation.Correlation does not imply causation.Bloody hell...

author: baljemmett
author_rank: Unknown
timestamp: January 03, 2015, 10:15:19 pm
content: Quote from: LvW on January 03, 2015, 04:05:25 pm… which, you might notice, has nothing whatsoever to do with the actual question asked. You took a question which could have been - and in fact was - answered within a couple of posts and turned it into yet another tedious rehash of a discussion that has been had many times already, and now you're spitting the dummy because you don't like the behaviour of some of the local argumentative bores. Congratulations.

author: free_electron
author_rank: Super Contributor
timestamp: January 03, 2015, 10:17:51 pm
content: Quote from: c4757p on January 03, 2015, 09:49:12 pmi throw a frozen pea at a jar of peas and nothing happensi throw accelerate the same frozen pea supersonic at the jar and it makes hole allowing other pea's to roll out.i apply a field to a base emitter of 0.1 volt. nothing happens.i apply a field sufficiently large to get an electron to flow from emitter to base. i have now created the hole for an electron to go from emittor to collector.you can sit there until doomsday : as long as that initial pea does not make the inital hole you have no flow of pea'swithout that initial base current , no collector current . yes you need a field to get electrons to flow. fortunately for us we only need about half a volt. if the force required to get electrons in valence band was a million volts it would have been bloody hard to make bipolar transistors.correlation indeed does not imply causation.case in point : base resistor 0 ohms. i apply 0.1 volt and nothing happens. i apply 0.5 volts and get collector current. i apply 0.6 volts and i get a larger collector current. i plot this out as a curve.Vbe vs Ic. cause- effect corellation. great.now. put a teraohm resistor in that base path. apply same 0.5 volts. see how much current you get in that collector. oops... your carefully plotted vbe /ic curve just went to snot.you will undoubtedly argue that the 0.5 volts stands over the teraohm resistor... fine. Ohms law says that for that voltage to be there there must be a current flowing through it.Do two things :1) figure out where that electron that creates a voltage across the base resistor came from.2) figure out why , the drift speed of that electron ( which is what resistors do ) has an impact on the flow in the collector. according to you it was purely field induced...

author: G0HZU
author_rank: Super Contributor
timestamp: January 03, 2015, 10:30:44 pm
content: QuoteAn electric field is defined as an electric force per unit charge. eg it has units of newtons per coulomb (or volts per metre)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 04, 2015, 12:40:39 am
content: Quote from: c4757p on January 03, 2015, 09:49:12 pmNow now. Back in the days of germanium transistors,negative base biaswas occasionally required. And yes, I mean negative in the sense of turning off, not just a technicality because germanium transistors were almost exclusively PNP.Quote from: free_electron on January 03, 2015, 10:17:51 pmCurrent still flows, even if you can't sense it; for 1pA = IS (typical within orders of magnitude for a 2N3904), one should expect Ic ~= 2nA at Vbe = 0.1V.It is true that hFE is severely reduced at low Ic (due to recombination), but nothing causes Ic to reach zero unless Vce also goes to zero.Leakage current stops at Tabs = 0 K, but so do the dopants, so you can't have a functional non-leaky BJT.Tim

author: G0HZU
author_rank: Super Contributor
timestamp: January 04, 2015, 01:18:45 am
content: In my day the recommended learning method for stuff like this was to listen to an experienced lecturer at college/university and to also read the (recommended) literature on the subject. This was a fairly safe and reliable way to learn.Two pages back we had both of these sources of knowledge on this very thread. Now they are gone...Today we also have the internet and the casual and curious googler can gain access to all kinds of subject matter in a few seconds. It is a truly remarkable place to share knowledge. However, the risk here is that the subject matter they find (and believe) may be controversial or at the very least unreliable and I'm not sure it is the safest method of learning or passing on information to beginners.Trying to debate and learn stuff on a hobby forum like EEVblog is also fraught with risk because the experts are massively outnumbered and not easy to spot amongst the pretenders or 'google professors' who are very keen to use their keyboard and show off.A little bit of knowledge can be dangerous

author: edavid
author_rank: Super Contributor
timestamp: January 04, 2015, 01:57:03 am
content: Quote from: T3sl4co1l on January 04, 2015, 12:40:39 amNow too - it's also true for silicon, at a high enough temperature.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 04, 2015, 02:15:27 am
content: The democracy of information is at once both liberating and stifling.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 04, 2015, 02:16:22 am
content: Quote from: edavid on January 04, 2015, 01:57:03 amIndeed, or at a low enough current one would suppose (Iceo vs. Icbo should be different by approx. hFE times, for whatever hFE is at that current).Tim

