jonk

@PortaL No problem. All of us start at zero and have to walk a path by taking the first step. We aren't born with any of this! Just to get your brain in the right frame of mind, consider the meaning of position when there is only one particle in the entire universe. It's only when you add a 2nd particle that, at least, the idea of a relative position exists. That still doesn't provide the concept of distance. It takes 3 particles to do that and even then you now only have relative distance.

@PortaL I'm mostly trying to suggest that you need to think more carefully. Don't complicate the situation by adding in iron. It's just a bunch of matter which tends to reduce the magnetic path length (by a large factor) while also adding still more complications, making any explanation far longer and more complicated to attempt. But some basic ideas are still there without the iron. You seem to be seeking a deeper understanding. To do that, remove the distractions and focus on the essence.

@PortaL Einstein quickly realized that magnetic fields must transform into electric fields under a Lorentz transformation. 4x4 antisymmetric field tensors now capture all the electric and magnetic fields (\$\pm E_x, \pm E_y, \pm E_z, \pm B_x, \pm B_y, \pm B_z\$) -- these 12 plus four zeros along the diagonal. The use of this field tensor to the situation described above would see certain summations collapse to single terms in just such a way that a pure magnetic field with no electric field components transforms, via a moving frame, so that it now possesses electric field components.

@PortaL Now, let's say you are in the inertial frame of reference of the wire, not the laboratory or its magnet. Now the electrons are at rest. But the magnetic field moves instead to the left. Tell me. The electrons must still feel a force. Yet in your new frame of reference they are not moving, so \$\vec{v}=0\$. And yet there is still a force acting on the electrons. This force on electrons here can only be due to an electric field. Then the only possible explanation is that a moving magnetic field creates an electric field.

@PortaL An electron will experience a Lorentz force equal to the charge \$e\$ times the cross-product with the magnetic field, \$\vec{B_z}\$: namely, \$\vec{F}=q_e\cdot\left(\vec{v_x}\times\vec{B_z}\right)\$. (This is where Mr. Franklin gets involved.... by the right-hand rule we know that the electron will experience a force upward so that an electron current in the wire points up. But because of Franklin, we say that the positive current points down in this case. Luckily, this isn't the point of where I'm going, so I don't care right now.)

@PortaL Suppose something simpler: a moving conductive wire (in its own inertial frame of reference) as observed from the inertial frame of reference of a laboratory using a fixed magnetic. The wire is oriented parallel and along the y-axis here and is moving along the x-axis. The magnetic field (created by a stationary magnet) in the laboratory has only one component for the purposes of this experiment, \$B_z\$ (pointing towards you out of the 'page' here.) And the laboratory has ensured that there is no electric field.

@PortaL I wasn't asking for a simpler question. (Your added comment.) I was asking for some clarification which likely required better writing and possibly, failing better writing, then perhaps a more elaborated question. Looking over your comments as they've been added over time, there's no longer any single thing to answer -- in fact, you keep moving the goal posts around. For example, an hour ago you asked, "... is the Lorentz force really the dominant force considering a coil with iron core?"

I wish I understood the question.

@user1850479 I've been known to use (insert into the connector) nearby and cheap 'wax paper' to mix modes in desperate cases where I had to do something "quick and now." And yes, that's another option to consider for the OP. But even with mode-scrambled fibers there are variations at the exit from the fiber to the target. So it's still a matter of trial and error to get where one needs to be. There's no magic wand. Just ideas that may improve a situation a bit.

@mkeith Agreed about specs. I wanted NIST-traceable LED standard candles which could be used to cross-calibrate instruments like the (now ancient) disappearing filament pyrometers. It wasn't necessary to be accurate, just precise and stable, since NIST would provide calibration data for accuracy. It was important that they "didn't move around" once I got them back from calibration. LEDs flicker. Even good ones do. Useless for this purpose as I needed something that would stay true to their calibration table long enough to be practical. Only very rare, hand-selected LEDs met that challenge.

