Enhance oscillation duration of tank-circuit  - Page 1
Forum section: electronics / rf-microwave

author: petert
author_rank: Regular Contributor
timestamp: December 25, 2020, 08:32:30 pm
content: Hello,A tank circuit will exhibit a damped oscillation due to parasitic resistances.What are some good components or other tricks, that don't involve using an amplifier, for prolonging the duration of a tank circuit's oscillation?I assume low ESR might be one factor, but I am looking for a set of "known good" components.I am aware that without an amplifier the oscillation will remain damped, but I would like to have it oscillate longer so it's easier to determine the frequency, and see how stable it is, etc. just with the tank circuit (or minimal modification that don't affect it much).

author: Ian.M
author_rank: Super Contributor
timestamp: December 25, 2020, 09:03:24 pm
content: Don't use a tank circuit unless the power level is too high for a ceramic resonator or crystal, or you require tunability!A ceramic resonator can have a Q an order of magnitude higher than a good tank circuit, and a crystal several orders of magnitude higher.If you have to, most of the losses are from the inductor due to its DC resistance, skin effect, core losses, and coupling to nearby conductive objects. The best you can do short of exotics like superconducting coils, is probably a large area minimum length and winding thickness air cored coil, ten or more diameters from any significant areas of nearby metal, and no nearby loops except normal to the coil, wound with heavy gauge OFC copper Litz wire, with only enough capacitance to swamp its self-capacitance so the tank operates at high voltage and relatively low current.

author: fourfathom
author_rank: Super Contributor
timestamp: December 25, 2020, 09:19:36 pm
content: Also, don't let your your measurement connection degrade whatever "Q" you do have -- at least no more than absolutely necessary.

author: petert
author_rank: Regular Contributor
timestamp: December 25, 2020, 09:34:57 pm
content: Would someone have an example of a simple LC-circuit setup connected to a scope with a reasonably long oscillation (and affecting the Q factor as little as possible by probing)?I am mostly into experimenting, not building a perfectly stable oscillator (and would like to use an inductor/capacitor-tank-circuit).What you say all makes sense, but a working example would make it simpler to get started.

author: bob91343
author_rank: Super Contributor
timestamp: December 25, 2020, 10:33:51 pm
content: You could trigger a parallel LC with a pulse and measure the result at a one-turn step of the inductor. Use a large inductor, maybe a couple of feet in size, made with silver plated tubing. If you know the Q you can estimate the decay rate.As mentioned above, a crystal or ceramic resonator has potential for this.

author: The Electrician
author_rank: Frequent Contributor
timestamp: December 25, 2020, 11:48:20 pm
content: https://en.wikipedia.org/wiki/Helical_resonatorhttps://coil32.net/design/helix-resonator.htmlhttps://www.bing.com/images/search?q=helical+resonator+design&qpvt=helical+resonator+design&form=IGRE&first=1&tsc=ImageBasicHover

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 26, 2020, 07:42:50 am
content: Quote from: petert on December 25, 2020, 08:32:30 pmWell, what are we to assume stays constant while we make our selection? Low damping is simply high Q. You need high Q components to begin with, and anything you hang off of it (like a scope probe) can only reduce it further (unless it's an active element/circuit).Good inductors simply have high Q at the desired frequency. Good capacitors are most of them besides the large ones (electrolytic, tantalum), in particular PP film and C0G ceramic.Tim

author: Ian.M
author_rank: Super Contributor
timestamp: December 26, 2020, 08:48:39 am
content: In pursuit of high-Q coils, this article on basketweave and honeycomb coils is interesting:http://www.pa2mrx.nl/UK/PA2MRX-UK.htm?coil%20homebrewing_uk.htm

author: petert
author_rank: Regular Contributor
timestamp: December 26, 2020, 02:28:49 pm
content: A simple solution seems to be to make a parallel RLC-circuit, and chose R as high as possible.https://en.wikipedia.org/wiki/RLC_circuit#OscillatorsQuoteWhich makes sense, given that the Q-factor for a tank circuit/parallel RLC-circuit is:Q = R * sqrt(C / L)I built a simple circuit in Falstad's circuit simulator, which seems to perform ok. I'll build it in real and see if probing or anything might affect it.

