John Larkin's LC oscillator
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From: John Larkin on 10 Jun 2010 18:47 On 10 Jun 2010 13:23:55 -0700, Winfield Hill <Winfield_member(a)newsguy.com> wrote: >dagmargoodboat(a)yahoo.com wrote... >> >> On Jun 8, Winfield Hill wrote: >>> was Re: Twin T circuit wanted >>> >>> John Larkin wrote... >>> ftp://jjlarkin.lmi.net/Ships_Bell.JPG >>> >>>> On Mon, 7 Jun 2010 George Herold wrote: >>> >>>>> I loved your bell circuit John! I didn't quite get how the >>>>> inductor was working. But still I got the idea. Thanks. >>> >>>> This is kind of a cute circuit. I first designed it when I needed >>>> a very frequency and amplitude-stable sine wave to drive a Talyvel >>>> LVDT-like inclinometer, part of the Boresight Alignment Kit for >>>> the C5A. We had to measure level to arc-seconds of accuracy. >>> >>>> It's a transformer with a resonant tank in the collector and a >>>> positive feedback drive winding into the emitter. The emitter >>>> feedback is just a couple of tenths of a volt p-p. >>> >>>> The cool thing is that the collector swing is almost exactly 2xVcc >>>> peak-to-peak. As the amplitude builds up, at the negative swing peak >>>> the emitter goes a little bit negative, to get out of the way, and >>>> the collector swings to just about ground. That forward-biases the >>>> c-b junction and discharges the base cap, reducing transistor base >>>> current hence gain. So it has a built-in peak detecting AGC amplitude >>>> leveling loop with close to zero TC. All from 5 parts. Or sometimes >>>> six. >>> >>> With your explanation, the oscillator is much more attractive. >>> Our old friend, Tony Williams, R.I.P., would have been pleased. >>> >>> John Larkin's LC oscillator, supply-V sets amplitude > > [snip in favor of the improved ASCII drawing below ] > >> Here's a quick implementation: >> >> Vcc = +5v >> --+-------+------+-- >> | | | >> | | |_ || >> | | _)|| >> .-. --- L1a _)|| >> Rb | | C1 --- 1mH _)|| >> 100k | | 1uF | _)|| >> '-' | _)||7 >> | | *| || >> | .-+------' || >> | |/ || >> +---| Q1 || >> | |>. 2n3904 || >> | | * || >> C2 --- +---------. || >> 1uF --- | L1b _)|| >> | | 25uH _)|| >> | | | >> --+---- | -------+-- GND >> | >> '-------------> output >> >> (5KHz values shown) >> >> collector swing ~= 2*(Vcc+0.6v) >> emitter swing = Vc * sqrt(L1b/L1a) >> >> 1 >> f(out) = ------------------- >> 2*pi*sqrt(L1a * C1) >> >> That's a 1rst cut -- I may have left L1b a little hot... >> Simulates really nicely--sine waves. > > Jim pointed out elsewhere, in the old Twin-T portion of the > thread, that I had misunderstood how the oscillator works. > He gave a hint: not CE. Yep, of course, now it becomes a > bit more clear. Thought of as an amplifier, it has to be > operating in common-base mode. > > First, given the transformer winding, the average DC voltage > on the emitter always has to be zero volts. Then, if the > transformer turns ratio is high enough, high enough to force > class A operation, with very little signal on the emitter, > the base capacitor voltage will be at roughly Vbe. > > Being class A, the gain includes 1/re in its terms, and the > gain has some proportionality to emitter/collector current. > The LC tank amplitude and collector-voltage swing can also > grow, proportional to this current. As oscillation starts, > with increasing current and gain, the tank amplitude grows > until the collector approaches saturation on each cycle, > causing two things to happen. > > First, the base-collector diode conducts and robs some base > current, controlling the base voltage and DC current at the > right value for the class A operation we're observing. > Second, this base-collector diode conduction directs some > of the emitter signal current to the base, away from the > LC tank, reducing the common-base gain, the tank amplitude, > and stabilizing the oscillator. > > Operating this way, it seems fair to characterize the level > control as an AGC common-base amplifier gain type of control, > but you can also think of it as a type of current-steering > amplitude control. > > A high transformer turns ratio may not be a healthy way to > run this oscillator, what if it's too high, and stops? So > it's likely that our happy turns ratio will be well below > a class-A operating limit. If so the oscillator emitter > amplitude will increase, and it'll be running in a kind of > switching or class C mode. I recall a pretty small AC voltage across the emitter winding, like 0.1 or 0.2 volts p-p maybe. Not pure linear class A, but not too radical either. > > In class C the resonant tank is excited by short pulses of > collector current. At the maximum emitter current, at the > bottom of each tank cycle, the emitter voltage will be well > below ground. The forward-biased base-collector diode will > get serious about stealing collector current from the tank, > when it has a chance, thereby regulating the AC RMS current > and the LC tank voltage amplitude. It appears Jim wants to > call this ALC, not AGC, action. The regulating function > may be better thought of as signal current control than as > a gain control, although the right term to use seems a bit > semantic to me. Unless, of course, there's some guiding > literature that we should be following. > > I'm wondering if it wouldn't be a good idea to have an > emitter resistor to help control the currents and stretch > out the pulses. Maybe John had one in his old version. > > In