What exactly is emitter degeneration anyhow? - Page 1
Forum section: electronics / beginners

author: Analog Kid
author_rank: Super Contributor
timestamp: November 25, 2024, 03:00:17 am
content: We see this term used all over the place in reference to common-emitter amplifier circuits.But I really don't know what it means.And don't tell me to look it up in Horowitz & Hill'sThe Art of Electronics: that book is not much help.In an example of what can only be described as hand-waving, the first time this term is mentioned is this:QuoteNowhere else in the book do they define this term; you're supposed to get it by osmosis or something.Now I understand that this has something to do with the fact that there's an emitter resistor, rather than having the emitter connected to ground. But I still don't really know what "emitter degeneration" means (nor why it's "degenerate" in any way). So an explanation would be appreciated here.

author: Smokey
author_rank: Super Contributor
timestamp: November 25, 2024, 03:18:14 am
content: As a current flows through that degeneration resistor it develops a voltage across it. If the gate voltage relative to ground stays the same, then the voltage across the resistor is subtracted from the gate voltage which results in a lower Vgs, and the device being "less on".It's a common way to make parallel FETs share current better since if one FET takes more current than the others it's degeneration resistor will cause it be driven less hard and take less current.

author: ejeffrey
author_rank: Super Contributor
timestamp: November 25, 2024, 03:24:05 am
content: Quote from: Analog Kid on November 25, 2024, 03:00:17 amThen look it up on Wikipedia:https://en.m.wikipedia.org/wiki/Common_emitterQuoteThat's literally all it means. You put a resistor there which reduces gain but increases linearity and reduces the sensitivity to gm variation.The name doesn't matter. It could just as well be called "emitter frobotization". But apparently "degeneration" is an old fashioned word for what is now called negative feedback, decreasing or "degenerating" gain to improve other performance. Doing this goes back to the earliest days of tube amplifiers before the development of external feedback networks or modern feedback theory.

author: TimFox
author_rank: Super Contributor
timestamp: November 25, 2024, 03:51:23 am
content: The old-school terms for positive and negative feedback are "regeneration" and "degeneration", respectively.

author: Analog Kid
author_rank: Super Contributor
timestamp: November 25, 2024, 04:47:11 am
content: OK, feedback: I understand one benefit of "degenerating" your emitter is negative feedback, which improves circuit performance (linearity, freedom from variations due to beta, etc.).I'm still trying to wrap my head around two things: one, how this works, and two, why it works. That second one is a deep subject that I'd rather leave for later.So taking this circuit fragment:Help me to understand how that resistor causes feedback. My understanding (and this is just me sitting here without looking up anything online or in a book, because I want to really grok this) is:o As current through the transistor increases (or temperature increases?), more current flows through RE, which in turno Causes a (higher? lower?) voltage drop across the resistor, which in turno Decreases? the current through the transistor, thus stabilizing operation.Or something like that.Please don't mock me or tell me to go read some book: I have some kind of mental block that makes it difficult for me to apprehend such dead-simple things. I'd like to be able to look at a circuit and intuitively understand these sorts of mechanisms.So can someone please correct what I wrote above, and explain, in simple terms (please, no Ebers-Moll!) how exactly this works? This seems to me to be a quite important and fundamental concept to learn. I know how crucial a thing negative feedback is, so I'm trying to understand it more fully.

author: Bud
author_rank: Super Contributor
timestamp: November 25, 2024, 05:37:06 am
content: As current flows through Re it creates voltage drop across Re which reduces the Base-Emitter voltage differential, forcing the transistor to reduce Collector-Emitter current You need to draw voltage polarity across Re to see this. The more current flows through Re the more positive Emitter becomes relative to Base, or equivalently the Base voltage decreases relative to Emitter. Reduction in B-E voltage reduces Ic , reducing voltage drop across Re, until an equilibrium is reached.

author: Analog Kid
author_rank: Super Contributor
timestamp: November 25, 2024, 05:54:38 am
content: Quote from: Bud on November 25, 2024, 05:37:06 amOK, back up a second: this is what I have trouble with, and your explanation doesn't really help, sorry.So what you're saying is that asincreasedcurrent flows through REthe voltage drop across itincreases, is that right? Meaning that the emitter is now more positive than it was before? And this reduces VBE? For some reason it seems to me that this would reduce VBE, not increase it, but apparently that's wrong.See, I need a really, really simple and clear explanation for this to sink into my poor brain. This happens, which causes that to happen, which causes ...

author: KE5FX
author_rank: Super Contributor
timestamp: November 25, 2024, 06:09:23 am
content: Because the resistor is in the signal's return path, it drops some voltage. Voltage that appears across the resistor is voltage that can't appear across the base-emitter junction. Ignoring for the moment the whole religious argument about whether a bipolar is voltage- or current-controlled, the overall gain of the amplifier stage is lower as a result.Often, when you can make a transistor circuit's operating characteristics depend more on the characteristics of linear passive components than on the the transistor's own characteristics, it's considered a Good Thing. That's what the resistor does.

author: Analog Kid
author_rank: Super Contributor
timestamp: November 25, 2024, 06:18:24 am
content: OK, more pictures time:Prizes to the most simple, clear, direct explanation of how REacts to create negative feedback here.Sorry, statements like "Voltage that appears across the resistor is voltage that can't appear across the base-emitter junction" , while no doubt true, are about as clear as mud to me.

author: Xena E
author_rank: Frequent Contributor
timestamp: November 25, 2024, 06:30:34 am
content: Quote from: Analog Kid on November 25, 2024, 06:18:24 amBecause it subtracts from the base bias which you do not seem to be considering? Hint: forget Vbefor the moment, consider it fixed.(Or this thread is just a troll ?)

author: magic
author_rank: Super Contributor
timestamp: November 25, 2024, 07:13:47 am
content: I think I'm becoming tggzzzQuote from: Analog Kid on November 25, 2024, 04:47:11 amI might suggest looking up Ohm's law in Horowitz&Hill or even Wikipedia and paying particular attention to signs of the numbers involved.Quote from: Analog Kid on November 25, 2024, 04:47:11 amThis cannot be explained for as long as one wishes tocompletelyignore the effect of Vbe on Ic.

author: RJSV
author_rank: Super Contributor
timestamp: November 25, 2024, 07:24:31 am
content: Yeah, like Bud says, but restated a little different:Any extra current, through that emitter resistor will RAISE UP your voltage right at the emitter.That then causes a reduction in the base to emitter input voltage, in an active fashion, meaning that the base input (voltage) is actively reduced. Not just the input current, but also active by way of the amplified current (collector to emitter current as output of transistor.So you could say it causes the gain to work against itself, with a self-limiting result.Intuitively, maybe a 'less wild' gain could be seen, as potentially 'more linear'.Otherwise, left unchecked, the temperature variations, during a day, could be causing noticably fluctuating output levels, aside from the linearity considerations.

author: Xena E
author_rank: Frequent Contributor
timestamp: November 25, 2024, 07:43:00 am
content: Quote from: magic on November 25, 2024, 07:13:47 am

author: Analog Kid
author_rank: Super Contributor
timestamp: November 25, 2024, 09:53:57 am
content: Quote from: Xena E on November 25, 2024, 06:30:34 amNo, it is most definitely not a troll. I'm sincerely trying to understand this concept.So when you say that an increase in the voltage drop across RE"subtracts from the base bias", is that because VBEreally is fixed, so anincreasein VE(because of the voltage drop puts it closer to the base voltage? (Because the base is, what? about 0.7 volts higher than the emitter?)I hope you can see my problem here: what I really need is a really, really simple explanation, perhaps with a diagram, showing which values move in which direction because of other values changing. No hand-waving: I'm not a dummy, but for some weird reason this is a difficult thing for me to grasp, although it may be second nature to you.Help a guy out here.

author: janoc
author_rank: Super Contributor
timestamp: November 25, 2024, 10:07:55 am
content: You will want to watch these two videos:https://youtu.be/Lr7MZQ6GInghttps://youtu.be/vtHCTJz0jr4The entire deal with the biasing of the transistor and the negative feedback/degeneration resistor is well explained there.

author: tggzzz
author_rank: Super Contributor
timestamp: November 25, 2024, 10:12:52 am
content: Quote from: magic on November 25, 2024, 07:13:47 amI admire your choice of role modelCaution: I'm considering changing my moniker to "Cassandra", in homage to the Greek legendsQuote from: Xena E on November 25, 2024, 06:30:34 amYou're not the first to have such thoughtsHowever, it could also be explained by someone who - for unknown reasons - can't understand Ebers-Moll, won't use Spice, and won't read the directly relevant parts of references (e.g. TAoE) people give him.His learning curve will be long and slow, and will take a lot of time and effort by other people.

author: Analog Kid
author_rank: Super Contributor
timestamp: November 25, 2024, 10:33:53 am
content: Quote from: janoc on November 25, 2024, 10:07:55 amWhy did you delete your explanation? It was the best that was offered so far.YouTube videos? Thanks but no thanks; I prefer text mode, like we're using here. I want the folks here who know a lot more than I do to explain things.So my guess was correct: increased IEcauses increased voltage drop across REwhich causes decreased VBE(contrary to what someone else posted here) which causes decreased IE(and IC). Good. I think my poor brain has finally absorbed that.(And of course the B-E junction has to be forward-biased in order for the transistor to operate in the active region, correct?)The upshot of all this being that this effect doesn't totally cancel the effects of amplification (as Horowitz and Hill put it, "a wiggle at the input gets turned into a larger wiggle at the output"), but it moderates it by reducing gain, which has all kinds of good effects.Which is something else I still don't understand: how exactly negative feedback acts to improve things like linearity and dependence on beta. But as I said, I don't even want to get into that whole mess until I understand the mechanism thoroughly.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 25, 2024, 11:00:36 am
content: Follow this step by step, and make sure you fully understand each step before proceeding.1/ Current flowing out of the emitter causes a voltage drop across the resistor, such that the top of the resistor is positive with respect to ground.2/ Current flowing into the base terminal travels through the transistor innards to the emitter terminal, and then through the resistor to ground.3/ When current flows into the base, it must overcome the base-emitter voltage drop (0.6V)plusthe voltage drop across the resistor on its way to ground.4/ When the preceding circuitry drives the base more positive, more current flows into the base.5/ The increase in the base current causes a bigger increase in the collector-emitter current.6/ This increase in the collector-emitter current increases the voltage drop across the resistor, making the top of the resistor more positive than it was before.7/ In 2/ above we said that the current flowing into the base must overcome the base-emitter voltage drop (0.6V)plusthe voltage drop across the resistor on its way to ground. But now the voltage drop across the resistor is bigger, so the total voltage drop the base current must overcome is bigger.8/ The bigger voltage drop from the base to groundreducesthe current flowing into the base. This in turn reduces the collector-emitter current.Therefore:9/ For a given voltage on the base, the collector-emitter current is less than it would be without the resistor. This effect is known as negative feedback, or degeneration.(I've skimmed over and simplified various details.)OP: that's my best effort, which I'm adding to the other explanations. If you still haven't got it, I suggest you do two things. Firstly, go back to Ohm's law and study it until you have it totally nailed. Not just the equation, I mean you have a complete, gut-level understanding. Secondly, come back to us with specific questions about which detail of the circuit operation you still don't understand.

author: Analog Kid
author_rank: Super Contributor
timestamp: November 25, 2024, 11:06:43 am
content: ^^^^^ Now we're getting somewhere. Yes, I think I've finally got it. Thanks for the detailed, step-by-step explanation.(Unlike some other folks here who can't be bothered to take the time to do that, but instead diss me for asking questions and not following their protocols ...)

author: Bud
author_rank: Super Contributor
timestamp: November 25, 2024, 01:33:17 pm
content: Quote from: Analog Kid on November 25, 2024, 05:54:38 amSeems you were lost in your Increased/Decreased word saladMore current through Re Increases voltage drop across Re, that voltage's Negative side is applied to the transistor Base through the bias circuit, Reducing the bias voltage at the Base relative to Emitter.

author: Xena E
author_rank: Frequent Contributor
timestamp: November 25, 2024, 01:38:08 pm
content: Quote from: Analog Kid on November 25, 2024, 11:06:43 amDont think anyone is trying to 'dis' you as you put it. I personally was intending to do a step by step walk through but as you'd gone off line its unfair to make these comments.If contributors with questions are blatantly not prepared to read and digest standard textsand say soI'm sure you will agree it will appear to others they're either arrogant or fucking lazy.My first post seemed to have you making a reply that was on the way to an understanding, however these are all mental models of how things actually work, you do need to understand the fundamentals to have even a remote chance of how apply them, and trying to take each part in isolation won't work.The easiest way to use BJTs as linear amplifiersisby employing emitter degenerationandfixed base bias voltageandproportioned colector load resistance, to provide predictable gain.Steve Thackery's explanation of the specific question you ask is excellent, (apart from the expression "current flowing"), but thats only part of the picture.In a previous employment I was mandated to help students study electronics but the ones who had the attitude that they wouldn't read the texts but just wanted someone to do their homework for them got shut off short order.As for the troll comment, it is a good troll question, not in itself a bad thing, as even if you understand what your asking already it can help other readers who don't.RegardsX.

author: ejeffrey
author_rank: Super Contributor
timestamp: November 25, 2024, 01:55:10 pm
content: Quote from: Analog Kid on November 25, 2024, 09:53:57 amNo. For the purposes of analyzing circuit gain, consider Vb fixed. Thats the input voltage to the amplifier referenced to ground.Vbe is just Vb - Ve. So increasing Ve decreased Vbe.