author: free_electron
author_rank: Super Contributor
timestamp: January 04, 2015, 06:28:33 am
content: Of course there is thermally induced current.That leakage is caused because temperature accelerates electrons. If they start movi g you get current.It's the same thing with photodiodes or transistors. Expose the base region (actually the base collector region, the emitter is not sensitive) to light and every photon will knock of an electron creating an electron hole pair. The recombination creates a base current and off we go.Every transistor is light sensitive. Every junction is light snesitive for that matter. Photons produce free electrons that start moving. Moving electrons is charge displacement . charge displacement is current. How much more clear does it need to be made ?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 04, 2015, 08:45:34 am
content: I'm not sure offhand how sensitive (or not) the emitter is, actually; I know it's not usually targeted (for obvious structural reasons). I would think it would be just as effective, as long as the light is absorbed within a few diffusion lengths of the B-E junction. Photodiodes are made with a PIN structure to enhance the drift / depletion zone width, making absorption more likely to yield useful photocurrent at the terminals.But be careful what you say; earlier you said "no chance of current", which I took to mean, NO current period, zero. Yet you just said "of course there is thermally induced current", which means nonzero current. As with the majority of this thread, it's a perfectly legitimate approximation to make in many circuits, but there are many others which will fail if this factor is forgotten. And if you go on wording such statements as absolute fact, when they are actually highly conditional approximations, you will find many more problems, both technical and interpersonal, in your life!Tim

author: free_electron
author_rank: Super Contributor
timestamp: January 04, 2015, 01:05:38 pm
content: Photodiodes are pin co struction indeed. (Most of them, the ones that are not are photovoltaic cells)That is the reason you need to reverse bias those things. The intrinsic layer creates a larger depletion zone. If i remeber ot right it is the depletion zone that absorbs the photons to get a current going. That intrinsic lump of semiconductor makes a 'thicker' junction thus it becomes more sensitive.Same goes forregular pin diodes used for switching rf. You bias those things in reverse to turn em off. The thick junction caused by the intrinsic material means that the plates of the parasitic capacitor are very far away . Instead of a few tens of atoms in a regular p-n recombination zone, they are now the width of the intrinsic material. That means these diodes have extremely low capacitance so no rf energy will leak through. They are susceptible to charge injection.In the harddisk preamps we use an array of pin diodes to switch gain and bandwidth of the amplifier (switching resistors and caps essentially) i once had a problem whereby after a few seconds the gain would suddenly drop. If i turned off a certain block in the chip this would not happen. We speculated there was leakage somewhere. I pushed the setup i to the probe station , turned off the lights in the probe room and targeted the laser, at very low setting, to the bank of pin diodes. After a few minutes iof prodding with my 'light probe' i could reproduce the effect. I had found the diode that produced that exact drop. Turning on the illumination of the microscope i looked around at adjacent components and noticed that there was no via ring around a control structure. Turn off the light again and aim at the control structure. By modulating the laser i tensity i could create charge accumulation and let it dissipate again . This caused the pin diode to react and show the gain behavior we saw on bench.It turned out that, under certain operating modes (these are bicmos products) the base of a bipolar transistor went floating as the cmos driver tristated when the control block was turned off.That foating base picked up stray electrons and became conductive enough to toggle the pin diode.So we put a few gigaohms of resistance between the base and ground and the problem was solved. (We actually made the bottom drive mos 'leaky')