author: petert
author_rank: Regular Contributor
timestamp: December 26, 2020, 04:11:40 pm
content: Ok, I tested it, in a circuit built on a breadboard. But the chosen resistor value makes no difference, I tried 10k, 1M and no parallel resistor at all.The oscillation is visible on the scope and at about 167kHz, but the amplitude does not depend on the resistor at all.Correct frequency, given the individually measured components (L and C values) would be: 162.7198 kHz, so the frequency is in the right ballbark, given that I built it on a breadboard and there are several wires.Theschematic in falstad.The power source is a 9V-battery. The scope is set to 20us and 500mV per division, 10x probe (which is connected across the capacitor/resistor/inductor (since they are all in parallel). At beginning the oscillation is at about 4V peak to peak.Why would the resistor have no effect on the breadboard, though it has in the simulation?

author: fourfathom
author_rank: Super Contributor
timestamp: December 26, 2020, 05:18:26 pm
content: Quote from: petert on December 26, 2020, 04:11:40 pmThe reactance of the inductor at 167 KHz is about 30 Ohms (same as for the capacitor, but negative). Let's assume the inductor has a Q of 100 at 167 KHz, that makes the inductor's series resistance about 3000 Ohms. You will need to reduce the parallel resistor to something near that in order to see the effect.In your simulation, is that an ideal inductor (zero resistance and zero capacitance)? This may be why your simulation doesn't match the actual circuit behavior. Inductors are *never* ideal. Neither are capacitors, but they can get pretty close.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 26, 2020, 05:25:24 pm
content: Looking at the sim, between values of 10k and 100k, it looks consistent. What are you referring to? The peak value is slightly lower in the 10k case (the damping acts during all phases of the waveform, even during the rise, so we do expect the peak to be strictly lower for lower resistances, just not by much at this value).31nF and 31uH has a resonant impedance of sqrt(31/0.031) ~ 32 ohms, so we expect from an initial current of 9V/100ohm = 90mA, a peak voltage of 0.09*32 = 2.9V. 4Vpp is in the right ballpark.If the inductor and capacitor are ideal, Q is the ratio between the resistor and the resonant impedance. Evidently, 10k is already so much higher than the tank's unloaded Q that it makes no difference.Relevant ESR is reciprocal, i.e. 32 ohms divided by Q. Probably your inductor has more damping from DCR than the 10k produces, therefore the 10k does little if anything.Evidently the time constant is around 5 cycles, or a Q of 5. If this all manifests as DCR in the inductor, it is apparently 6.4 ohms. Not unreasonable for a tiny RFC I suppose, but seems rather poor otherwise. If it's a cored (especially powered iron) type, that will provide additional ESR. Also, type 2 ceramic capacitors have pretty significant ESR (typically having a Q of 20 or so).We can also calculate the equivalent circuit of the 10x probe, given some component values. I don't have a model of your probe, but assuming it's around 10M and 10pF, tapering to 500 ohms at high frequency, then at 160kHz it will have an equivalent of 100kohm reactance in parallel with 10M and in series with 500 resistances, or a parallel equivalent around 6M. Even if I got this off by a factor of 10, it should have little effect.Tim

author: petert
author_rank: Regular Contributor
timestamp: December 26, 2020, 06:15:50 pm
content: Quote from: fourfathom on December 26, 2020, 05:18:26 pmThe Q of the inductor (when measured alone, not in-circuit) is 0.459, the DC resistance is 0.42 Ohm.So if my parallel resistor is not less than 0.42 Ohm, it wont have any effect?