author: tggzzz
author_rank: Super Contributor
timestamp: November 25, 2024, 02:18:55 pm
content: Quote from: Xena E on November 25, 2024, 01:38:08 pmThere was a tradition on usenet, where it was easy to spot such questions in October and November. The responders had gentle fun by gave answers that weresuperficiallyreasonable, but anyone with the slightest knowledge would realise was wrong. The lazy and/or clueless submitted the answers, and their tutors instantly spotted what was going on.Nowadays that technique is redundant, having been superseded by ChatGPT "answers".I'm more than happy to help people who are prepared to put in the work to understand the subject more deeply.I'm more than happy to point people in the right direction so they can avoid known blind alleys.I'm not prepared to spoon feed lazy people who aren't prepared to do their homework.The OP might care to see CharlotteSwiss' attitude;manypeople weredelightedto help her. This threadhttps://www.eevblog.com/forum/beginners/what-an-oscilloscope-recommended-for-a-woman-passionate-about-electronics/msg3100087/#msg3100087had almost 1000 responses over 4 years.

author: coppercone2
author_rank: Super Contributor
timestamp: November 25, 2024, 02:27:05 pm
content: Lol someone bothering to ask a question about their home work... I Have a really bad feeling this is now considered work ethic. That would assume they pre-plan for at least 1 week before the assignment is due and bothering typing. You don't see that kinda thing anymore. That is the kind of initiative that gets you on the work management team. Refuses to work but DOES plan. manager

author: janoc
author_rank: Super Contributor
timestamp: November 25, 2024, 02:37:15 pm
content: Quote from: Analog Kid on November 25, 2024, 10:33:53 amI have deleted it because it was redundant to what has been posted above.I am not fan of Youtube for something like this either but these videos are very good, step-by-step walkthrough of a common emitter amplifier calculation. The first one focuses on biasing the transistor and the second one addresses the small signal behavior, including using the simplified transistor model to calculate the gain and the various component values and the various "rules of thumb" and simplifications that are commonly used but rarely explained.So well worth the time.Quote from: Analog Kid on November 25, 2024, 10:33:53 amThat's correct.Quote from: Analog Kid on November 25, 2024, 10:33:53 amOf course. Without that you have no collector current and no amplification either.Quote from: Analog Kid on November 25, 2024, 10:33:53 amYes, it can't totally cancel it - for that you would need to increase the emitter resistor value so that the voltage drop would drive the transistor to unity gain.Quote from: Analog Kid on November 25, 2024, 10:33:53 amThe beta independence isn't really the work of the emitter resistor but the way you bias the base. The resistor's role is to set the gain of the amplifier stage, then the beta only needs to be larger than the final gain of the stage you want. The exact value won't matter because the base bias network + the emitter resistor will ensure the gain is fixed to the value you want.The main role of the emitter resistor is to compensate for DC bias/collector current changes due to temperature. That's why it is commonly bypassed by a capacitor too - we don't want it to affect the AC signal amplification, only (undesired) changes in the DC bias point.The linearity improvement comes from this as well - when you bias the transistor in an amplifier, you typically put the DC bias point on the linear part of the characteristic because you want to minimize distortion. If the bias point moves due to the transistor getting warm or ambient temperature changes (or part tolerances/variations), it could land in the non-linear part of the characteristic. Same thing could happen even if the bias point remains on the linear part of the characteristic but the added input signal could end up on the non-linear bit already -> distortion. So you stabilize the bias point by adding this negative feedback to prevent this.But do watch those two videos - Nick explains it really well and in an accessible way. All you need is Ohm's law, really.

author: tggzzz
author_rank: Super Contributor
timestamp: November 25, 2024, 03:18:07 pm
content: Quote from: coppercone2 on November 25, 2024, 02:27:05 pmI have absolutely no problem with people asking about homework,providedthey mention that, are prepared to work to understand, and then create their own answer.Asking sensible questions is the precursor to being able to work towards understanding. Asking interesting and subtle questions is even better

author: rstofer
author_rank: Super Contributor
timestamp: November 25, 2024, 03:42:54 pm
content: w2aew has a great series of videos on BJT amplifiers including this one on the omon emitter configuration:

author: TimFox
author_rank: Super Contributor
timestamp: November 25, 2024, 04:13:03 pm
content: When I learned about feedback about 45 years ago, I was taught that there were four configurations.1. Either the output current or the output voltage drives the feedback network (current or voltage feedback).2. The feedback network either puts a voltage in series with the input, or applies a current in parallel with the input (series or shunt, respectively).3. Current feedback increases the output impedance, since it tries to keep the output current independent of output voltage.4. Voltage feedback decreases the output impedance, since it tries to keep the output voltage independent of output current.5. Series feedback increases the input impedance.6. Shunt feedback decreases the input impedance.The four elementary cases are drawn below, illustrated with simple BJT circuits, with negative feedback. (Of course, multiple feedback paths may be applied to a given amplifier.) The "characterized" comment is the most useful way to treat each amplifier. Note that the load is not part of the elementary amplifier in these drawings.1. Series current feedback (as in emitter degeneration). Feedback taken from voltage across emitter resistor that is approximately proportional to output current. Increased input and output impedance. Characterized by transconductance: output current from input voltage.2. Shunt voltage feedback. Feedback taken from output voltage (across load). Decreased input and output impedance. Characterized by voltage gain: output voltage from input voltage.3. Series voltage feedback. Requires second transistor to get negative feedback. Increased input impedance and decreased output impedance. Characterized by transimpedance: output voltage from input current.4. Shunt current feedback. Feedback from emitter resistor on second stage. Decreased input impedance and increased output impedance. Characterized by current gain: output current from input current.

author: magic
author_rank: Super Contributor
timestamp: November 25, 2024, 04:43:50 pm
content: This is an amazing flamewar fuel.Explain how current feedback opamps end up having low closed loop output resistance anyway...

author: TimFox
author_rank: Super Contributor
timestamp: November 25, 2024, 04:45:10 pm
content: Because the feedback is driven by the output voltage.OTA circuits, taking feedback from the output current, have high output imipedance.With op amps, sometimes "voltage feedback" or "current feedback" refers to the input of the amplifier, rather than the output, as in my description.

author: magic
author_rank: Super Contributor
timestamp: November 25, 2024, 04:51:42 pm
content: But it'scurrentfeedback, come on...Yeah, this isn't gonna work hereBut I've seen it blow up on DIY Audio several times, when a few people assume different definitions of "current feedback" and argue about supposed properties of "c.f." amplifiers without bothering to establish basic definitions.

author: TimFox
author_rank: Super Contributor
timestamp: November 25, 2024, 04:59:41 pm
content: Nonsense.I definedmyterms early in the post: voltage feedback isfromthe output voltage, current feedback isfromthe output current.Note that I write "voltage feedback", not "feedback voltage" in that sentence: they are different.The analysis I showed is relevant to the original topic: emitter degeneration.In op-amp circuits, normal and "current-feedback" devices always take feedback from the output voltage, so discussions in that area concentrate on the feedback as applied to the input.

author: adeuring
author_rank: Contributor
timestamp: November 25, 2024, 08:10:52 pm
content: Quote from: tggzzz on November 25, 2024, 02:18:55 pmThis thread is indeed a great example how a polite and friendly conversation can lead a long way.The OP's attitude reminds me in a twisted way of an anecdote about the famous physicists Niels Bohr and Evgeny Lifshitz:Quote(source:https://en.wikiquote.org/wiki/Lev_Landau)The odd and in a weird way perhaps even funny twist here is that it is the student who tells people trying to explain things to him that they are fools.AnalogKid, please keep in mind thatmost people trying to help you in this forum are, I assume, probably not professional teachers but engineers, technicians and so on. So some answers may be a bit more complicated than they need to be.most people contribute to this forum voluntarily. It is not fun when you simply snap back "I prefer text mode" to somebody suggesting a Youtube video. Do you expect janoc or somebody else to write up a summary of the video they recommended or even to transcribe it for you? You expect quite some patience from others... Many people who might have interesting contributions will simply stay silent because they are not very keen to risk an unnecessarily rude answer.Let's try to follow Niels Bohr's example rather than that of Landau and Lifshitz. Makes life more pleasant.

author: RJSV
author_rank: Super Contributor
timestamp: November 25, 2024, 08:54:23 pm
content: Busy. Sorry. So, a brief hopefully helpful:Transistor here operates at NOMINAL 0.7 V which is why it doesn't hurt to say, '0.6 V'.That Nominal voltage isn't fixed, it moves as a small variation, according to input current condition.But by now any arguments about the emitter resistor voltage pretty much follow OHMS LAW.If there is higher voltage on top of that emitter resistor, then that reduces your Vbe...Please don't follow that statement with a rigid, statement that it's always 0.7 it's not.If you have raised the emitter voltage then there is less to go to Vbe. Ohms law.That's because this happens while the voltage input stays the same. For any given drive, into your transistor input (base), the Vbe will be lower, if Ve goes up. And it does go up the harder you drive the transistor.Thus, negative feedback, causing an opposing voltage, to reduce Vbe.If you wish to keep Vbe static,...why even use a transistor at all, if input stays at one value.Remember that the transistor only sees the world at its 3 terminals, not any of the surrounding circuit.No Vbe change ? Then make the sucker out of cement...cheaper....

author: TimFox
author_rank: Super Contributor
timestamp: November 25, 2024, 09:21:35 pm
content: A small change inVbemaking a change inIcwith respect to a larger quiescent (static) value is calledtransconductance.A typical approximate expression for the transconductance isgm= dIc/dVbe=(Ic static)/(26 mV).(Depends on temperature and geometry of the device: this expression is reasonable for small-signal BJTs.)

author: Xena E
author_rank: Frequent Contributor
timestamp: November 25, 2024, 09:46:42 pm
content: Quote from: RJSV on November 25, 2024, 08:54:23 pmHey Rick why do you think it would be cheaper to make them from cement?They're made of sand already, that's as dirt cheap as you can get. 🤣

author: Analog Kid
author_rank: Super Contributor
timestamp: November 25, 2024, 10:01:27 pm
content: So one upshot of all thisSturm und Dranghere is that I did come to an understanding of how this whole deal works, which comes down to the most basic law of all:E = IRmeaning that voltage is directly proportional to current, so as current increases (across RE), so does V. Which is what kicks off the whole chain of events here.Thanks to all those who offered explanations instead of admonitions. I really do not need the latter. Life is too goddamn short as it is.And @coppercone, where did you get the crazy idea that this is "homework"? I'm in my 70s and have been out of school for a long time. I guess you could say it's "homework" in the sense that I'm trying to self-educate here, but that's it.Yes, I know all about "September" from Usenet days ...

author: TimFox
author_rank: Super Contributor
timestamp: November 25, 2024, 10:07:58 pm
content: Sturm und Drang. (I know you appreciate grammar corrections.)Be careful aboutV=I R.Many people apply it to any object with two wires.Feedback theory is about where the voltage goes.

author: Xena E
author_rank: Frequent Contributor
timestamp: November 25, 2024, 10:42:00 pm
content: Quote from: Analog Kid on November 25, 2024, 10:01:27 pmWell... it's not actually a law, even if its called that. It only applies in specific circumstances, but for steady state DC and ohmic devices, (which transistors are not BTW), then, yeah we'll let that slide.Quote from: Analog Kid on November 25, 2024, 10:01:27 pmCurrentacrossReOh boy!Quote from: Analog Kid on November 25, 2024, 10:01:27 pmYou reap as you sow,they say.Quote from: Analog Kid on November 25, 2024, 10:01:27 pmPerhaps your general manner and possibly your username?70! FFS! Good job I didn't offer to spank your bottom for you, you naughty boy.And if you want to self educate, read some "gaddamn" books.Regards,X.

author: G0HZU
author_rank: Super Contributor
timestamp: November 25, 2024, 10:52:39 pm
content: If you add an emitter resistor to create degeneration, you should also find that the input impedance at the base will go up.However, up at RF frequencies, it's also possible to add a small amount of inductance at the emitter to create emitter degeneration. You don't have to use a resistor.If you add inductance at the emitter instead of resistance, you should see the resistive part of the input impedance of the BJT go down (at RF frequencies). The same degeneration trick (using inductance) can be used with FETs (called source degeneration) and it can help achieve a good compromise between noise figure and the input impedance. It can sometimes help with stability as well.

author: tggzzz
author_rank: Super Contributor
timestamp: November 25, 2024, 10:58:23 pm
content: Quote from: Xena E on November 25, 2024, 10:42:00 pmOh, the number of times we've had to explain that! Usually coupled with an outright rejection of the concept of negative resistanceQuoteWell, he already has TAoE, " ... it's a copy given me by a friend who's also a friend of Paul Horowitz"https://www.eevblog.com/forum/projects/current-mirror-question-hybrid-mirror-(pnpnpn)/msg5701395/#msg5701395However I have to admit that, as I get closer to 70, I begin to appreciate Victor Meldrew more and more

author: Analog Kid
author_rank: Super Contributor
timestamp: November 25, 2024, 11:12:41 pm
content: Quote from: tggzzz on November 25, 2024, 10:58:23 pmIt's true; I have the book and I've read quite extensively in it.That doesn't mean I understand all of what I read, as good as H&H are at 'splaining things. Some things either go over my head, or as I've pointed out several times here, I have trouble "grokking" even if they're simple, fundamental things.Which is why I ask questions here.Don't like it? So sue me.

author: tggzzz
author_rank: Super Contributor
timestamp: November 25, 2024, 11:43:57 pm
content: Quote from: Analog Kid on November 25, 2024, 11:12:41 pmManyof the things in TAoE "go overmyhead". That's why I like it and take the time and trouble to read and re-read it.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 26, 2024, 03:38:55 am
content: And to be fair, whilst TAoE is great, there are many other books which are a great learning resource. In particular you can choose ones with a strong mathematical, theoretical approach, or others with a more intuitive, practical approach, to suite however your brain likes to work. Also, you might really struggle with one author's explanation, but "click" immediately when you read another's.Books are a fantastic learning resource. So are (certain) YouTube vids; so are forums like this. But neither of them replace books: they supplement them.