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 04, 2015, 02:10:54 pm
content: Well, quantum efficiency doesn't change much with bias (having the N/P encroaching on the intrinsic section doesn't really prevent charges from finding their way, there's still a built-in potential). Photocurrent is generally written as independent of bias. This seems to be reasonable, even in the forward direction (PV panels have a squarish V-I curve; basically the only reason current drops is because it's shunted internally by forward I = Is*exp(V/Vth) current). It does do *wonders* for capacitance though!Heh, nice story. Lesson learned: don't leave nodes floating at undefined voltages! They'll inevitably do something unexpected or undesirable. Hope it wasn't a gotcha, like, from a bad appnote... those are just mean.Surprised light did much around ICs, unless there was, like, glass body diodes or something. Are PIN diodes usually in glass format? Were those..?Think I measured this before, a 1N914 is so-and-so, I think around 1nA at room temperature and ambient light levels; rising to 2-4nA under the influence of my (admittedly pretty intense) LED flashlight at point blank range. An effect not attributable to temp rise, because it doesn't vary over time (a valid concern, because I can physically feel the warmth from the LED when turning it on and off near a sensitive region, like my lips).On the other hand, I've got a circuit with 2N3904/6 collectors driving a capacitor; when both are fully off, leakage is in the ~pA range, for a voltage drift in the 10mV/min range. At least... at room temperature it is.If you need a really good diode, I've heard a plain old 2N3904 is at least as good as those so-called picoamp diodes (PAD-1 etc.?). Go figure, putting the tiniest bit of effort into making something other than a diode and you get a wonderful diode...Tim

author: c4757p
author_rank: Super Contributor
timestamp: January 04, 2015, 02:22:42 pm
content: Quote from: T3sl4co1l on January 04, 2015, 02:10:54 pmI get the feeling that these were allinthe IC - perhaps with a transparent top for just this purpose?

author: SeanB
author_rank: Super Contributor
timestamp: January 04, 2015, 02:35:50 pm
content: Under the right circumstances silicon junctions emit light, as they are basically a poor LED. They thus can emit light inside the package which, will affect the other sensitive junctions nearby in some cases.As well most packages these days are thin enough that light can penetrate, though this is likely to be very little in typical use. Might be an issue if you have the package in direct sunlight out in space, but otherwise hardly an issue.However Vince was likely using a development chip, made in a ceramic package and likely without a lid, so that they could probe it electrically during development. In final use it likely was in a QFP chip on a board, but there it was a LCC carrier. Having an onboard photodiode is used in some secure chips to defeat probing after decapping, if current flows through the photodiode the chip will kill itself.Glass encapsulated diodes do however make very good if poor quality photodiodes. This is a problem in high impedance use where you might find high leakage or noise if the case is opened.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 04, 2015, 02:41:31 pm
content: Quote from: SeanB on January 04, 2015, 02:35:50 pmMmm, it's not like an LED (carrier recombination) -- actually, I don't remember what the mechanism is! It's an avalanche phenomenon, as far as I know, on the rare occasions when this actually occurs.In a dark room, a 2N3055 with the top cut off is just visible with ~100mA E-B (it'll drop about 6V and start to heat up..). It's kind of a yellowish green. I wonder if it's a continuum spectrum or if it has peaks.One of the more esoteric bits of semiconductor lore, indeed; supposedly, Pease had a riddle, whereby, in a circuit with absolutely no oscillation and no negative voltage to start with -- just resistors and transistors -- a negative voltage, small and feeble, but negative nonetheless (or if you prefer: a voltage beyond either supply rail -- same thing, upside down), can be produced.Tim

author: SeanB
author_rank: Super Contributor
timestamp: January 04, 2015, 02:46:05 pm
content: Try your 2N3055 using a phone camera which is sensitive to IR light..........

author: dannyf
author_rank: Super Contributor
timestamp: January 04, 2015, 03:55:31 pm
content: Not even 3055: it is true for most power amps. And if you google, you will find those pictures of glowing emitters.It is a well known phenomenon.