a ringing bell application, as the supply voltage sags, > and the amplitude drops, I imagine the circuit will move > from class C back to class A operation, before stopping. It is a simplification to say that the c-b conduction, at the bottom of the sine wave, is the only thing that pulls the base voltage down. Some base current goes into the emitter, too. It's sort of academic, since the collector and emitter essentially collide, and *both* help to discharge the base capacitor. Two interesting variants are James's suggestion of a Baker clamp, and your idea of an emitter resistor. Both get the emitter out of the way as the discharge/AGC path, and make the c-b clamp idea purer. I'm not sure if the DC drop in an emitter resistor is a problem. As far as the idea that there is *not* an AGC effect, consider: A lot of current comes in through the base resistor. The collector voltage is above saturation 98% of the time. So the collector current should be about the current through the base resistor times beta. But it isn't. It's much less. The reason is that the AGC effect has stolen most of the base current. Just enough, in fact, to make nice oscillations. One picks the base resistor to give a healthy margin for starting and running, but not so low that the flat-top (bottom, actually) on the sinewave is gross. The temperature thing is interesting. Small-signal, the base/capacitor voltage is one jd above the emitter voltage, with the usual tc. When the collector swings down, it has one jd worth of tc in pulling down the base capacitor. So the tc of the c-b junction compensates the remembered tc of the b-e junction. John
From: George Herold on 10 Jun 2010 22:26 On Jun 9, 11:01 pm, Jon Kirwan <j...(a)infinitefactors.org> wrote: > On Wed, 9 Jun 2010 19:02:56 -0700 (PDT), George Herold > > <gher...(a)teachspin.com> wrote: > >Hi Jon, so good to see a post from you. Everything OK? > > Nah. Still horribly busy around here -- probably going to be > killing myself through the summer... too much to do (or, at > least, I hope I can force myself to keep at it without giving > up at some point.) > > >I'm going to have to go an reread Fenyman on this... (I thought I got > >it the last time, but that's how it is with Feynman.) I think there's > >a magnetic field everywhere in the magnetic circuit. In an electro- > >magnet it starts with the current in the coils and is transfered to > >the gap by the magnetic material...iron. But the field is everywhere > >in the circuit. It may very well be that the gap determines the field > >strength. But the magnetic energy is stored over the entire volume > >of the field. > > I _imagine_ aligned atoms as 'short circuits' that bridge > vacuum. Vacuum is where the energy gets stored. (I think a > tiny amount of energy is 'used' to align an iron atom, for > example. But it is tiny and doesn't represent recoverable > energy... I think it gets converted to heat in the end.) I'm > very much open to being wrong, of course. > > With all this in mind, it's easy to imagine the concept of > permeability arising. For air core, it's vacuum all the way > down, so to speak. All of the loop length is, in effect, the > magnetic path length where energy gets packed away. But when > you insert iron atoms chained end to end they act as short > circuits, each of them bridging across a small bit of vacuum > to make the effective length of vacuum shorter (when aligned > -- the molecules of O2 and N2 don't align and don't bridge so > in effect they simply don't count and don't short the > vacuum.) So if you imagine a core material inserted there, > what counts is the interstitial vacuum that remains when > computing the magnetic path length. But what remains?? Well, > that depends. You can take a ruler and measure that core > piece, of course. But that ruler won't help you figure out > what the effective remaining vacuum length is and that is > what you need to know. So in order to come up with a number > that represents the percentage of the measurable length, you > need a factor of some kind. This is the permeability figure. > A permeability of 1000 would mean that if you had a toroid > with a mean centerline path length measurable with a ruler of > say 100mm, that the actual vacuum path length would be 100 > microns and the other 99.9mm would be shorted out by the iron > that is present there. A permeability of 100 would mean that > 1mm would be vacuum and 99mm would be iron. Since all of us > can only use observable measures, we need to have some idea > about what it looks like from the magnetic field's point of > view, vacuum wise. At least, this is how I like to imagine > all of this... energy gets stored in vacuum and the rest is > just fudge factors to get around how we practically measure > lengths. > > Jon Hi Jon, I hope youre not offended if I say that the first time I read this I thought it sounded crazy. I took vol. 2 of The Feynman Lectures of Physics down to the creek when I got home this evening. In chapter 36 he does ferromagnetism. (do you have a copy?) He does an electro-magnet near the end. (36-5). Early in the chapter he says that if you are willing to accept the fiction of magnetic monopoles, you can think of ferromagnetism as the magnetic analog of the polarization of dielectrics. And so I totally accept your picture though I understand it differently. Thanks! Oh Feynman says the energy loss is the area enclosed by the B/H curve. George H.