author: tggzzz
author_rank: Super Contributor
timestamp: November 26, 2024, 09:21:07 am
content: Quote from: SteveThackery on November 26, 2024, 03:38:55 amAgreed on all counts, with one proviso.With yootoob vids you can't quickly skim them to find parts that might be useful. Given that many are talking heads where any information could be equally well given in the form of text+photos, yootoob wastes a lot your time.

author: CaptDon
author_rank: Super Contributor
timestamp: November 26, 2024, 01:48:53 pm
content: Some YouTube videos are very mis-leading and factually incorrect. It is a slippery slope. One of the things I like about the emitter resistor in the common emitter configuration is that the 'apparent' increase in base resistance allows the circuit to more closely resemble a triode vacuum tube amplifier following the signal voltage applied to the base and less like following the signal current applied to the base although the latter is truly what is happening. One thing to mention is that in some forms that emitter resistor or resistance could actually be the load impedance in an emitter follower circuit which again takes on an appearance of a voltage follower although it still is truly a current follower with current gain. You got to love some of the old transistor sync separator circuits and oscillator circuits where the measured voltages indicate the transistor appears to be reverse biased!!

author: tggzzz
author_rank: Super Contributor
timestamp: November 26, 2024, 02:04:43 pm
content: Quote from: CaptDon on November 26, 2024, 01:48:53 pmFTFYAnd most of those that aren't turn out to be a waste of time. Nowadays there is so much stuff that the key skill has become how toquicklydetermine whatnotto look at. Given that I can speed-readmuchfaster than people can speak (especially with pauses, ums, ahs), that makes most yootoob vids a poor use of my time.

author: TimFox
author_rank: Super Contributor
timestamp: November 26, 2024, 05:32:10 pm
content: Quote from: CaptDon on November 26, 2024, 01:48:53 pmActually, the BJT with emitter resistor looks more like a vacuum pentode, rather than a triode, due to its high output impedance.The emitter follower into a load resistor is still a voltage follower, since the output voltage follows the input voltage (both with respect to ground) closely. A “current follower”, where the output current into a higher load resistance (more power) follows the input current would be a common-base configuration, not a common-collector circuit.A convenient way to approximate a circuit with emitter or cathode resistor is to consider the BJT or tube as an ideal device (with infinite transconductance) in series with an internal resistor between the ideal emitter or cathode and the external circuit (for small-signal use around a quiescent DC current).For a tube,rk= 1/gm. For a BJT,re=hib= (26 mV)/IEis a useful approximation for small-signal transistors at room temperature.The external emitter or cathode resistor is then in series with that resistor: ifRe>>hib, then the linear resistorRedominates the circuit and reduces the distortion caused by the variation in the device's small-signal internal resistancerewith current.Incidentally, for a tube, adding the external cathode resistorRkdoes not affect the voltage gain {mu} of the tube, but with the calculation above it increases the plate resistance by an additional value due to the resistor (and a corresponding decrease ingm).rp' =rp+ ({mu}+1) xRk.

author: mawyatt
author_rank: Super Contributor
timestamp: November 26, 2024, 07:37:26 pm
content: Along the lines of Bipolar Emitter Degeneration is Source Degeneration for a JFET. If the Input JFET Source Degeneration is alike(matched) JFET with its Gate to Source shorted and its Drain connected to the Input JFET Source, then the Gate to Source Voltage Vgs of the Input JFET is also the same as the lower JFET Vgs, which is Zero.In this configuration the Input JFET becomes a Source Follower with very high Input Impedance and ~Unity AC and DC Gain with ~Zero Offset Voltage (Vgs ~0). Old Analog Scopes utilized this configuration as the Input Buffer Amplifier, often they placed a 50Ω resistor in series with the Input JFET Source and the lower JFET Drain, which had its Gate connected to the bottom of another 50Ω resistor from its Source.The Output Signal was taken from the Resistor side opposite the Input JFET Source, and has an effective impedance looking back into the Buffer as ~50Ω which is desirable for higher frequency internal scope use. One can compensate for the finite JFET channel conductance by using a smaller Resistor so the sum becomes ~50Ω.Anyway, recall the 2N5911 Dual JFET as a popular JFET in these old analog scopes.Best

author: TimFox
author_rank: Super Contributor
timestamp: November 26, 2024, 08:07:59 pm
content: Quote from: mawyatt on November 26, 2024, 07:37:26 pmWith matched JFETs, that circuit also works with equal resistors in each source circuit (to the relevant gate), giving equal negative gate-source bias voltages.WithIdssbiasing, one must be careful about excessive positive input swings.

author: shabaz
author_rank: Frequent Contributor
timestamp: November 27, 2024, 03:34:56 am
content: If a book has not helped, it's not that using books doesn't work; it's that you actually need additional books.AoE is a fantastic book, but it's unrealistic to rely on it for all theory, because it covers a broad spectrum which results in it being a different type of book.I try to have a couple of books, so that if I cannot understand a topic very well in one book, then I'll swap to the other.I like "Circuits, Devices and Systems" by Ralph J. Smith, it's one of the simpler/easier-to-follow books (not as practically-oriented as AoE however).Another bonus is that the calculations in that book are not too intensive (and often, explanations don't necessarily require the ability to follow all the equations).It's a typical first book for students and covers this topic in a biasing chapter, and there are circuits and diagrams to help explain things throughout the book. Anyway, this is just one suggested book, everyone will have their preferences, and you could take some suggestions and just try a few books until you get to what you're comfortable with.That particular book is out of print, but that's not an issue; old editions of books are very low-cost, sometimes sub-$10, from Amazon or eBay, and you could probably re-sell some books for the same price they are purchased for, if you find some purchases are not as ideal as desired.

author: Analog Kid
author_rank: Super Contributor
timestamp: November 27, 2024, 03:46:26 am
content: Quote from: shabaz on November 27, 2024, 03:34:56 amI have that book! Have it by me right now.Got it out of a dumpster in the 2000s after some student chucked it, along with a bunch of other textbooks.I've looked through it a good deal, but I have to admit, it makes my head hurt.That's because I know just enough math to get into trouble.I took one course in calculus in college back in the 1980s. Learned just enough to get the two important concepts in that field: the derivative (the slope of the line of a function at a certain point) and the integral (the area below the curve of a function). But beyond that I'm lost: I can't calculate a derivative to save my life.So Smith's book is largely Greek to me.Too bad, because it is a very well-written and comprehensive text.If I had it to do over again ...

author: shabaz
author_rank: Frequent Contributor
timestamp: November 27, 2024, 03:58:07 am
content: Ah, I see. Oh well, at least that book can be eliminated! I'll think some more, in case I can dig up a better book suggestion. Another path to investigate could be old magazine/journal articles, they are often written to be easy-to-follow. I'm speculating a lot though; maybe they are no good for theory, I have not looked in old magazines for this topic in particular so this idea might be completely unhelpful : (

author: Analog Kid
author_rank: Super Contributor
timestamp: November 27, 2024, 04:07:05 am
content: Well, the other book I got out of that dumpster dive--and I may have mentioned it here before, can't remember--is another very good one that I do understand a lot of:Electronic Principlesby Albert Paul Malvino. Written in 1979, it's a little out of date regarding things like ICs, but other than that it's a solid beginning electronics text. Covers transistor biasing, current sources, feedback, diff amps, all that good stuff.Unfortunately, this book has very little to say on the subject of emitter degeneration.One big difference beween it and Horowitz & Hill is that Malvino writes an extensive section on semiconductor physics; H&H explicity eschew that, saying it's not really relevant to understanding electronics. Which is true, but I do find the physics of solid-state devices fascinating, so it's appreciated.

author: tggzzz
author_rank: Super Contributor
timestamp: November 27, 2024, 09:30:17 am
content: Quote from: Analog Kid on November 27, 2024, 03:46:26 amOne of the aspects of becoming older is that we realise that there are some things we will never be able to do. There is a fine line between meekly accepting that and raging against it.Modern teachers seem to think that calculus is difficult. They refuse to believe that in my state school everybody learned integration and differentiation of polynomials (except 1/x) before taking maths O-levels at 15. They continue to refuse to believe even when they see the textbook.I still remember the teacher taking two double periods (i.e. 160 minutes) to teach us differention from scratch. A couple of months later he repeated that with integration. Also taught us complex numbers in a couple of lessons, when we reminded him that our A-level (18yo) exams were in a couple of monthsHe also said that after taking maths at university, he realised he would never be a mathematician, and that's why he became a (very good) teacher.OTOH, a good mathematician (invented a new way of finding Pythagorean triads) was a poor teacher.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 27, 2024, 11:28:00 am
content: Quote from: tggzzz on November 27, 2024, 09:30:17 amI was at a UK secondary school in the late '60s to early '70s (aged 11 to 15) and we were taught differentiation from first principles, but not integration or complex numbers.

author: tggzzz
author_rank: Super Contributor
timestamp: November 27, 2024, 12:27:09 pm
content: Quote from: SteveThackery on November 27, 2024, 11:28:00 amComplex numbers were definitely A-levels.Integrationmighthave been Further Maths O-level, but it is so long ago I can't remember! Certainly before A-levels. Everybody in my state school took maths (and English Language) O-level a year in advance, on the principle that if you failed then you could have another go next year. Nobody needed to.I also have three maths A-levels: pure, applied, and pure+applied; the last was taken a year early to make it look good on university application forms

author: Gyro
author_rank: Super Contributor
timestamp: November 27, 2024, 12:32:30 pm
content: Yes, complex numbersmay have beenwas in the A level syllabus back then, certainly not O level. Irrc, Integration and differentiation were in the O level syllabus. I dropped maths after O level (16) where I managed to scrape a C grade - it was never one of my strong points. Luckily I had the 3 sciences to keep me occupied for A levels.

author: shabaz
author_rank: Frequent Contributor
timestamp: November 27, 2024, 07:18:11 pm
content: I happened by chance to find the Sylvanus P Thompson book on a classroom bookshelf back in school days, and found it intriguing, and was surprised that some form of that wasn't suggested to us to read. Anyway, I read it at the time, and it helped. Also helped that I had good teachers. I do feel sorry for kids who are not as lucky with that, since teaching isn't a well-paid profession.

author: Analog Kid
author_rank: Super Contributor
timestamp: November 27, 2024, 11:58:40 pm
content: Quote from: tggzzz on November 27, 2024, 09:30:17 amI can't speak for all humanity, of course: everyone has their own capacity for learning. But I found calculus not difficult when I took it, once I "got" those two great concepts (the derivative and the integral). I also had a good teacher.One thing I've always said about calculus is that once you take it, you understand why you took algebra in high school. Algebra by itself seems kind of, well, useless for the most part, and mainly an intellectual exercise. But once you take calculus, you realize that algebra is just a tool used in calculus and other parts of mathematics. Just like polar notation, matrix math, etc.One of my fantasies is that if I ever go to prison for some reason, I'd take my calculus textbook with me and go through it assiduously from the start, really learn calculus. But out here, I find I don't really have the appetite to do this.

author: tggzzz
author_rank: Super Contributor
timestamp: November 28, 2024, 01:00:45 am
content: Quote from: Analog Kid on November 27, 2024, 11:58:40 pmThere's a better equivalent over here. There is a weekly radio programme called "Desert Island Discs", which has been running since 1942 (i.e. 82 years). Each week a guest, called a "castaway" during the programme, is asked to choose eight audio recordings (usually, but not always, music), a book and a luxury item that they would take if they were to be cast away on a desert island, whilst discussing their life and the reasons for their choices. To avoid boring non-choices, they are given the Bible and the complete works of Shakespeare.I'd take TAoE+X-Chapters. Only 2480 (out of ~3400) episodes are available for streaminghttps://www.bbc.co.uk/programmes/b006qnmr/episodes/player

author: Xena E
author_rank: Frequent Contributor
timestamp: November 28, 2024, 01:24:34 am
content: Quote from: tggzzz on November 28, 2024, 01:00:45 amDesert Island Discs, the pinnacle of BBC programming!I was listening to that as a kind even before moving to the UK.TAoE...Couldnt agree more, It's generally highly regarded simply because is just such a good example of clearly explained content.In addition, for Mathematics the penny dropped for me finding an old book called "Basic Mathematics for Radio Students" (ILIFFE Books, F.M.Colebrook 1946), in a second hand book shop. (I still have it.) Other students laughed at it because it was such an old book.I think it saved my skin as far as understanding algebra, complex numbers, trig, and calculus. I found it relevant because it wasnt abstract, it was applied to an actual discipline, without it I'm sure I would have struggled to even get an entry grade in maths.