@matcha Those incandescent bulbs, once calibrated, are only good for 100 hrs of operation before you buy another. So $2000 divided by 100 means they cost $20 per hour of use. I figured I could offer an alternative. And LEDs "last forever" right? hehe. So wouldn't need to re-calibrate so often, either. Lots to write home about. But the stupid things are annoyingly bad as irradiation standards. It can be done. And yes, I made some that worked. But the investment in baking them out and throwing so much time and LEDs away just made it not nearly so viable. Now it's your turn at the wheel! ;)

@matcha They are all bad news, so far as I know. Some day I will go identify a viable market, buy 10,000 of the best LEDs I can find from the best manufacturer, set up another heated testing room, build the geometric testing setup and calibrated spectrophotometers, spatial filtering, and run through all this again. Maybe. Or you can do it for me and let me know how it goes! ;) I was involved in using NIST traceable incandescent standard bulbs (about $8-$20 each, uncalibrated, but about $2000 each calibrated) and considering the use of LED candles as an alternative. No go.

@matcha You probably also want to characterize the stupid, darned things for emission versus angle because they won't be the same on that score, either. You will need to work things out to carefully integrate irradiance over area at your calilbration distance and geometric arrangements. LEDs are a real pain to use as calibrated sources. I started out thinking "Wow! These can make very nice, cheap sources. I can sell a million of these and get rich." How so very ignorant I was.

@matcha You will definitely want to control the current. But you will want to calibrate each LED. And you will either need to carefully stabilize their temperature when operating them (thermal control by heating, likely) or else you will need to calibrate the darned things over a wide range of operating temperatures and include a temperature sensor of some kind, as well. Also, most LEDs drift over time. And I don't mean smoothly. Most of the darned things flicker. (Short time scales.) So you will want to throw away a lot of them, too -- those that flicker go in the garbage.

@matcha So you do want these to present the same irradiation, then. This gets into a host of issues. Are you looking for irradiance in watts per m^2 given a known distance and alignment with the radiator, independent of the samples being irradiated? Or some knowledge of what's absorbed by specific at the biological site? I think I'd like to know a lot more about what this is all about.

@matcha We'd throw all the rest of them away as useless.

@matcha, Do I understand correctly that these are 780 nm LEDs? And that therefore this has nothing to do with illumination but is only about irradiation? (Since human perception isn't an issue.) If you are looking to make "standard irradiance candles" out of these LEDs, you are going to be working VERY HARD to get there. Out of 1000 LEDs held in a stable oven to make sure they all operated at the exact same temperature, and operated with 0.5% precision current sources, and operated over a monitoring period of 3 days, I'd find about 5-15 of them are stable enough to use as candles.

@Antonio51 Even that won't help. Even LEDs cut from the exact same wafer won't be the same as each other driven by the same current.

Until later.

I'm also glad you could help! ;)

I'm following. (You are taking the emitter degeneration 33 ohm resistors together with the dynamic resistance in parallel with the AC leg of 37.4 ohms.) And again, thanks. I will get more familiar with it over the next few days. Best wishes.

That's all the questions I have. Just mostly thanks for setting me straight, right now.

In the amplifiers I've built (not this kind) that makes sense. I just need to see why this works as a dominant pole comp here. I'll look closer at Q1.

What was the point of R24 and C1? I've some thoughts about why, but when I look at the values all those thoughts go away. Do you see its point?

And with your lower beta values I calculate closer to 1200. So I'm feeling better given your needed re-alignment of my head!

So thanks for taking a moment.

And today, I didn't alter my view.

And thanks for the kick in the head. Last night when asleep I was 'seeing different things.'

@JonathanS. I've ignored emitter degeneration, etc. By the way, I broke the loop and disconnected one end of R1 from the output. I then applied two different but very tiny voltages to R1 to see what changes happened at the output. LTspice says it is about 880 V/V, if that technique isn't totally flawed.