author: petert
author_rank: Regular Contributor
timestamp: December 26, 2020, 06:40:01 pm
content: Quote from: T3sl4co1l on December 26, 2020, 05:25:24 pmThe sim looks fine, to me, my issue was with the circuit built on the breadboard.Quote from: T3sl4co1l on December 26, 2020, 05:25:24 pmSince L and C are in parallel, Q = R * sqrt(C / L).I assume "resonant impedance" is X = sqrt(C / L), so Q= R * X and not Q= R / X. So why ratio?Given Q = R * X, a larger R would always improve Q, no?QuoteIf DCR means DC resistance, then it is rather low, I measured it to be 0.42 Ohm. Measured ESR of the inductor is 0.4 Ohm, and 0.67 Ohm at 100 kHz.Where did you get that from: "Relevant ESR is reciprocal, i.e. 32 ohms divided by Q."?QuoteSorry, not following anymore. How are 5 cycles (of the 167kHz wave) and Q related?QuoteCouldn't find what RFC means.QuoteIt's a simple air core self-wound inductor.QuoteA single ceramic capacitor has a Q of 52, several connected in series (to reach the desired total capacitance) gives me a Q of only 0.230. Maybe that's part of the issue?But I have to admit there are too many variables, and I would need a more specific/"walked" example. Starting to loose the overview.QuoteGreat that the probe is not the issue (and yes your values are about right).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 26, 2020, 07:08:45 pm
content: Quote from: petert on December 26, 2020, 06:40:01 pmTry that again -- Zo = sqrt(L/C) has units of resistance, sqrt(C/L) has units of conductance.QuoteIf the explicit R is the only R in the circuit, yes. There may be series, or more complex (equivalent) losses besides the physical component attached to it.QuoteUnder suitable assumptions/approximation, one can transform between the series and parallel tanks.Q = Zo / ESR = EPR / ZoConsider an inductor of some Q. It has the impedance,Z = w L / Q + j w LThis is simply the impedance of a resistance ESR = w L / Q in series with an inductive reactance Z_L = j w L.Suppose we had a parallel resistor network with the same impedance. That would be:Z = R || X_L= j R X_L / (R + j X_L)We need to rearrange this to separate the real and imaginary parts. We do this by multiplying top and bottom by the complex conjugate: (a + j b) * (a - j b) = a^2 + b^2Z = j R X_L (R - j X_L) / (R^2 + X_L^2)= j R^2 X_L - j^2 R X_L^2 / (R^2 + X_L^2)= j R^2 X_L + R X_L^2 / (R^2 + X_L^2)= j R^2 w L + R w^2 L^2 / (R^2 + w^2 L^2)Equating this to the original Z = w L_eq / Q + j w L_eq, we have a pair of relations:w L_eq / Q = R w^2 L^2 / (R^2 + w^2 L^2)j w L_eq = j R^2 w L / (R^2 + w^2 L^2)L_eq / Q = R w L^2 / (R^2 + w^2 L^2)L_eq = R^2 L / (R^2 + w^2 L^2)L is transformed to an equivalent value L_eq, depending on frequency and resistance. Oh, I left this in fixed resistance form, I should've used Q. I'll fix this up in a moment. We likewise have,Q = L_eq (R^2 + w^2 L^2) / (R w L^2)Q = (R^2 L / (R^2 + w^2 L^2)) (R^2 + w^2 L^2) / (R w L^2)Q = (R^2 L) / (R w L^2)Q = R / (w L)Whereas the original had ESR = w L / Q, we now have EPR = w L Q.Finally,L_eq = (w L Q)^2 L / ((w L Q)^2 + w^2 L^2)L_eq = w^2 L^3 Q^2 / (w^2 L^2 Q^2 + w^2 L^2)L_eq = w^2 L^3 Q^2 / ((Q^2 + 1) w^2 L^2)L_eq = L Q^2 / (Q^2 + 1)Now we can easier see how L is transformed, given the assumption of constant Q (which isn't a very realistic assumption, as it happens; this just serves as example). If Q is very small, Q^2 / ((very small) + 1) dominates, and so L_eq becomes asymptotically small. That makes sense, as the R is then dominant. Likewise if Q is very large, (Q^2 + (very small)) dominates, Q^2/Q^2 cancels out, and L_eq --> L.(Note: I'm writing w instead of lowercase omega for expedience. j = sqrt(-1).)This is easy enough to show, and I hope you can follow along on a pad and pen -- I have shown most all the algebra steps to facilitate checking progress. Complex algebra isn't any different from regular algebra, the j symbol is manipulated just like any other variable; we simply apply the identity -1 = j^2 when handy, and when we can't reduce it, just leave it be.QuoteQ reflects many properties of a resonant tank, or networks (or parts thereof) in general. The bandwidth, and impulse response decay time constant, are among these.QuoteRadio Frequency Choke -- typically a component used to carry DC current, such that its impedance is "high enough", without needing to be accurate. So, an RFC might be forgiven for having a crummy Q, as long as it passes DC. The complement of the series coupling capacitor (which conversely blocks DC current).(You don't give dimensions of the air-core coil; but it does seem unlikely its Q would be entirely this bad, given the DCR.)Quote? ? ?Tim