author: MrAl
author_rank: Super Contributor
timestamp: November 28, 2024, 02:07:52 pm
content: Quote from: Analog Kid on November 25, 2024, 03:00:17 amHi,By now I think you have a better idea about how this works. I had to comment because it's been years since I have heard or read the phrase "degenerative feedback" (haha). That's such an awful choice of words because it sounds like something bad. If it wasn't for negative feedback we would have audio amplifiers that sound like total garbage.The reason this took a bit longer may be because you were given some information that was not exactly true. For one, the voltage Vbe is not constant, it can change when the current through the base emitter changes. It's an exponential relationship, but for small changes it is easy to think about it just being a resistor which has a linear characteristic.If the equivalent resistor value was 600 Ohms for example, then when the BE current was 1ma, the voltage Vbe must be 0.600 volts. If the current goes up to 1.1ma, then the voltage Vbe goes up to 0.660 volts. That's actually an approximation that is used sometimes. The other way of looking at it is that as the voltage Vbe goes up, the current through BE goes up. Since we often look at the collector current as the base current iB times Beta, the collector current rises also. If the Beta was 100 then the collector current goes up from 100ma to 110ma.Ok, so now we have the emitter resistor Re and its voltage to look at.As the collector current rises, the emitter current rises, and so the voltage across Re rises, and this means the voltage at the emitter rises also. We next look at Vbe again. If Vbe was 0.600 volts with a constant base voltage of 1.600 volts, then the voltage at the emitter was 1 volt and that means the resistor Re must be 10 Ohms (100ma through 10 Ohms gives us 1 volt, ignoring the smaller BE current).Next let's say that the input voltage increases to 1.700 volts. What that means is that the current through BE goes up (emitter at 1 volt) to about 1.17ma and so the collector current would go up to 170ma, but now 170ma through that Re of 10 Ohms means the voltage across Re would go up to 1.7 volts, which would mean the voltage across Vbe would decrease to zero, meaning the current through the collector would drop to zero. So you can immediately see that the changes causes less current through BE. In reality though, as the collector current rose and the emitter voltage rose, it would reach a point where the current through BE and the collector levels off and becomes stable and non zero. If there was no Re though this would not happen was there would be no feedback.In short, this boils down to this expression:Ibase=(Vbase-Vemitter)/600and since the collector current is Ibase*Beta, we have:Ic=Beta*(Vbase-Vemitter)/600and since Vbase would be considered the input, we can write:Ic=Beta*(Vin-Vemitter)/600This shows the relationship of the feedback mechanism, which is Vin-Vemitter. Vemitter is the feedback voltage.We don't need to here, but taking this one step further, we can replace Vemitter with Ic*Re and then we end up with:Ic=(Beta*Vin)/(10*Beta+600)This was achieved by noting how the negative feedback mechanism works. We can have an equation that only needs the input voltage (once we set the Beta to some constant value which is a usual approximation).Some of these steps were found using approximations and that's a usual way to proceed with these kinds of problems.The approximation of Ib=Vbe/Rbe is a linearization of the voltage and current relationship of the base emitter diode. It is valid over a small range of base currents.Another thought is that even without an emitter resistor there is still some negative feedback. This is sometimes viewed as a small internal resistor that is internal to the transistor in series with the emitter. We can't get at that resistor though only the external one, but that internal resistor causes some reduction in gain also. This one is often written as "re" lower case letters. You can read more about that on the web if you like.

author: magic
author_rank: Super Contributor
timestamp: November 28, 2024, 02:23:18 pm
content: Quote from: MrAl on November 28, 2024, 02:07:52 pmI don't know who uses this "approximation" and for what, because the ratio of change in current in your example is off by an order of magnitude.

author: RJSV
author_rank: Super Contributor
timestamp: November 28, 2024, 02:38:27 pm
content: Ahhh, Kirkoff ! That's what I missed.Reading about using an inductor, for your degeneration voltage, I've realized that I was mentally applying Kirkoff's Law, but then erroneously claiming Ohm's Law applies !Gawd, at least THAT mistaken advice can be corrected. 'Everything in the circuit, going around, MUST add up the zero.' (To paraphrase the Kirkoff Circuit topology law.That's why you want to 'add up' things, in this case a transistor is the upper box, and your impedance device is the lower box, with both on a DC battery or power supply.

author: TimFox
author_rank: Super Contributor
timestamp: November 28, 2024, 02:54:46 pm
content: The apparent internal resistor in series with theemitteris given by the usual approximation ashib= 26 mV/Iefor small-signal transistors at room temperature.Its reciprocal is the transconductancegmof the transistor in small-signal common-emitter amplification.This is a simple derivative of the Shockley diode equation.

author: RJSV
author_rank: Super Contributor
timestamp: November 28, 2024, 02:55:50 pm
content: This business of proceeding with a transistor input model that is a '600 Ohm' resistor equivalent is garbage, looks SCHMART andeaves the poor OP reeling, at the shear bulk of the explanation.Heck, you could do 'models' all day, if you know how to multiply and divide. The repulsive effect can be so bad, the dude perhaps would wish to change careers.The Vbe model could be said to have messy exponential math, skipping the contorted details. It's a direction thing, meaning that any increase in base drive current has a reaction voltage, that also goes up.Same for any input decrease, as a decrease in the base drive current will cause a decrease in Vbe.For a resistor on emitter, or anywhere really, it's a straight linear, as ohms law applies.

author: RJSV
author_rank: Super Contributor
timestamp: November 28, 2024, 03:31:36 pm
content: Many respondents here are missing something...the TITLE and context being as a category called 'BEGINNER', and so the simple wins, over these over-wrought, drawn out 'explsnations' complete with stuff that even Senior level Engineers.Then layer on a bunch of criticisms pointed at the OP. He has the context correct, such as rejecting YouTube and many of the other suggested study.'his learning curve will be long and slow'.Really? That's a nasty teacher, having no room for sincere desire to go the simple route.That's not LAZY, that's practical, and I see this mis-applied 'complexity' all over the place in big governments for sure.But several responders are getting it right, when they give a good, simple, non-exponential discussion, skipping the insulting claims, that the OL won't cooperate, and go off to study 47 YouTube videos. Probably exasperating videos, by the way.Simple, BEGINNER, Simple, BEGINNER.Heck, I could be claiming that responders here are not, uh, READING MY PERFECT REPLIES, because they are LAZY ass.I would add only that anyone should have some basic math that describes what an exponential is, in simple terms.Exponential stuff is not straight lines, that's a simple start. Plus they are a difficult thing, when exact details are needed.Beginner. Simple. Exact not necessary.

author: RJSV
author_rank: Super Contributor
timestamp: November 28, 2024, 04:32:11 pm
content: Suggested way forward:1.). Stop this thread, and create / re-start the abandoned topic.2.). I would copy in a couple of acceptable replies, especially from janoc, who graciously delved into the linearity advantage. That, actually, is part of what I wished to read.3.). Make it clear, that multiple references, and nasty words like 'FUELING LAZY' are clearly OFF-TOPIC,...as OP set forth originally.4.). Also copy over, maybe 3 or 4 good responses, from this, abandoned thread.OP has responded approx. 14 times, so doesn't fit the type, that posts and never comes back online.I'm of senior Engineer status, UC Berkeley EECS, 1976, and even I cringe and skip any complex, exponential math, especially cause I know that violates the OP 's topic, at least in spirit.Looks like OP has courteous tolerance, for those directly opposed to the topic.Beginner Section, please don't involk YouTube videos, keep it simple pleases.Beginner, Simple.Too bad we can't have OP issue ratings, or grades, on our replies.

author: Xena E
author_rank: Frequent Contributor
timestamp: November 28, 2024, 06:52:07 pm
content: Quote from: RJSV on November 28, 2024, 04:32:11 pmLooking through some of the OP 500+ posts made within the two months since joining, one can conclude they are niether a beginner or stupid, in fact come across as generally well educated.Quote from: Analog Kid on November 27, 2024, 11:58:40 pmThe above comment from the OP is not one that is expected from a person who has also said:QuoteAnd:QuoteFucking Lazy, no probably not.Troll? Fire starter? Time vampire?More likely...Best wishes.X

author: tggzzz
author_rank: Super Contributor
timestamp: November 28, 2024, 07:10:09 pm
content: Quote from: RJSV on November 28, 2024, 04:32:11 pmOr on weird spellings, punctuation and formatting.Firstly OPs can issue "ratings": all they have to do is quote and reply, giving reasons.Secondly upvoting/downvoting is aseriouslybad concept, since it encourages cliques and - when examined - offers no benefits. Somebody suggests it every couple of years; fortunately Dave Jones understands why it would be disastrous here. FFI, see the arguments made in the previous threads on that subject; I'm not going to rehash them.

author: tggzzz
author_rank: Super Contributor
timestamp: November 28, 2024, 07:13:01 pm
content: Quote from: Xena E on November 28, 2024, 06:52:07 pmClearly it is quite possible to be a time vampire without being a troll or fire starterThe only (very imperfect) prophylactic is to hide the contents of a member's posts. That gives you a big effin reminder that they are likely to waste your time.

author: TimFox
author_rank: Super Contributor
timestamp: November 28, 2024, 08:03:34 pm
content: Re: useful elementary books about BJTs.In my long-distant youth, I found books by Lawrence G Cowles quite useful.There were several, published by Prentice-Hall from about 1968 to 1976.Long out of print, but available used from the usual sources.The first book,Transistor Circuit Design, is especially useful about bias networks for small-signal BJT amplifiers, as well as use of degeneration to stabilize the circuit performance.

author: Xena E
author_rank: Frequent Contributor
timestamp: November 28, 2024, 09:49:43 pm
content: Quote from: tggzzz on November 28, 2024, 07:13:01 pmIndeed. I've been accused of all individually and in combination before now myself.Suggested as possible alternative,or, not necessarily, and, options.Quote from: tggzzz on November 28, 2024, 07:13:01 pmThe penny has just dropped on this one.Yes, it is something I do, and should have done it in this instance too, though I often don't clock the author of posts when reading a thread through.It would save bad feelings on the part of the correspondent as well; most people don't get that I'm only responding to a post or point that's been made, unless it's blatantly obvious that they are being a dick.X

author: MrAl
author_rank: Super Contributor
timestamp: November 29, 2024, 04:00:42 pm
content: Quote from: magic on November 28, 2024, 02:23:18 pmWow, you've never studied linearization of a nonlinear component? At the very least, I helps unravel a much more complex model so the understanding can be more intuitive. And remember, this "600 Ohms" comes with an assumption about the operating point.The intelligence level of this group can't be that low, so I assume you had something else in mind. Maybe you could make it more clear.

author: MrAl
author_rank: Super Contributor
timestamp: November 29, 2024, 04:05:04 pm
content: Quote from: RJSV on November 28, 2024, 02:55:50 pmThere are big books written on how to turn even whole systems into linear systems, with a given set of assumptions of course.But if you are so knowledgeable that you can argue with a veteran engineer who has worked in the field for years and studied in this field for years, then explain your rendition of how to linearize a nonlinear component. I'll be waiting.

author: magic
author_rank: Super Contributor
timestamp: November 29, 2024, 04:19:42 pm
content: Your idea of modeling Ib as flowing through a resistor across Vbe and then multiplying it by β to get Ic only produces correct numeric result if β=Vbe/Vt, or about 20~25. On a typical small signal transistor with β=200 it's off by an order of magnitude.The model is based on wrong principles, misleading and plain wrong. It's also more complex that the actually correct and well known equation for transconductance.

author: mawyatt
author_rank: Super Contributor
timestamp: November 29, 2024, 04:45:27 pm
content: Linearization is about calculating the desired result about a fixed operating point, or mathematically taking the derivative wrt such. Mathematically taking the derivative means using an infinitesimally small increment, practically this means a small increment (delta) wrt the fixed point.Best

author: TimFox
author_rank: Super Contributor
timestamp: November 29, 2024, 04:51:12 pm
content: When doing an ".AC" simulation, SPICE first calculates the quiescent conditions, then linearizes the operation around that point to calculate small-signal gain, etc.In such simulations, one usually sets the input voltage AC generator to 1 V, so that other simulated voltages give the voltage gain, even if the actual output voltage could not reach, for example, 100 V due to the power supply.

author: shabaz
author_rank: Frequent Contributor
timestamp: November 29, 2024, 06:26:32 pm
content: I had a look, and couldn't find any book I owned with a simpler and reasonable explanation. There's another classic first-year book called Microelectronic Circuits by Sedra and Smith, but it's not very concise, they take 300 pages to get there to emitter degeneration. Perfect for students with time on their hands, however. That could be an option if one doesn't mind wading through things a bit more slowly.Googling around reveals university notes that are not bad, but I'm unsure about their copyright status so won't link them here.Quote from: TimFox on November 28, 2024, 08:03:34 pmNice, I found that for sale for about the price of a coffee, on Amazon. Ordered it, since I like looking through older texts occasionally.

author: tggzzz
author_rank: Super Contributor
timestamp: November 29, 2024, 06:44:47 pm
content: A Sedra and Smith book was one of my first year university textbooks.It was not concise, but it did lead us gently through understanding fundamentals of how to think about circuits and systems and to describe and analyse their behaviour. It did require a lot ofslowthought, and in my case a realisation that I didn't understand as much as I thought, nor as deeply as I needed to understand.That has been crucial during my career.The requirement for slow thought makes it difficult for some people, which is a shame because it reduces their chances of gaining more than a superficial understanding.

author: shabaz
author_rank: Frequent Contributor
timestamp: November 29, 2024, 07:01:40 pm
content: Same here, I still have my old uni copy of it. And perfect for self-learning since it doesn't miss much.Currently available on Amazon US for $14, if the OP doesn't already own it.

author: MrAl
author_rank: Super Contributor
timestamp: November 29, 2024, 08:11:58 pm
content: Quote from: mawyatt on November 29, 2024, 04:45:27 pmHi,Yes, that's almost right, very good.There is a difference and I hate to point this out because I don't want to sound overly critical, but when you say "practically this means a small increment..." that is right, but it's not the same as the derivative, which as you noted, is an infinitesimally small increment.The theory is to use an infinitesimally small increment to get to the linearization, but then we can use a larger increment to determine the overall operation. I used the example of 1ma and then an increment of 0.1ma, which led to 1.1ma, but if you are not comfortable with that then think of it as 1.000ma to 1.01ma (0.01ma increment which is 1 percent) or even 1.001ma which is an increment of 0.001ma and about 0.1 percent.However, the idea was not to explain how linearization works in general, it was just to show how some quantities relate to each other. Another way of saying this is that it was simplified so that it should be more understandable.It might get down to semantics here I'm afraid of that, which leads to long threads about simple stuff

author: TimFox
author_rank: Super Contributor
timestamp: November 29, 2024, 08:19:56 pm
content: You can use the second derivative to estimate how far from quiescent is too far for using the first derivative to obtain an approximate response to a "small" increment.