author: free_electron
author_rank: Super Contributor
timestamp: January 04, 2015, 04:31:39 pm
content: yes this was all inside the same chip. this was a prototype for a new harddisk preamp we were designing. remember that in my previous life ( up till 3 months ago ) i spend 23 years mucking around in silicon ... i have SEEN electrons flow.. ( e-beam prober . you could measure electron density with that thing. discarge an internal cap through a node and it would show you how many electrons were now in the touched area. you say the number decay over time. )I'm nowhere near the levelof the peopl that actually design the circuits or design the transistors , but i can find my way around a chip and debug the darn things. gimme a probe station ,a few picoprobes, a green / uv laser and a darkroom and i'll prod around in there like the best. i've had setups where is nipped off gainstages ,b rought out the inputs, outputs and control lines and measured a buried diff amp with programmable gain. 7 or 8 needles in contact. requireing drilling through multiple metal layers, cutting passivation , metal , poly. i've done it all. i do understand the principles how electrons travel , how to create effective dams , know they have a tendency to roll where we don't wan t them. i can spot a floating well a mile away ... i can show you where the latchup occurs by smearing liquid cristal on the bare die , exposing it to polarised light and pulsing the chip in latchup so i create a hotspot. liquid cristal changes its spin under temperature. so the polarized light would not bounce back up when cold , but come back when hot. by carefully lettingthe chip go in latchup and cut the current before thermal runaway let the whole thing heat upyou would get a flashing beacon of light showing you where exactly the parasite sat that caused latchup. and then you could go sniff out the well that was not contacted.the chip was in standard black epoxy body and had the problem that after a few seconds of operation it's gain would suddenly drop. this coincided with the turning off of the configuration block. turn off tha particular section and you could time it with a stopwatch...so i decapitated the chip shoved it under a probestation , made the room dark and started prodding with a 'light probe' in the form of the laserbeam under the microscope ( the probestations have a resolution down to 200 nanometer so i can see the structures clearly )there was metal over the structures but a few blast of the laser drilled a hole down to the junction area. turning of the microscope illuminator but leaving the laser illuminator running gave me a spot . by playing with the laser aperture array i can create a rectangle. setting the rectangle slightly larger than the size of the intrinsic area ( i had a sun workstation next to me with the chip layout so i knew what i was looking at ) i could toggle the diodes on and off trying to find out which one was toggling ( i could not turn the control block on because the control block powered solved the problem. the problem only starte a few seconds after the control block went in standby. so i used light to toggle the diodes. once i foudn the one that caused exaclty that drop in gain : microscope light on , look at image , look at layout on the workstation : click the node and up comes the schematic. follow the drive patway for that diode and find it hits a bipolar transistor.. the base drive was in standby. so you have a floating base at that point. that stuff is so high impedant ( teraohms ) in combination with the base capacitance that any stray electron will tickle the transistor in conduction.so we could do mulitple things :- eliminate the stray electrons by building a wall of ground (substrate) via's- add a pull down resistorall of which would require layout changes to make room for that stuff to be placed.. that meant full maskset . at a million dollar a pop , 12 masks .. ain't gonna happen.so we altered only one mask to alter the doping of the lower mos so he became 'leaky' essentially a mos with a resistor path in parallel. you can't really turn the mos off completely. that fixed the problem.this was an oversight of the people designing the control block (pure digital , generated from verilog and plonked down) and the people doing the analog block. when the circuit was connected the analog guys did not know the digital block would go in standby and become tri-stated. the digital guys. well those only know about ones and zeroes . if you show them a resistor they panic ...so the transistor had a base series resistor , but no pull down ...not a true problem a , no current means no transistor effect. except in this case there was a leakage current. if that had been trapped by substrate contacts no problem. in this case the electrons went into the base and the damn thing turned on... it took a while to reach the correct biassing but once there wer eneough going in all hell broke loose.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 05, 2015, 04:57:44 am
content: Cool stuff!

author: atferrari
author_rank: Frequent Contributor
timestamp: January 05, 2015, 09:41:49 am
content: Not sure if cool...but close to sci fi for me!Wondering what somebody would say to those that commited the "mistake"? Is it? Can you actually blame them?

author: free_electron
author_rank: Super Contributor
timestamp: January 05, 2015, 04:41:54 pm
content: Quote from: atferrari on January 05, 2015, 09:41:49 amno. it is an oversight that wasn't even caught during design review. shit happens. deal with it. that is life. only when playing with 'refined sand' mistakes become very expensive ... but that is life

author: SeanB
author_rank: Super Contributor
timestamp: January 05, 2015, 04:57:59 pm
content: Even Intel has them..... Look at the errata and you will see functions that are marked as "do not use" with various steps and such.

author: atferrari
author_rank: Frequent Contributor
timestamp: January 05, 2015, 05:38:56 pm
content: Quote from: free_electron on January 05, 2015, 04:41:54 pmI see.BTW, how a mask comes to cost 1 million?What is the essential reason for it? Aren't they just implementing what somebody else (you in this case) has designed already? Shouldn't their work be somewhat an ordered sequence of well defined steps? Sorry but you see I know actually nothing of all that.