From: Jon Kirwan on 11 Jun 2010 02:20 On Thu, 10 Jun 2010 19:26:20 -0700 (PDT), George Herold <gherold(a)teachspin.com> wrote: ><snip of what I wrote> >Hi Jon, I hope you�re not offended if I say that the first time I >read this I thought it sounded crazy. Not at all. Cripes, can you imagine what would have happened if Rutherford had gotten all offended when meeting with his young students who might have proposed some idea? I had loved reading, somewhere, that he would listen to any and all if they thought they had _any_ idea, no matter how modest or odd it may have been. They would take a break every afternoon for tea, cake and buttered bread and just sit on stools talking. Nice. 11 of his students went on to receive Nobel Prizes, not to mention some of the collatborators who also attended and also won Nobels. Not that I think what I wrote is anything much. I'm just saying... >I took vol. 2 of �The Feynman Lectures of Physics� down to the creek >when I got home this evening. In chapter 36 he does ferromagnetism. >(do you have a copy?) I do have a copy and have carefully read the first 5 or 6 chapters of the first book, only. I have not read chapter 36. >He does an electro-magnet near the end. >(36-5). Early in the chapter he says that if you are willing to >accept the fiction of magnetic monopoles, you can think of >ferromagnetism as the magnetic analog of the polarization of >dielectrics. I'll read it, tonight. Though I may need to _also_ go back and read on polarization to refresh that, as well. >And so I totally accept your picture� though I understand it >differently. Thanks! It's the way I imagine it and it works to solve any questions I've had about the concepts others teach differently (and to me with less clarity.) I am sure, someday, there will be some behavior that requires me to modify it. But so far for my limited experience, it works well. I like it better because I don't have to think about how recoverable EM energy (which in my imagination can ONLY be stored in a vacuum) might be one figure for vacuum, another figure for atom A, another figure for aligned atom A, another figure for atom A in an excited state, and so on. I just have to keep one thing in mind... energy is stored in vacuum, only, and permeability is simply a fudge factor used to estimate the effective ratio of the vacuum path length and the vacuum path length less the magnetic short circuit length (once alignable atoms are aligned.) L_vacuum / (L_vacuum - L_shorted). It's a fictitious parameterization so that we humans can use basic measuring tools and simple observations plus that figure to estimate the effective vaccum psth length remaining for energy storage. But the recoverable energy really only goes into vacuum, I believe. >Oh Feynman says the energy loss is the area enclosed by the B/H >curve. Which makes complete sense. I mentioned before that I also imagine aligning the atoms takes energy, but it isn't recoverable. The work involved in aligning is largely ('all' probably, unless there is a wound up spring recoil effect which I don't at all imagine exists) converted to heat. If you reverse the alignments, you do more work. And so on. But you don't get it back. That goes into heat. It's only the energy stored in vacuum (energy cannot be created or destroyed) where there is no possibility of "heat" that you can recover. (Would be interesting to imagine the case where you could actually lose energy as work converted to heat in a vacuum... hmm... could you cause the vacuum to heat up enough to create 'vacuum ash?' Enough letting my imagination get away with me... ;) Jon
From: Jon Kirwan on 11 Jun 2010 02:31 On Thu, 10 Jun 2010 23:20:32 -0700, I wrote: >The work involved in aligning is largely ('all' >probably, unless there is a wound up spring recoil effect >which I don't at all imagine exists) converted to heat. If >you reverse the alignments, you do more work. I should add that the way I imagine re-aligning taking place is with each atom as little aligned gyroscopes. When a force is imposed on them to alter their orientation, they twist in the field and this twist bumps them against others also being turned and this results in some of the energy going into lattice/material vibrations that amount to 'heat' in the end. I'm not sure how to make that quantifiable, except that if that imagination has any predictive value, it would suggest that very fast changes would cause a great deal of heat as the little gyros would twist quite strongly then. Jon
From: John Fields on 11 Jun 2010 08:02
On 10 Jun 2010 13:23:55 -0700, Winfield Hill <Winfield_member(a)newsguy.com> wrote: > In a ringing bell application, as the supply voltage sags, > and the amplitude drops, I imagine the circuit will move > from class C back to class A operation, before stopping. --- Why imagine? Here's a circuit list that'll show it all _and_ it'll create a .wav file, "bong.wav" in whatever folder the cicuit's in, so you can hear it. Enjoy! :-) |