author: mawyatt
author_rank: Super Contributor
timestamp: November 29, 2024, 08:26:18 pm
content: Quote from: MrAl on November 29, 2024, 08:11:58 pmSorry it's EXACTLY Right!!!Maybe you need a serious review of Electronics 101As mentioned in SPICE, when engaging AC Analysis, which BTW isLinear Analysisjust in case you don't know, everything is Linearized and results computed and displayed are based uponLinear Circuit Analysis.Give it a try, LTspice is free, you might find it enlightening, and please show the results you get with your examplesBest,

author: MrAl
author_rank: Super Contributor
timestamp: November 29, 2024, 09:20:28 pm
content: Quote from: magic on November 29, 2024, 04:19:42 pmHello again,It's not misleading, nor based on wrong principles, nor wrong. It could be off by an order of magnitude, but we do not have to use it to that extreme either. It's not meant to be a perfect description of the diode action. To do that we would have to use the exponential model, which we don't want to get into right now.It's well known that when we do a linearization of a diode, we often replace it with a resistor in series with a voltage source. This must be all over the web. It's no mystery, look it up.Suppose we have a diode with these exact numbers for voltage and current:v1=0.60v, i1=0.00016275379av2=0.65v, i2=0.00044241239av3=0.70v, i3=0.00120260328aThat's roughly a 3-fold increase in current for every 0.05 volts in voltage. If we wanted to, we could use that information, but this is about an actual linearization so we move on.We can linearize around the point with voltage 0.65v:0.65/i2=1469.217 Ohmscheck:0.65/1496.217=0.00044241239 (OK)Linearized model with voltage increment:inc 0.001v: 0.651/1496.217=0.00044309316 (about 2 percent off) exact: 0.0004513497inc 0.005v: 0.655/1496.217=0.00044581570 (about 10 percent off) exact: 0.000488941This is not meant to be 100 percent accurate, and we can see the limits of the model like this.The simpler explanation does not have to go into any of this.

author: MrAl
author_rank: Super Contributor
timestamp: November 29, 2024, 09:25:16 pm
content: Quote from: mawyatt on November 29, 2024, 08:26:18 pmYou are being overly critical, you are not trying to understand the point of view, you are trying to mock the analysis, which gets nobody nowhere.THIS ANALYSIS has been used for years and years, and if you don't understand it, then perhaps you should try next time instead of repeatedly trying to dismiss it. It's up to you.I've partly agreed with the post but you cannot recognize that fact. It's true that dv/dt or whatever is the theory, but that's to obtain the linearization. When we use this linearization model, we DO NOT keep linearizing. We do that once, that's it, unless it really becomes necessary to do it a second time.AFTER we do that, we USE the model, but that's long after the dv/dt calculation.TRY AND UNDERSTAND THIS or just go on being ignorant and stating that everybody else is ignorant. Read it on the web if you don't trust this.My previous post shows a real example. If you have a problem with it, then show your example.

author: MrAl
author_rank: Super Contributor
timestamp: November 29, 2024, 09:27:27 pm
content: Quote from: TimFox on November 29, 2024, 08:19:56 pmIt's starting to look like some people do not want to understand the point of view, they would rather state errors due to their misunderstanding.if there was some error I would have never said it. Also, how hard is it to look it up on the web.Linearization of a diode and it's use.

author: TimFox
author_rank: Super Contributor
timestamp: November 29, 2024, 09:34:33 pm
content: Using your example numbers for voltage and current, one could model the diode as a battery in series with a resistance.With those numbers, the battery voltage would be 0.65 V, and the resistance would be (0.70 - 0.60) V / (1.20260 - 0.00016)x10-3A = 83.2.However, with the current being such a strong function of voltage (as diodes do), your increment in voltage (0.60 to 0.70) is too high for an accurate linearization on both sides of 0.65 V.I prefer the usual approximation:Rac= (26 mV)/Iqui, which is rather simple.For inherently non-linear devices, the ratioV/Iisnot equal tothe derivative dV/dI, especially for exponential devices.

author: mawyatt
author_rank: Super Contributor
timestamp: November 29, 2024, 09:51:17 pm
content: Quote from: MrAl on November 29, 2024, 09:25:16 pmThe only one here that'signorantis you my friendWe're not doing your homework assignment for you, these are your examples, you need to show your work here in a single post for others to review.Many folks have tried to show you how a bipolar transistor behaves and how your general "assumption" of "linearity" can quickly lead to considerable errors. This is where you need to pay attention as to "why" and understand the reason behind the errors, otherwise you will stay forever in the dark wrt basic bipolar transistor behavior!!For example, you do know that achangein Vbe of 18mv causes achangein collector current of 2X, and 60mvchangecauses a 10X currentchangewith a transistor that's not in saturation. This is fundamental bipolar transistor behavior and a basis for just about every competent bipolar transistor design ever done!!BTW this is why Emitter Degeneration (agree this is an awkward term) works, it reduces the effects of Base Voltage variation wrt collector current, as usually change in Vbe is swamped by the voltage across the Emitter, and it somewhat "Linearizes" the effect.Best

author: tggzzz
author_rank: Super Contributor
timestamp: November 29, 2024, 10:20:46 pm
content: Quote from: MrAl on November 29, 2024, 09:20:28 pmNope.Let's start with a point that is specific to diodes: the exponential model is useful in many circumstances, but it does not tell the whole story and does not match all observed behaviour. See, for example TAoE3 1x.7.1.D Plus using "all over the web" as a basis for understanding is naive in the extreme!Secondly you appear to misunderstand the fundamental advantages of linear models in analysis. (Aside: that is y=mx+c linear, but only with c=0). The point is that linear models allow you to use the superposition principlehttps://en.m.wikipedia.org/wiki/Superposition_principleThirdly, there is no definition of "small" or "large" w.r.t. signals in linear system analysis. By definition the linear model analysis is valid for all signal levels. Yes, we know that doesn't match reality, but that's missing the point. Models != Reality. Never forget that "all models are wrong, but some are useful"That's all a long winded set of points indicating that mawyatt and TimFox are correct, and that your understanding is simplistic and limited to the point of naivety.

author: tggzzz
author_rank: Super Contributor
timestamp: November 29, 2024, 10:30:25 pm
content: Quote from: TimFox on November 29, 2024, 09:34:33 pmI do hope we aren't going to have to rehash that, as it was beaten to death in this threadhttps://www.eevblog.com/forum/beginners/bulbs-connected-in-series-physics-doesnt-add-up/msg5657963/#msg5657963One example is tunnel diodes, which have negative differential resistance over part of their range. Another is batteries; what is the "R=V/I" for this curve

author: TimFox
author_rank: Super Contributor
timestamp: November 29, 2024, 10:35:21 pm
content: A hint: when considering devices which are approximately exponential,I(V) =Ax exp(+V/B) , whereAandBare constants, it is better to analyzeVas a function ofI.

author: MrAl
author_rank: Super Contributor
timestamp: November 30, 2024, 03:55:38 am
content: Quote from: mawyatt on November 29, 2024, 09:51:17 pmMaybe I should not have called anyone ignorant, but you still are not getting my point and you do not seem to want to, that's the real problem here.But what I can do is give one more example, and if you do not 'get' this one, I'll have not choice but end this particular conversation with you. However, I seriously doubt you will be able to knock this example.BTW how the heck did you reject the previous example when the errors were clearly demonstrated. I believe it is because you are 'stuck' on one kind of analysis and can't accept anything else. I am not calling you an idiot though, just saying for this one topic you are not getting the point I am making.I will present this next example graphically because that's almost impossible not to 'get'. You can take a look and see if it matches your idea of a linearization. This is PROBABLY what you are more used to seeing, and that would be understandable, but there are other ways.I'll post the example in the next post or two.

author: MrAl
author_rank: Super Contributor
timestamp: November 30, 2024, 04:04:30 am
content: Quote from: tggzzz on November 29, 2024, 10:20:46 pmThe funny thing here is that you said my "understanding" is simplistic, but that's because your understanding of the main point is too simplistic.My example is SUPPOSED to be simplistic, THAT's the WHOLE idea. It's for a beginner.Now here is the more detailed analysis. I am presenting this one graphically because it's dang near impossible not to understand for anyone. You can't possibly disagree with this example.In the attachment, the blue plot is an actual diode plot.the red plot is the linearization about the point where I=0.500 amps (near the center). If we look to either side, we can see that the red plot closely follows the blue plot for a decent variation in current. In fact, the voltage between 0.250 amps and 0.750 amps is almost a straight line, which indicates a resistor. The only thing is, the starting voltage is around 0.750 volts, so we have to include a voltage source. Once we do that, the same current in both the actual diode and the linearized model is the same, and the voltage is nearly the same between those two points and even beyond. There are certain areas where it is not valid, but that's considered to be places where the model will never be needed to be used.If you disagree with this then you'll have to look around the web because it's impossible to disagree with this example. It's the more typical linearization. It's a constant voltage source of 0.768v in series with a resistor of 0.2 Ohms.Also note that we could get the red line to match up with the blue line even better, but I did not want it to match up too well because then it would be hard to tell that they are almost the same ... the red line would cover the blue line in too many places.

author: magic
author_rank: Super Contributor
timestamp: November 30, 2024, 09:28:47 am
content: You didn't originally mention any voltage sources in series with the "equivalent" resistor and you made it look like one can simply divide Vf/If to get the dynamic resistance, which is bollocks.That's the part which we described as "simplistic", "plain wrong", etc.Generally, it's exactly this kind of "SUPPOSED to be simplistic, THAT's the WHOLE idea, it's for a beginner" mis-explanations which cause people like OP (and many before him) to come up with all sorts of nonsensical ideas and then take lots of work to straighten out. You leave out important details, you don't say what was left out and why, your model is pitifully limited, but somebody is going to use it anyway and then argue with us that something is supposed to work even if it really is not.And lastly, base current doesn't cause collector current to flow - mawyatt mentioned transistors so leaky that they can have their bias set simply by leaving the base open circuit at DC. Where is your god now?

author: TimFox
author_rank: Super Contributor
timestamp: November 30, 2024, 03:05:13 pm
content: The simple model with a voltage source (battery) in series with a resistor is an elementary example of when the ratioV/Idoes not equal the derivative dV/dI.

author: tggzzz
author_rank: Super Contributor
timestamp: November 30, 2024, 03:14:09 pm
content: Quote from: TimFox on November 30, 2024, 03:05:13 pmIt is entertaining to get people tonumericallydefine the R=V/I of a battery.In normal operation, that R is negative, i.e. V/-I (negative since the conventional current is flowing out of the positive terminal).When open circuited, that R=infinity, i.e. Voc/0When short circuited, that R=0, i.e. 0/-IscWith no strictly positive value of RSome people still try to argue

author: mawyatt
author_rank: Super Contributor
timestamp: November 30, 2024, 03:51:30 pm
content: Quote from: MrAl on November 30, 2024, 04:04:30 amHere's aquotefrom your earlier post #62 responding to the OP Analog Kid.For one, the voltage Vbe is not constant, it can change when the current through the base emitter changes. It's an exponential relationship, but for small changes it is easy to think about it just being a resistor which has a linear characteristic.If the equivalent resistor value was 600 Ohms for example, then when the BE current was 1ma, the voltage Vbe must be 0.600 volts. If the current goes up to 1.1ma, then the voltage Vbe goes up to 0.660 volts. That's actually an approximation that is used sometimes.The other way of looking at it is that as the voltage Vbe goes up, the current through BE goes up. Since we often look at the collector current as the base current iB times Beta, the collector current rises also. If the Beta was 100 then the collector current goes up from 100ma to 110ma.Note your lack of a Series Voltage Source as mentioned by magic, thisomissioncompletely changes the concept of the linearization discussion and will likely confuse others. Linearization of the Vbe is about the static DC Bias Point to be meaningful, for AC Analysis the Voltage Source is replaced by a Short Circuit (all Voltages Sources become Shorts, and Current Sources become Opens for AC Analysis), for other discussions/analysis the Voltage Source along with the Resistor should be included.In your example above inBold, that analysis isn't correct and gives a significant error and that approximation for Vbe is never used by anyone skilled in electronics. At ~300k the correct Vbe result is ~0.602 volts not 0.660 volts as you derived!! What do you think the 0.660 volt Vbe will actually produce with this transistor, we'll leave these solutions for your discovery (clue it's not even close to 1.1ma!)Best

author: magic
author_rank: Super Contributor
timestamp: November 30, 2024, 04:53:36 pm
content: It is actually a well known rule of thumb that Vbe increases 60mV for each 10x increase in emitter current (at least until plain old ohmic resistance begins to dominate at high currents), which makes this particular example (600mV, 660mV) quite ironic.Also,Quote from: MrAl on November 30, 2024, 04:04:30 amLet's not forget that BE junction currentincludesthe collector current. That's the whole trick of the BJT, and the reason why calculating Ic as a function of Vbe is particularly straightforward. Again, at least until base spreading resistance begins to matter.

author: temperance
author_rank: Super Contributor
timestamp: November 30, 2024, 07:26:49 pm
content: It is not the first time for MrIA to get into heated debates about analog circuit design topics where the final argument thrown in is mentioning years of experience.I have the Sedra Smith book and dynamic resistance is discussed in the very beginning of the book in the chapters about diodes. I can scan the pages for those who want that.Some people state that text books are skipping this. But this is not true. Perhaps every book I've seen on the subject explains the dynamic resistance of a PN junction (du/di) right away in the beginning of the book. It is up to the reader to understand that a BE junction is just the same thing and what has been written about diodes applies to a BE junctions as well. But the subject can be a little intimidating at first.On small signals: it has been something I've been struggling with myself at the age of 16 when I bought a second hand copy of Sedra Smith microelectronics. I've searched high and low for a definition of what was meant by a small signal (long before the internet came into existence) and searched trough every book I could find in libraries but couldn't find an answer at the time because of my limited understanding.A small signal in my minds aquarium is a signal small enough to not cause gigantic errors when analyzing a circuit. Calculate a diodes dynamic resistance in a few places and see how this resistance changes with applied forward voltage. That's my intuitive understanding of what is meant by a small signals. Maybe someone has a better definition or explanation?