author: atferrari
author_rank: Frequent Contributor
timestamp: January 05, 2015, 05:42:09 pm
content: Quote from: SeanB on January 05, 2015, 04:57:59 pmAt my level, when I print the datasheet of a new micro I intend to use, I look first for the last errata. Learnt how convenient it is, the hard way.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 05, 2015, 06:32:38 pm
content: Masks have always been expensive because they are engraved, etched, deposited or etc. in a very slow sequential method. Once made... they're easy to use, because they print a complete layer for an entire design in one shot, and that shot is exposed and stepped over the wafer very quickly.I'm guessing, masks have always been in high demand so production has always been expensive, but the capital equipment to produce it has always been even more expensive, hence the scarcity of supply as well. (And the supply of THAT in turn, well... same idea?)Tim

author: photon
author_rank: Frequent Contributor
timestamp: January 05, 2015, 09:37:17 pm
content: Quote from: free_electron on January 05, 2015, 04:41:54 pmEngineers definitely do get blamed for mistakes like this. For a chip with million dollar mask costs, more than half the engineers on the team are doing verification. More than half the engineering cost of the chip is spent to prevent bugs like this from happening. If there were a floating net this would be very embarrassing for the verification team to explain how their test plan missed it.On the other hand, no chip is 100% bug clean on first silicon. I have heard engineers brag about how their chip was perfect, but I tend not to believe them. I have been on the team of about 20 large asics and none of them were perfect on first silicon.When the budget of a chip is made, a part is set aside for one metal-only spin. First silicon is a declared to be a success if the number and severity of the bugs is small enough that parts can be given to first customers. You will see the terms used like "the part is now sampling, production shipment in the next quarter."

author: free_electron
author_rank: Super Contributor
timestamp: January 05, 2015, 11:10:56 pm
content: Quote from: T3sl4co1l on January 05, 2015, 06:32:38 pmthe mask is an optically flawless quartz plate with chrome sputtered on.an ion-mill then traces the shapes where UV light has to get through. ion milling is done in a vacuum chamber. they ionize some corrosive material ( i don't know what they use to 'eat' chrome ) and then steer a beam of the ionized gas much like a photoplotter would do. this happens at a scale larger than what is ultimately needed.only a few companies in the world have the equipment to do that. Dupont is by far the largest maskmaker in the world.

author: photon
author_rank: Frequent Contributor
timestamp: January 08, 2015, 09:11:19 am
content: Quote from: Dave on December 25, 2014, 05:44:43 amYes, so interesting I calculated the closed solution as learned in EE circuits class using a ti89 calculator.I wrote current equations rather than voltage equations since the arithmetic is easier. Using KVL and KVC I get:(I1 + I2)/sC1 + I2/sC2 + RI2 = 1/s(I1 + I2)/sC1 + I3/sC3 + RI3 = 1/sI1 = I2 + I3Substituting:I2(833333 + 1000s) + I3(500000) = 1I2(500000) + I3(1500000 + 1000s) = 1Solving for I2 and I3:I2 = (s + 1000)/(1000s^2 + 2333333s + 9999999500)I3 = (s + 333.333)/(100s^2 + 2333333s + 9999999500)Expanding in partial fractions:I2 = .000639/(s + 1767.59) + .000361/(s + 565.471)I3 = .001193/(s + 1767.59) + .000193/(s + 565.471)Finally, inverse Laplace transforms gives the solution:i2 = .000639(e^(-1767.59t)) + .000361(e^(-565.471t))i3 = .001193(e^(-1767.59t)) - .000193(e^(-565.471t))Switching over to voltages and doing more of the same gives:v1 = .51822(1 - e^(-1767.59t)) + .148478(1 - e^(-565.471t))v2 = .120503(1 - e^(-1767.59t)) + .212695(1 - e^(-565.471t))v3 = .67493(1 - e^(-1767.59t)) - .341146(1 - e^(-565.471t))vn2 = .639e^(-1767.59t) + .361e^(-565.471t)vn3 = 1.193e^(-1767.59t) - .193e^(-565.471t)I attach the plot below to show that they agree with your plots and that the arithmetic is correct.The good people at Mathworks want about $3000 for a Matlab license. One must have a good business using it in order to justify that kind of expense. Yes, LTSpice is an unbelievably good and free tool. And so easy to use. The above calculation is long and tedious, as you mentioned.I took the circuits class a long (35 yrs) time ago. Back then it was one of the killer classes you had to pass to get your degree. Is it still so important today? It's old (goes back to Heaviside and about 1890), not very helpful in understanding the circuit and not used much, if any, in practice.