@JonathanS. This is then multiplied (roughly) by the two \$\beta\$s of \$Q_4\$ and \$Q_8\$ (ignoring for now that it is really the VBE of each that controls where I'm headed), which increases the current sunk at the output and, through \$R_{27}\$, thereby \$V_{_\text{OUT}}\$ declines in response. NFB! If I estimate the two \$\beta\$s at 200, I get about \$-3200 \frac{\text{V}}{\text{V}}\$ response to drive back. I'm looking for quantification here. Do you find similar results?

@JonathanS. So allow me to walk it through at DC to see how a bias error is compensated. (So no AC paths!) Suppose \$V_{_\text{OUT}}\$ isn't zero and is positive. Then \$I_{R_1}\approx \frac{V_{_\text{OUT}}}{R_1}\$ is sourced into IN 1. This has to sink through \$R_{16}\$ (who cares right now?) but then \$R_4\$ where we do care. This is a \$\Delta V\$ change across \$R_{17}\$ leading to a \$\Delta I\approx \frac{V_{_\text{OUT}}}{R_1}\frac{R_4}{R_{17}}\$ at \$Q_{11}\$'s collector.

@JonathanS. That still fits my earlier view of it. Nothing has changed yet. So IN 1 can be taken as essentially "near ground" at all times, even with AC applied. Though the current feedback will 'adjust that' by multiplying 30+ ohms times the current feedback through R1. So it will move around a little bit. That means R1 really can be taken as having one end near ground and the other end at the output voltage. What I need to do is to calculate the loop, now. It's off to some paper.

@JonathanS. I'm starting at the front end. Variations in the current source/sink won't matter too much about the voltage difference between the two emitters of the first two BJTs as most of it is just dumped to the rails. The input node without a signal will be 2.2k times the base current differences (not much) so should be near ground (within a few mV, probably.) And the node between the emitter resistors of the next two BJTs of the input stage (IN 1 as I labeled it) shouldn't be more than some tens of millivolts away from that input voltage. So a few tens of mV away from ground.

@tobalt yeah. That's the misdirection part. It looks just like a standard NFB that you see all the time. Just not here!! Here, I'm sure the OP pasted it in just for the same reason it looked right to you. Then the OP does create an AC tie between the two groups. But the damned tie is AC and doesn't do any good to set the DC quiescent point. Probably starts to work kind of right with high enough AC frequency at the input because that series cap plus resistor begins to do something then. It annoyed me at first. Something subtle was nagging me. Finally it popped out to mind and the whole thing

@tobalt that the input group does get set by the input resistor to ground so one end of R1 will be somewhere around ground, we hope. But the output group can just drift away causing a drop across R1 which doesn't bother the output group one iota.

@tobalt Look. You've got four bjts that are arranged to move up and down with the input. Fine. And at the midpoint of that section sits one end of R1. Full stop. Switch views. You have an output stage that includes a Vbe multiplier and four output bjts. If you drive that somehow then it works. But it is isolated from the input group. So two groups floating on collector's everywhere. The other end of R1 goes to the output end of the output group. At DC, you can't haul the output group around by the nose that way to get a nice quiescent point. All that happens is

@tobalt No idea about chat. I'm as clueless as you, it seems. And anyway since I'm in bed I'm using Android app for stackexchange that has been deprecated and isn't supported now. No idea what would happen.

@tobalt No! Start at the input. Two bjts floating between current sources. Then two more bjts split apart by the vbes and resistor drops, to develop a current through their emitter resistors. That too also floats with the first pair. The Vbe multiplier is floating yet is drives the output group. R1 tethers across but doesn't change the floating and it's insane anyway -- it looks like what is used with a diffamp but that's misdirection. I'll have to wait to tomorrow to rub noses or else beg forgiveness.