author: KE5FX
author_rank: Super Contributor
timestamp: November 30, 2024, 07:30:24 pm
content: 'Small signal' often refers to circuits where signal voltages are low in comparison with prevailing bias voltages. It'd be interesting if there is a more formal definition somewhere, but that's what I've always heard.

author: mawyatt
author_rank: Super Contributor
timestamp: November 30, 2024, 07:55:21 pm
content: WRT to a good text on this subject and other Electronics Topics, our old Grey & Meyer from undergrad was pretty good and covered a wide range, including tubes!!Yes we are that oldBest

author: tggzzz
author_rank: Super Contributor
timestamp: November 30, 2024, 08:58:29 pm
content: Quote from: KE5FX on November 30, 2024, 07:30:24 pmThat is often true, but the adjective hides as much as it illuminates.The key point is that small signal linear models allow you to use the superposition principlehttps://en.m.wikipedia.org/wiki/Superposition_principleThe extent to which the model's analysis adequately matches reality determines the acceptable magnitude of the signal. If you have a piecewise linear model the signal could be "large" - provided it stays within one "piece".

author: G0HZU
author_rank: Super Contributor
timestamp: December 01, 2024, 09:33:31 pm
content: I don't know if this helps, but see the plot below. This shows a crude estimation of the dynamic resistance of a 1N4148 diode vs bias current.This was generated by measuring the IV curve in small increments and then calculating the dynamic resistance for each tiny step change.This is why the curve is a bit wiggly at each step change, but if you ignore the wiggly nature of the curve, you can see that the dynamic resistance is about 25 ohms at 2 mA bias current and it is just over 10 ohms at 5 mA bias current. If I had chosen smaller step sizes the curve would have looked less wiggly and the curve would have looked like a smooth line. This was carried out at DC but the dynamic resistance should be quite constant up at RF frequencies spanning up into VHF.It's an old plot, and the aim at the time was to make a piece wise model for the Genesys simulator. I modelled the diode as a non-linear resistor in Genesys using the IV data. This worked quite well although it needed lots of IV data points.The other way to estimate the dynamic resistance is to measure it (at small signal levels) with an impedance analyser at various bias currents.In this case, I think small signal can be defined as a signal that makes only a minor change to the bias current throughout its cycle. So signals ideally need to be smaller when operating at lower bias currents.

author: G0HZU
author_rank: Super Contributor
timestamp: December 01, 2024, 09:44:02 pm
content: You can also use degeneration to help define the input and output impedance of RF amplifiers.For example, a simple BJT (common emitter) amplifier can be configured to have a very good match to 50 ohms across low frequencies up into VHF using degeneration. The gain will also be quite flat across this frequency range.Some of the degeneration can be achieved using resistance and some can be achieved using inductance at the emitter. Usually a combination of negative feedback from collector to base and more feedback (degeneration) can be applied at the emitter. I've designed lots of wideband 50 ohm amplifiers like this over the years.

author: tggzzz
author_rank: Super Contributor
timestamp: December 01, 2024, 09:52:45 pm
content: Quote from: G0HZU on December 01, 2024, 09:33:31 pmUseful plots. It would be nice to see the corresponding dynamic resistance graph predicted by the standard equations.As for what "small" means, see my post before yours. Consider that with a piecewise linear model, the delta w.r.t. the bias point could be large scale - provided the delta doesn't cross a boundary into the adjacent linear piece.

author: G0HZU
author_rank: Super Contributor
timestamp: December 01, 2024, 10:27:24 pm
content: See below where I've had a quick go at predicting the dynamic resistance based on a simple model where Is = 5nA and n =1.9 and Vt =0.026VIt's not going to fit perfectly with the measured data as the 1N4148 diode only (closely) follows the equation at currents below about 5mA. This can also be seen if I compare the classic IV curve of the real 1N4148 against the equation based IV curve.Edit: I've reloaded a new image below as there was a slight offset in the data in the first plot.There's very little difference though.

author: G0HZU
author_rank: Super Contributor
timestamp: December 01, 2024, 11:04:53 pm
content: For another cross-check I put the 1N4148 diode in an RF test fixture and measured it with a VNA at 1 MHz through to 30 MHz. This setup uses a bias tee so only the diode gets measured at each bias current as the bias tee is part of the VNA. The source power of the VNA was set quite low at -30dBm.At 1 mA the resistance of the 1N4148 diode was 49 ohms across this frequency range.At 2 mA the resistance was 25 ohms.At 5 mA the resistance was 11 ohms.At 10 mA the resistance was 6 ohms.This agrees quite closely with my initial (wiggly) plot of dynamic resistance measured with a PSU and a couple of bench DMMs.

author: TimFox
author_rank: Super Contributor
timestamp: December 01, 2024, 11:30:03 pm
content: Addendum to G0HZU's graphs for 1N4148.I started with the "typical at Ta = 25oC" curves from the OnSemi datasheet for the 1N4148 diode.I combined the curves from 1 uA to 100 uA and 0.1 mA to 10 mA to get the "forward voltage" curve (left axis) that plots very close to a straight (heavy) line on linear V vs. log I axes.From the forward voltage curve, I numerically calculated the incremental resistance dV/dI, plotted as a dashed line on log R vs log I.Superimposed on that dashed line is a solid line representing a resistance given byRac= (49.8 mV)/I.These DC results are very close to those he obtained with RF voltages in the VNA test.Suitable choices of linear and log axes make the data easier to understand.

author: Xena E
author_rank: Frequent Contributor
timestamp: December 02, 2024, 09:23:03 am
content: Quote from: G0HZU on December 01, 2024, 09:33:31 pmActually, I'm putting this (and your subsequent posts in this thread), up for a special useful content award.Thank you Jeremy !X

author: MrAl
author_rank: Super Contributor
timestamp: December 02, 2024, 04:54:49 pm
content: Quote from: TimFox on November 30, 2024, 03:05:13 pmI am not sure which model you are referring to here.If you have a voltage source in series with the resistor, dv/dt must be the same for both that and the exponential model because that's how we get the resistor value, but is only exact at the chosen operating point. In the last plot I posted this should be clear.This is if you mean the voltage source in series with the resistor presented in that post as a plot.Maybe you meant the lone resistor model vs the dv/dt ?Those can't be equal because the 'dynamic' resistance is different.

author: MrAl
author_rank: Super Contributor
timestamp: December 02, 2024, 05:06:50 pm
content: Quote from: temperance on November 30, 2024, 07:26:49 pmHello,I don't think it was heated. People disagree, that's life.I have many years of experience in both design, testing, and study, in many areas. I've seen many models and how they are to be used. The problem here is that someone wants to reject one of the models. It's not the best model in the world, but the main point was not to talk about an exact or nearly exact model, just to illustrate how this negative feedback works. It did illustrate that, AND I even gave the region of operation and stated the errors, and it was still rejected by same person.For any model we have, there will always be errors. Some models are better than others because they stem from the physics of the device. Others are just behavior models and are not meant to be super accurate. Heck, they almost always leave something out that is not 'modeled' in that particular model.That's the basic idea about models.The base emitter diode modeled as a resistor OBVIOUSLY is not a great model, but it's much simpler than the dv/dt thing we got into. However, since there was some discontent with that model I did present the dv/dt thing model also, complete with a graph. The question is, did the original poster find the simpler one simpler or the dv/dt model better for their understanding. That's really what it is all about. It does not matter who does not like it, as long as the OP finds it useful, keeping in mind the original question was about the negative feedback not the model and not how a base emitter diode works.Also, there's now been some mention of misleading the OP with a model that is not that good, even though it illustrates the main point, and that misleading the OP could cause years of problems (ha ha). However, that's the way knowledge is acquired, one step at a time.[Sarcasm ON]Yeah, we all MUST learn EVERYTHING all at once[Sarcasm: OFF]

author: MrAl
author_rank: Super Contributor
timestamp: December 02, 2024, 05:22:46 pm
content: Quote from: magic on November 30, 2024, 09:28:47 amFirst, the acquisition of knowledge takes time, it's incremental. We never discuss the entire details of something all at once, especially not on forums like this.The base current 'causing' collector current is an accepted way of viewing the bipolar transistor, it's been used throughout the ages. It's expressed as:iC=ib*BetaI find it hard to believe you don't know this, really.If instead you are arguing details again, then you are again trying to put everything into one paragraph. Do you really, really, want me to describe everything about a transistor in one post? I can post some equations but it's going to take a lot of room, and the OP will be blown away with too much information all at once.When discussions like this come up, you have to be able to acknowledge what level the discussion is on at the time, and what the main point is.For example, when we talk about a resistor we do not talk about Johnson noise. We do not even talk about thermal perturbations which we all know happens in most devices. We leave out all kinds of stuff, even how we consider it to be a lumped element model most of the time.We only consider those things when they become necessary. Otherwise, we just make it too complicated.Modeling is just part of approximation theory, but the approximation is subject to the desires of the author. You can always reject a model or accept a model. If you would like we can present the same explanation using the voltage source + resistor model I presented with the graph in a previous post. I do not think you have a problem with that one.

author: MrAl
author_rank: Super Contributor
timestamp: December 02, 2024, 05:27:26 pm
content: Here's an interesting quote...I won't say who quoted this.What is the philosophy of approximation?Abstract: Approximation involves representing things in ways that might be close to the truth but are nevertheless false. Given the widespread reliance on approximations in science and everyday life, here we ask whether it is conceptually possible for false approximations to qualify as knowledge.

author: TimFox
author_rank: Super Contributor
timestamp: December 02, 2024, 06:17:32 pm
content: Quote from: MrAl on December 02, 2024, 04:54:49 pmThe elementary model of battery and internal resistor in series:The terminal voltage as a function of current through terminals isV=Veq+ (IxReq)The ratio of (voltage across terminals) / (current through terminals) =V/I= (Veq+IxReq) /I= (Veq/I) +ReqThe derivative dV/dI=ReqThis is an elementary model with the simple result that the ratio and derivative are very different.

author: tggzzz
author_rank: Super Contributor
timestamp: December 02, 2024, 06:17:49 pm
content: Quote from: MrAl on December 02, 2024, 04:54:49 pmHe is referring to this:Some people (q.v.) think R=V/I. While r=dV/dIhas a single value, R does not.

author: mawyatt
author_rank: Super Contributor
timestamp: December 02, 2024, 07:52:14 pm
content: Here's a simple circuit that folks might be interesting as it helps plot the Vbe transistor and diode dynamic impedance vs bias current.The concept uses a current source to bias the DUT at a fixed bias current and measure the small signal dynamic impedance about that bias point. We used a LCR Meter (TH2830) with a small test level of 10mv @ 10KHz to measure the dynamic impedance (R + jX).We've attached our notes and a plot of a 2N3904 connected as a diode (CB shorted) and swept the collector current from 10ua to 20ma, note the dynamic impedance at 1000ua agrees well with theory of ~26Ω at 300k. Vcc was set to +4V and Vee varied along with R to achieve selected bias current from 10ua to 20ma. We'll include some diodes later for comparisons when we have time.Edit: added 1N4148 Diode, 1N5817 SBD, and Setup Images. Later added the LED plot.Anyway, hope some folks find this circuit and concept useful.Best

author: TimFox
author_rank: Super Contributor
timestamp: December 02, 2024, 09:29:00 pm
content: Note that for the 1N4148 diode, the dynamic resistance is about (50 mV)/I, as in G0HZU’s and my graphs, but for the 2N3904 BJT it is closer to (26 mV)/I.The 26 mV value is typical for BJT base-emitter behavior.

author: temperance
author_rank: Super Contributor
timestamp: December 03, 2024, 03:49:45 pm
content: Quote from: mawyatt on December 02, 2024, 07:52:14 pmPeople doing real experiments with bare semiconductors. Very nice. I can encourage anyone in electronics do to just that.I did a similar experiment except I didn't use an LCR meter to measure re (I never thought of using an LCR meter.). I measure the gain of a common emitter amplifier with a cascode in it's collector to cancel the Early-effect.

author: mawyatt
author_rank: Super Contributor
timestamp: December 03, 2024, 06:45:59 pm
content: Using same circuit shown above just added the LED Dynamic Impedance plot.Best

author: MrAl
author_rank: Super Contributor
timestamp: December 03, 2024, 07:40:38 pm
content: Quote from: TimFox on December 02, 2024, 06:17:32 pmHi,Yes ok, I agree with that, and I believe you are correct.However, I do not believe that is enough of a comparison to fully understand the two (or three) models and their applications.Since there has been some misunderstandings about the models I felt that a more comprehensive comparison would be more illustrative.In the attachment, we can see the plots for the three models. Since the single resistor model (green line) was the only one that came into question, I mapped out the region of validity with the light gray horizontal and vertical lines. The best place is within the square formed by all four of the light gray lines.We can note that outside that range it's not going to be very good, but inside, it's not too bad. It has the benefit of not having to explain what the calculus definition of dv/dt is which may not yet be part of the coursework. It can certainly be called an oversimplification, no doubt there. The question is, did it have any uses. Well, the answer to that has to be that it has very limited use, that's probably why the complaint came inWhat I did not show was the more common oversimplification, which is the constant voltage model (no resistor). From experience we know this is taken to be between about 0.6 to 0.8 volts DC, with the most common being 0.7 volts. That's a huge simplification (ha ha) yet we still find it very useful.When circuits get complicated this is not an unusual thing to do. We oversimplify just to get a rough idea how the circuit works, then later if we want to go deeper, we go deeper with the theory or use spice. A spice model for a bipolar is fairly complex so we often avoid that except in a simulator.