author: rsjsouza
author_rank: Super Contributor
timestamp: January 08, 2015, 12:34:38 pm
content: Quote from: photon on January 08, 2015, 09:11:19 amThat reminded me of how many equations and programs I had in my HP48SX back in university... These are pheonomenal calculators.Quote from: photon on January 08, 2015, 09:11:19 amInstead of Mathworks' package, there is alwaysOctave.

author: Syntax_Error
author_rank: Regular Contributor
timestamp: January 08, 2015, 11:52:27 pm
content: Circuits 1 is a common "first class" after digital circuits in many university curricula. It is presented early on, with frequent assertions that this is simple; just wait until you get to E&M field theory later, etc.It's interesting to hear the different perspective on this, coming from "the industry" rather than academia.

author: IanB
author_rank: Super Contributor
timestamp: January 09, 2015, 12:59:18 am
content: I was taught how to solve differential equations analytically using Laplace transforms in the first months of my engineering degree, but it is not something I have found to have much practical use in the days of powerful computers. Many (most?) systems are non-linear, and for any non-trivial system a numerical solution by computer is quicker and more convenient.In fact we were taught many "hand calculation oriented" design methods that are no longer in common use by modern engineers. The days of charts, nomographs and tables for detailed design are largely history now, and find a use mainly for estimation and for getting a feel for a solution.

author: Syntax_Error
author_rank: Regular Contributor
timestamp: January 10, 2015, 04:15:25 am
content: Quote from: IanB on January 09, 2015, 12:59:18 amI did read all of what you wrote, but I wanted to key in on this to ask you a question: How on earth is this taught in the first few months of a college education? Did you learn the prerequisite mathematics (differential and integral calculus, and differential equations) prior to university, in high school/secondary school?This somewhat blows my mind. Not that this is unlearnable, not at all. But thinking back to my high school education, and what I was capable of as a teenager, oh boy...

author: IanB
author_rank: Super Contributor
timestamp: January 10, 2015, 07:33:20 am
content: Quote from: Syntax_Error on January 10, 2015, 04:15:25 amYes, I had been introduced to differential and integral calculus and differential equations in secondary school. I had studied A level mathematics, which is a prerequisite for acceptance into an engineering degree program. The A level syllabus in the link is representative of the ground we covered in high school before reaching university:http://www.cie.org.uk/images/92083-2014-syllabus.pdfIn the first year of university we were given a core mathematics course which was both a revision of what we were already presumed to know and the introduction of new material relevant to the study of engineering. The new material went into a lot more depth around differential equations than I had encountered before.

author: dannyf
author_rank: Super Contributor
timestamp: January 10, 2015, 12:42:41 pm
content: QuoteThere are differential equations, and then there are differential equations -> they come in different shape, form and complexity.The simplest differential equations you would have solved in a typical highschool math class, like dx/dt = C, or the maximization of a quadratic equation, or free fall of an object....Then the overwhelming majority of them have not been solved, or cannot be solved in closed form.So I wouldn't take that comment too seriously.

author: dannyf
author_rank: Super Contributor
timestamp: January 10, 2015, 12:58:11 pm
content: Too bad Ian didn't decide to stay.QuoteMy questions to him were trying get him to apply the same logic to a different (current) driving mechanism.Ian's argument sounds like "if I can cause collector current to flow by applying a voltage to the b-e junction, then a bjt must be a voltage controlled device". Well, if you apply a current to the b-e junction, you can cause collector current to flow too. By the same logic, a bjt must be a current controlled device.The fact that applying the same principles to the same device yields diametrically opposing conclusions should cause one to question the validity of such "principles".

author: Zero999
author_rank: Super Contributor
timestamp: January 11, 2015, 02:31:43 pm
content: Quote from: dannyf on January 10, 2015, 12:58:11 pmThe argument is repeating itself.The voltage controlled model stems from the fact that the collector current is more strongly linked to the base-emitter voltage, rather than the base current.We can argue about this until we're blue in the face but it doesn't change the fact that it's better to design an amplifier circuit based on the assumption that a BJT is voltage controlled. The dependence on base current current can quite easily be eliminated by ensuring there's adequate base drive.As far as the current controlled model is concerned: it's very good to design circuits assuming the worst case for the Hfe. If you're using a BJT as a saturated switch, it's best to assume it's current controlled but one still needs to be aware of the base charge when turning the transistor off.