@tobalt Not that I see. But what do I know? I'm asleep and in bed. Scrub all the capacitor stuff off the schematic. I do this in my head but you can use an eraser. See what is left!

mtx4, And don't tell me that C1 and R24 do. C1 is useless at DC. It's not a galvanic connection. It's an open. I guess I'm not surprised at your results.

mtx4, Let's assume you are correct about the offset moving so far out that it is 2 V at some point!!! (Even cold?!) No input, I gather. (And I may presume you measured the input as near ground and your rails are fine.) So just R6 to ground. Scrub your schematic of all capacitors and anything in series with them. Clear out the rubbish. Since there's no AC they will just charge up to something we don't care about anyway. What I see left is a VBE multiplier hanging in "empty" space between the collectors of Q1 and Q11. What's controlling the bottom or top of the VBE multiplier? What moves it?

@grandzello Professionals may often go for Chebyshev because it was developed to answer a very specific math problem: to approximate a given continuous function on a closed bounded interval with a polynomial of a specified maximum degree. Error can be defined different ways but the minimax criterion is the most suitable for filter design. So Chebyshev is kind of a go-to. You can specify Chebyshev with low passband ripple (lowest for Butterworth, but say 0.1 dB isn't all that bad often enough.) But, anyway, I don't want to answer something where I'm the one who doesn't follow things.

@grandzello Okay. What's confusing me is that there is a straight forward thing called a Butterworth. The two stages will have the same cutoff for this. The process and results would be different for Chebyshev. And if you really were doing a bandpass then my brain would also be elsewhere. Anyway, I don't know what textbook you are using. Do you have a single cutoff frequency as a specification? Or if not what exactly are the specifications. (I think I see you calling out some, but I don't want to assume anything right now.)

@grandzello In case you are wondering about my question, I get two analytic characteristic equations for 4th order Butterworth (getting rid of exact solution): \$s^2 + 1.84775906502257\cdot s + 1.0\$ and \$s^2 + 0.765366864730179\cdot s + 1.0\$. Both of these clearly have the same corner frequency.

@grandzello I just +1 your question, now. Your work addition deserves it. What I don't follow is why you appear to have chosen different \$\omega_{_0}\$ for the two stages. Can you explain your reasoning? (I do get about different gains, as you are talking about 4th order Butterworth and to keep it a Butterworth you will end up with some overall gain that isn't 1.)

@aconcernedcitizen Hehe. Got it. :)

@grandzello No applied mathematician would except a pile of conclusions known to be the result of incorrect axioms and invalid logic and then take a shot at seeing a way backwards through the invalid logic to perceive what might be sets of potentially useful axioms among which they could select one or two of them in order to move forwards again using valid logic. (There's no accounting for abstract mathematicians. They might see the whole idea as a way to explore some new impractical [the more impractical the better, they think] but as yet unexplored field to romp around in.)

@aconcernedcitizen Oh, I was already aware of that. Clearly if there's no current then the input sees the output. That's so obvious I didn't even think to mention it.

@grandzello Write down your filter specifications. We can't help you when the schematic isn't even connected up right and when the two stages appear to be unrelated to each other's design parameters (only similar in basic topology.) It's a mess, which suggests your thinking is muddled. You need to write out the goals because there is no way I'm accepting the job of sorting through the output of muddled thinking to somehow ferret backwards from there to some hoped-for light at the end of a twisty tunnel. Soundness comes from correct assumptions and starting values and valid applied logic.

@grandzello Ignoring the fact that Q determines gain here -- your design website does NOT magically ensure a design result it produces is Butterworth by the way -- and also ignoring the fact that your schematic is wired up wrong per eigenzero's answer: You don't even have the same \$\omega_{_0}\$ for your two 2nd order stages! This really puts the pressure on you. Do you want a Butterworth? If so, do you want a 4th order Butterworth? Or are you just chaining up two random lowpass filters taken from some website and unknown inputs to it? Write something more!!!