author: temperance
author_rank: Super Contributor
timestamp: December 03, 2024, 08:05:38 pm
content: Quote from: MrAl on December 03, 2024, 07:40:38 pmYou are beating around the bush and wagging your tail like a fish on dry land. Your oversimplification is wrong and can't be used in practice. How is the OP going to understand when your oversimplification is only valid over a limited range without understanding what has been simplified? It is to those who are trying to understand this very confusing. What if they believe you and start to apply this limited model in places or over ranges where it doesn't hold?I think you also have to admit that you changed your oversimplification a few times now to fit the real real model. You first presented the model as U/I (over a limited range). Later on you added a voltage source in series with a resistor and now you return to the U/I model to show us that your model is valid over a limited range. Besides, the model is valid over a limited range for any resistor and the range can be all the way close to the Y axis or far away to the right.QuoteI think you must consider turning in your degree.

author: mawyatt
author_rank: Super Contributor
timestamp: December 03, 2024, 08:44:38 pm
content: Here's a simple example of how the range of linearization can effect results. We'll use the Diode connected 2N3904 NPN Transistor mentioned above.The Dynamic Impedance of the Base Emitter junction is well known and Vt/Ic which is derived by taking the derivative of Vbe wrt Ic, where Vt is kT/q and ~26mv @ 300k. This equates to a Dynamic Impedance of 26Ω at Ic = 1ma, and the derivative is an infinitesimally small increment around the collector bias current Ic, 1ma in this case.To show this is actual use we've employed the circuit previously mentioned and set the collector bias current to 1ma and varied the 2N3904 Diode connected base to emitter excitation from the LCR Meter from 10mv to 100mv.Note as the LCR Meter excitation voltage becomes larger deviating from infinitesimally small (~ zero) the Vbe Dynamic Impedance rises well above the theoretical value of ~ 26Ω @ 1ma, and over 60Ω at 100mv. This illustrates the effects of "Small Signal" and "Increment Range" when considering linearization of a non-linear component about a fixed point (bias), such as a bipolar transistor.Anyway, hope this helps show how linearization details need to be carefully considered to achieve desired approximations and results.Best

author: iMo
author_rank: Super Contributor
timestamp: December 03, 2024, 09:07:17 pm
content: Quote from: G0HZU on December 01, 2024, 11:04:53 pm.. therefore we use the PIN diodes as the switches and attenuators till many GHz and kWatts..

author: TimFox
author_rank: Super Contributor
timestamp: December 03, 2024, 09:21:39 pm
content: PIN diodes are handy at RF because a reasonable DC current will store enough charge in the device to allow much higher AC current, so long as the period is short enough that the charge is not swept out of the diode.

author: mawyatt
author_rank: Super Contributor
timestamp: December 04, 2024, 12:36:42 am
content: PIN Diodes are great for RF and up, but severely limited for lower frequency use due to Carrier Lifetime.Best

author: MrAl
author_rank: Super Contributor
timestamp: December 04, 2024, 12:54:23 pm
content: Quote from: temperanceSee this is what it comes down to with people like you who would rather look for things that are wrong then try to understand what *could* be right. You just can't, for the life of you, resist the temptation to mock not only the person whom you have NO idea who it is, as well as the subject matter and its applications.You're stating something that is wrong and can't be 'used', when it has already been 'used' in practice.You can't even grasp the reason for the presentation of the different models, you even attribute that to some fault with the person who posted the models. This certainly shows you would rather seek problems with people and technical issues that you don't seem to understand, then try to grasp the information presented to you.You might want to check and see if I dotted the end of all my sentencesThere's no reason for us to have any kind of technical discussion until you get over this trait. You'll only keep hunting for more things wrong.When you mock out people and their ideas, new or old, you tend to piss people off, not have a meaningful discussion about the various issues. Now I sort of don't care what you think of myself or even others, or what you think about any technical information being presented here by anyone.Here's another interesting quote...START QUOTEphi·los·o·phy/fəˈläsəfē/nounthe study of the fundamental nature of knowledge, reality, and existence, especially when considered as an academic discipline.a particular system of philosophical thought.plural noun: philosophies"Schopenhauer’s philosophy"the study of the theoretical basis of a particular branch of knowledge or experience."the philosophy of science"END QUOTEwith attention to:"a particular system of philosophical thought"

author: tggzzz
author_rank: Super Contributor
timestamp: December 04, 2024, 04:44:31 pm
content: Quote from: MrAl on December 04, 2024, 12:54:23 pmThere's no point in considering whatcouldbe right when when it is known whatisright (and whatiswrong).Hint: it *could* be that Maxwell's daemons are determining the externally observed behaviour. Why don't you presentthatconjecture to the OP?

author: Analog Kid
author_rank: Super Contributor
timestamp: December 05, 2024, 12:07:32 am
content: Quote from: tggzzz on December 04, 2024, 04:44:31 pm OP here. All I've got to say at this point is that y'all have turned a perfectly innocent question, which BTW was pretty well answered long ago in this here thread, into a nearly-theological discussion akin to how many angels can dance on the head of a pin.So my reaction at this point is the canonical *yawn*. Not even worth getting out the popcorn for.So as Linda Richman (Mike Myers) used to say, talk amongst yourselves ...

author: shabaz
author_rank: Frequent Contributor
timestamp: December 05, 2024, 03:12:41 am
content: Quote from: TimFox on November 28, 2024, 08:03:34 pmI ordered the Transistor Circuit Design book (used, Amazon), and it covers more topics than I expected. It's concise and does make a few assumptions (the author mentions there is additional content in his other books) but is actually very clear to follow.I have briefly read through several of the chapters, and I'm going to enjoy reading them more. It will be a nice addition to the small collection of books that I use.I've scanned the table of contents (PDF attached) in case anyone wants to take a peek at what the book contains.

author: MrAl
author_rank: Super Contributor
timestamp: December 05, 2024, 11:13:51 am
content: Hi,What I find most interesting about a lot of people who talk in forums like this is that they tend to only believe what they have learned before the time they read something new. It does not matter to them where this new information came from either.Maybe this is because of the trend toward misinformation in many areas of life and news. We do have to be careful what we believe these days that's for sure, so it's hard to point a finger sometimes.I have talked to a great number of people on the web and on the phone and in emails and in real life. Most of them (not all though) will stick to a theory or idea that they have learned in the past no matter what proof you show them about something new. They even believe that physics itself is a settled argument for how nature works, which means that there is absolutely NO room for change. With that in mind it is interesting how physics has changed over the years, from Newton to Einstein, and from classical physics to quantum physics. The only thing that does not change is that everything changes.One more thing is interesting that came up in this thread too. That's about a case study of how new information affects a person's learning experience. it should be mentioned that this is something that has to be researched over a rather long period of time, and there is no way to form a conclusion based on one single case. When this is done in any subject area, there are always a lot of sample subjects involved, never just one.

author: SteveThackery
author_rank: Super Contributor
timestamp: December 05, 2024, 11:36:58 am
content: A person's knowledge and beliefs form part of their identity, so embracing new, contradictory knowledge requires giving up part of that identity. It requires you to accept that you were wrong about something for years. Both are extremely hard to do.

author: magic
author_rank: Super Contributor
timestamp: December 05, 2024, 04:25:10 pm
content: Not sure what this has to do with anything.Nothing novel was discussed in this threadOr are you describing your own journey to the realization that dV/dI ≠ Vf/If for a diode?

author: SteveThackery
author_rank: Super Contributor
timestamp: December 05, 2024, 04:42:34 pm
content: Quote from: magic on December 05, 2024, 04:25:10 pmI was responding to this:Quote from: MrAl on December 05, 2024, 11:13:51 am

author: KE5FX
author_rank: Super Contributor
timestamp: December 05, 2024, 06:29:57 pm
content: Type 1 Moron: Someone who believes the first thing he/she heard, and is difficult to convince otherwiseType 2 Moron: Someone who believes the last thing he/she heard, and is easy to lead astrayMost of us have played both roles at one time or another, but many others fall persistently into one category or the other.

author: Analog Kid
author_rank: Super Contributor
timestamp: December 05, 2024, 09:30:22 pm
content: As the OP here, I sincerely hope I fall into neither of those categories.It's true that I probably have some fixed ideas about things like transistor theory and operation that are just plain wrong.That's one reason I post here; I hope that by thrashing things out in a forum like this that I'll become better informed, that is, actuallylearn something. (What a concept!)At the same time, for my money I could do without a lot of the side issues that inevitably come up in a discussion like this. Not that I'm advocating that people just shut up here: but when someone like me asks a question that is pretty clearly requesting first- or second-order explanations, and then gets deluged with 3rd, 4th, 5th and beyond extrapolations, it just muddies the waters. So it'd be nice if people could tailor their responses to the level of information being sought, and not delve into subatomic physics or whatever (unless that was the question in the first place). Not that I'd really expect that to happen ...Just a thought ...p.s.: I really do appreciate those who take the time to explain things here clearly and with the original question in mind. Good stuff!

author: temperance
author_rank: Super Contributor
timestamp: December 05, 2024, 11:11:05 pm
content: KE5FX comment is surely not directed towards to you but those like me who commented on some questionable method presented where a diode can be replaced with a resistor by simple dividing Vf /If.You can do that but you must carefully select two points on the Vf If graph and calculate the difference in resistance between those points which is the same as taking the change in voltage divided by the change in current. Or you simply use rd= 25 mv / I.Understanding what the intrinsic or dynamic emitter resistance (denoted as rd) is all about is important and if you are learning this stuff on your own I can understand it can be intimidating.As tip: in a common emitter amplifier, the gain is equal to Rc (the resistor in the collector) / (Re + re). Re is the fixed resistor and re is the intrinsic emitter resistance. You can exploit this to create a variable gain amplifier by changing the transistor bias current.Don't stop asking question.

author: mawyatt
author_rank: Super Contributor
timestamp: December 06, 2024, 12:06:07 am
content: Quote from: Analog Kid on December 05, 2024, 09:30:22 pmThe Base Emitter Junction is such an important aspect of bipolar transistor behavior that folks want you to stop and think about what's going on, rather than just overstep it. You've already demonstrated a curiosity about transistors and folks just don't want you to miss out on a incredible learning experience that awaits one with these fascinating devices, basically the more detail the betterThe bipolar and FET devices are such incredible devices, probably the most important inventions of all time, even after being involved with them for over 60 years, we still get a kick out of playing around with them, and that includes sketches (back of the envelope), hand analysis, SPICE simulations and actually lab measurements.Today you can do all this right at home, no need to go to University with Fortran Cards to run SPICE, then travel to the University Labs to get the expensive transistor devices and then to the TE labs for all nighters like we did as kids. With transistors available for less than a dime, DSOs for few hundred bucks and a $50 AWG and PS one can have a really great time exploring the fascinating world of bipolar and FET transistors, oh and you can run SPICE right on your home computer for freeThis then becomes a lead in for the physics defying CMOS scaling that's going on and the incredible chips built with such, and one slowly begins to appreciate what's going on "behind the scenes" as their semiconductor knowledge base expandsBest

author: Analog Kid
author_rank: Super Contributor
timestamp: December 06, 2024, 12:11:48 am
content: Thanks. Just one thing, though:AWG???o American Wire Gauge?o Associated Wholesale Grocers?o American Welding & Gas?o Association for Women Geoscientists?(like something from that "pet peeve" thread)

author: mawyatt
author_rank: Super Contributor
timestamp: December 06, 2024, 12:51:39 am
content: Arbitrary Waveform Generator!!These are amazing generators that can produce all sorts of built-in waveforms, well beyond the usual Sine, Square and Triangle and have the ability to accept user created custom waveforms as files.Best

author: TimFox
author_rank: Super Contributor
timestamp: December 06, 2024, 03:43:48 am
content: Quote from: Analog Kid on December 06, 2024, 12:11:48 amI have found that an important thing to tell novices is the "term of art" for different concepts to facilitate looking it up.

author: Analog Kid
author_rank: Super Contributor
timestamp: December 06, 2024, 04:05:13 am
content: Or how 'bout just using the damn term itself instead of a TLA?

author: coppercone2
author_rank: Super Contributor
timestamp: December 06, 2024, 05:09:35 am
content: Quote from: mawyatt on December 04, 2024, 12:36:42 amIs it possible to make a impractical pin diode that would work for low frequencies? I.e. making it some stupid size or something like that?If the charge decays too quickly in the region, is it possible.. to some how add more charge? or make it bleed slower?For some reason this makes me wonder about semiconductors made with radioactive isotopes. I wonder what they would do.Gallium-67, Be-10, Zn-65 for P-type additives and P-33, As-76, Sb-125Substrate can be Ge-76 or Si-32Would making a pure radioactive material transistor do anything lol?

author: mawyatt
author_rank: Super Contributor
timestamp: December 06, 2024, 02:08:28 pm
content: Quote from: coppercone2 on December 06, 2024, 05:09:35 amWRT the PIN Diode, the limiting factor for lower frequency use is the carrier lifetime, which is the "Intrinsic Semiconductor" in the P-I-N junction where this becomes relatively slow since the "I" region has no doping for excess carriers and separates the P and N doped silicon but due to high level injection from the highly doped P and N regions becomes "flooded" with carriers when forward biased, and depleted when reverse biased. As frequency decreases these carriers in the Intrinsic region when forward biased become modulated and eventually this starts to behave like a regular diode, thus not useful as a "Switch" for supporting large signal AC currents riding on a junction DC forward bias.A better lower frequency semiconductor switch is the CMOS type which utilize both N and PMOS devices to improve the overall switch linearity and work well from DC and up in frequency. We did a study last century to find the best active semiconductor switch wrt to 3rd order IMD, and the NMOS device proved the best overall for a given active area, and later confirmed by others.Not sure what use a radioactive transistor would be, due to local ionization seems it would have really bad low frequency noise thoBest

author: iMo
author_rank: Super Contributor
timestamp: December 06, 2024, 02:38:34 pm
content: I like coppercone2's out of the box ideas..Re PIN diodes and radioactivity - years back I built myself a radiation detector with a PIN diode photodetector SFH.. something (pretty large surface, like 3x4mm, black), fet+2transistors, put into an old black film container covered by Alu foil. And yes, it worked. Cannot be compared with geiger tube, no way, but I got pulses from time to time (against WWII dials).PS: SFH205F infra

author: coppercone2
author_rank: Super Contributor
timestamp: December 07, 2024, 07:09:24 am
content: Well a radioactive diode might be the easier thing to try to make. I don't know if you can get all those isotopes together in a reasonable amount of time, you might need a special factory with multiple reactors to make all the materials needed for assembly before it decays too muchBe-10 and sb125 have a long one.Are isotope free transistors lower noise then?

author: iMo
author_rank: Super Contributor
timestamp: December 07, 2024, 07:23:07 am
content: Quote from: iMo on December 06, 2024, 02:38:34 pm..almost getting a feeling the AI is creating them..

author: MrAl
author_rank: Super Contributor
timestamp: December 08, 2024, 01:22:14 pm
content: Quote from: SteveThackery on December 05, 2024, 11:36:58 amHi,Well that is considered an "inductive" personality relating to how they can adapt to change. "Inductive" here is similar to how current cannot change quickly in an inductor. It is known that the higher the intellect, the easier for them to adapt to changes. They also tend not to whine about everything.People that can adapt easily see opportunity where others see failure.Sometimes we also view the non-adaptive personality as stubborn.What might be the most interesting aspect of this though is when we have ideas A and B, and person 1 claims A could be the right choice and person 2 says that both A and B could be good choices, yet person 1, even without definitive evidence, still claims that A is the best. Sometimes person 1 will even say that they acknowledge that A and B can be right, but they still go on to push A repeatedly, even though they still do not have any evidence that A is right.I guess it just comes down to how open minded a person is. Some experts are very open minded and will consider many arguments in detail before rejecting something, while others are very close minded and believe what has been found in the past must always be the right answer. This surprises me sometimes because we know from the historical record that physics changes.The best example here 'might' be archeology. The stuff found in the past dictates the theory, but it's held onto like it's gold or something. Then all of a sudden something new is found, and that changes everything. If you find something on day 1 and take it to be the ultimate answer, then on day 2 when you find something new it changes things that were taken as truth on day 1, you should be able to learn from that lesson.Anyway, back to the original questions ...

author: temperance
author_rank: Super Contributor
timestamp: December 12, 2024, 08:48:05 am
content: MrAI,take this post from you:https://www.eevblog.com/forum/beginners/what-exactly-is-emitter-degeneration-anyhow/msg5735353/#msg5735353Which I quote here:QuoteYou admit yourself that the oversimplification is only valid over a very limited range and it's use is very limited. And here I think you have to admit that the size of the rectangle for which the approximation is good enough becomes smaller and smaller with decreasing bias current to the point of unusable well before the diode forward voltage falls below the knee voltage and vice versa. Those are the things you fail to mention. Instead you choose a bias point where the approximation is reasonably valid in a tiny rectangle to prove your point. I think this is misleading and should only be used by people with experience who developed a solid understanding of what is going on to understand where the oversimplification falls apart. Or in other words: how is someone without experience able to tell when the model falls apart? They don't . When people confront you with that you revert to insulting people in here several times by calling them species of substandard intelligence who fail to lean something from what you have to offer. And till today you continue do so.You then go on to state that your Vf / If oversimplification avoids complicated math and is as such easier to understand while all that is required is calculating ΔVf / ΔIf at a few points. That's not difficult at all and doesn't require any calculus. Neither does the ΔVf / ΔIf simplification fall apart over a very wide range above the diode knee voltage. (You demonstrated this in an other post.)In the end, to validate their intelligence you get people like KE5FX and SteveThackery - who probably didn't read anything of what has been discussed here - supporting irrelevant visions backed up with some smart sounding quotes posted by MrAI.But maybe I've spend to much time reading "the art of being right" written by your favorite author.

author: SteveThackery
author_rank: Super Contributor
timestamp: December 12, 2024, 10:33:19 am
content: Quote from: temperance on December 12, 2024, 08:48:05 amWhat the HELL? This totally misrepresents my contributions to this thread. My first contribution was a lengthy post answering the OP's question in a way which I tried to make simple and intuitive explanation of emitter regeneration. It was thanked by the OP. My second contribution was a passing comment on the value of book-learning.I can only think you must be referring to my third post:"A person's knowledge and beliefs form part of their identity, so embracing new, contradictory knowledge requires giving up part of that identity. It requires you to accept that you were wrong about something for years. Both are extremely hard to do."That is a philosophical statement, based on my personal experience, about why humans seem to be resistant to change, especially resistant to changing their ideas. I was emphatically NOT taking sides in this pathetic willy-waving debate, about which I don't give a fuck.It was a passing comment, that's all. And by the way, I totally stand by it, even though it's an aside to the main debate.I detest people misrepresenting my words or my position. You owe me a big apology - are you big enough to give it?

author: magic
author_rank: Super Contributor
timestamp: December 12, 2024, 11:00:25 am
content: Quote from: temperance on December 12, 2024, 08:48:05 amIs it this plot?I can't help but notice that there is a much simpler and more accurate approximation of Vf inside that gray box: just assume it doesn't depend on If at all.

author: MrAl
author_rank: Super Contributor
timestamp: December 12, 2024, 11:57:12 am
content: Quote from: magic on December 12, 2024, 11:00:25 amHi,Yes sure, and for the model not shown you can simply draw a straight horizontal line that intersects the Y axis at v=0.7 volts. That's another model that is used time and time again. It assumes that the voltage is always 0.7 volts no matter what.There are better models, and even better than that exponential curve indicates. They are all approximations, and we use whatever approximation works, as long as it works for what we need it to. The exponential model is rarely used in hand calculations because it's much more complicated, and when you have more than one diode you have to do some nasty numerical solution to get the right result. It can be fun though, unless you don't like math.

author: tggzzz
author_rank: Super Contributor
timestamp: December 12, 2024, 12:29:32 pm
content: Quote from: SteveThackery on December 12, 2024, 10:33:19 amTemperance's post was very badly worded w.r.t. exactly who he was referring to at each point.However, my (speed?) reading of it was that he was showing MrAl a LART, not showing you one. In that view of the world, I can agree with both Temperance and your earlier posts.

author: MrAl
author_rank: Super Contributor
timestamp: December 12, 2024, 12:31:18 pm
content: Quote from: temperance on December 12, 2024, 08:48:05 amHello again,So, you explain to me that I've insulted some people, by you yourself insulting some people. You actually even mentioned them by name. Does that sound right to you?You have to be careful how you interpret things said, they are not always to be taken in the negative sense. For example, by saying that intelligence plays a part in understanding a certain concept, it does not mean that people who do not understand it yet are stupid. For one thing, we all have to learn when presented with ideas we never heard of before. That does mean that we are ignorant of that concept to start with, but that does not mean we boil it down to we are stupid as an absolute, never-ending phenomenon. Once we learn that concept we then have that under our belt.So in most cases we are just acting stupid, not entirely stupid for all time.Also, by stating something about intelligence it could be an attempt to get people to think about it more carefully, then they might attempt to learn the concept. A typical dialogue: "John-Boy, you are acting so stupid right now!"Actually, I don't think anyone here is stupid in the absolute sense, but some may be more stubborn than others. I can be stubborn sometimes too, but I've learned that it's best to consider new ideas than to knock them.This is a topic we should really take to PM's though it's not helping the OP I don't think. The OP wanted something simple and understandable and has little interest in the behavior of people in forums.

author: MrAl
author_rank: Super Contributor
timestamp: December 12, 2024, 12:33:02 pm
content: Quote from: tggzzz on December 12, 2024, 12:29:32 pmActually he seemed to be addressing a MrAI, and that's not me

author: MrAl
author_rank: Super Contributor
timestamp: December 12, 2024, 12:48:42 pm
content: Here's another interesting quote I found written by Frederic Crews:"General rationality requires us to observe the world carefully, to consider alternative hypotheses to our own hypotheses, to gather evidence in a responsible way, to answer objections. These are habits of mind that science shares with good history, good sociology, good political science, good economics, what have you. And I summarize all this in what I call the “empirical attitude.” It’s a combination of feeling responsible to the evidence that is available, feeling responsible to go out and find that evidence, including the evidence that is contrary to one’s presumptions, and responsibility to be logical with one’s self and others. And this is an ideal that is not so much individual as social. The rational attitude doesn’t really work when simply applied to one’s self. It is something that we owe to each other."I like the last line best.

author: temperance
author_rank: Super Contributor
timestamp: December 12, 2024, 01:38:12 pm
content: Quote from: SteveThackery on December 12, 2024, 10:33:19 amIf I'm wrong I'm not afraid to admit being wrong at all. In my interpretation I didn't see what you wrote as a passing comment but taking sides. So I apologize for accusing you of such.

author: SteveThackery
author_rank: Super Contributor
timestamp: December 12, 2024, 05:28:52 pm
content: Quote from: temperance on December 12, 2024, 01:38:12 pmThank you,@temperance. All is well. 👍😁

author: KE5FX
author_rank: Super Contributor
timestamp: December 12, 2024, 08:03:08 pm
content: Quote from: tggzzz on December 12, 2024, 12:29:32 pmIthinkhe insulted me, but if so, it was an unusually easy one to shrug off.

author: Analog Kid
author_rank: Super Contributor
timestamp: December 12, 2024, 08:08:54 pm
content: And who is "MrAI"?

author: KE5FX
author_rank: Super Contributor
timestamp: December 12, 2024, 09:01:03 pm
content: The object of MrFusion's sibling rivalry?

author: G0HZU
author_rank: Super Contributor
timestamp: December 12, 2024, 09:14:42 pm
content: If it helps demonstrate the power of using degeneration up at RF, have a look at one of my old reference LNA designs below. This is a low noise BJT RF amplifier that uses inductive degeneration at the emitter and some collector-base degeneration (negative feedback) to achieve a good compromise between noise figure and input and output match from 1 MHz to just over 100 MHz. This produces a flat gain of just over 15dB and a low noise figure and good port VSWR.This is one of the reference amplifiers I use to do sanity checks on my noise figure measuring gear. Because the input and output VSWR is very low, it produces low uncertainty. The noise figure is a flat 1.3dB all the way up to 100 MHz.Note that this is the original design simulation. The real version of the amplifier isn't quite this good. The noise figure of the real amplifier is just under 1.4dB most times I measure it and the input and output VSWR is about the same up to about 100 MHz. I can improve on the noise figure and the VSWR but this comes at the expense of bandwidth. So this design is fairly optimal up to just over 100 MHz.The degeneration helps produce consistent performance all the way up to about 100 MHz or so.It's a great way to make sure my noise figure meter is working correctly when measuring noise figures from a few MHz up to about 145 MHz.

author: KE5FX
author_rank: Super Contributor
timestamp: December 12, 2024, 09:41:39 pm
content: Nice. Have you posted a schematic of that one?

author: G0HZU
author_rank: Super Contributor
timestamp: December 12, 2024, 10:03:20 pm
content: Quote from: KE5FX on December 12, 2024, 09:41:39 pmNot yet. I'll dig out the real amplifier over the weekend and demo it on a VNA and a noise figure meter. It uses a classic 2SC3355 BJT from NEC and this BJT was very special in its day. In this application it only draws about 6mA so the large signal handling is very poor. Therefore, this amplifier isn't of much use outside its intended use as a reference amplifier for my noise figure system.There's nothing special or clever in the design, it just uses C-B resistive feedback and a small amount of emitter degeneration using inductance. This is achieved by not having the shortest possible emitter leg connection. So the original leg of the BJT can provide the degeneration here.

author: KE5FX
author_rank: Super Contributor
timestamp: December 12, 2024, 11:23:51 pm
content: Quote from: G0HZU on December 12, 2024, 10:03:20 pmMakes sense. It might be interesting to try duplicating it to see how it looks on my 8970B, if you do get a chance to document it. Most HF noise figure measurements I've made on that box have come out a bit worse than expected, and that might be a good excuse to look into it further.

author: MrAl
author_rank: Super Contributor
timestamp: December 13, 2024, 05:28:36 am
content: Quote from: Analog Kid on December 12, 2024, 08:08:54 pmHa ha, that's what I was wondering tooI got my name from a nice older Spanish lady who used to call me (in all caps) MR AL. She was really cool. Long time ago.

author: G0HZU
author_rank: Super Contributor
timestamp: December 13, 2024, 05:42:49 pm
content: Here's a VNA plot showing S21 S11 S12 S22 for the BJT amplifier with degeneration. I've also measured the noise figure using a 346A noise source and a E4440A PSA spectrum analyser.I used an N4431-600006 B Ecal to calibrate the VNA.You can see that the noise figure isn't quite as good as the simulation but it is typically just under 1.4dB according to the analyser.The return loss of port 1 and port 2 of the amplifier is > 30dB up to about 100 MHz. The aim here was to make the port match as good as possible whilst still achieving a low noise figure. It is possible to get the noise figure lower than this by changing the degeneration and the bias point. But when I did this the amplifier lost some bandwidth.