Q factor and resonant frequency - Page 1
Forum section: electronics / projects

author: ricko_uk
author_rank: Super Contributor
timestamp: December 09, 2020, 06:14:29 pm
content: Hi,with reference to this datasheet for a Qi charging coil:https://www.we-online.com/catalog/datasheet/760308101141.pdfIt shows self resonant frequency at 11MHz (table in the first page) but the Q-vs-F graph on the second page shows the Q peaking around 150KHz. I thought the highest Q was at the resonant frequency...If not, then what is that Q referring to?Thank you

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 09, 2020, 06:33:45 pm
content: What's the expression for misreading something? A reado?I see kHz on the axis. Which is pretty nice, it's not that easy to get a Q of some hundreds at as many kHz, let alone with no core. That's the power of litz.Tim

author: ricko_uk
author_rank: Super Contributor
timestamp: December 09, 2020, 06:42:53 pm
content: Hi Tim,it actually was a typoon my OP (just corrected it).But my original question still validThe highest Q is shown at 150KHz but the self resonant frequency is shown at 11MHz. Miles apart from each other. Aren't they supposed to coincide? That is what I was wondering about. Isn't the highest Q always at the self-resonant grequency?

author: bson
author_rank: Unknown
timestamp: December 09, 2020, 06:48:21 pm
content: It's not the highest Q. It's the Q at the test conditions; i.e., the Q at 125kHz and 10mA.

author: bob91343
author_rank: Super Contributor
timestamp: December 09, 2020, 06:55:35 pm
content: The Q and resonant frequency are related but not the same. Actually at resonance the Q is unity.

author: ricko_uk
author_rank: Super Contributor
timestamp: December 09, 2020, 07:27:37 pm
content: Oops... I just saw the conditions mentioned in the table.Thank you all

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 09, 2020, 07:29:33 pm
content: Ah, yes. Alright.Actually even worse than that, the Q at resonance is zero.Why?A typical definition of resonance is the point of zero reactance.Q is the ratio of reactance to resistance (or vice versa, depending on series/parallel resonance). If reactance is zero, well, so is Q! (Or for the parallel case, reactance is infinite, same idea.)If you have a model for the resonance (it might be a simple 1st-order parallel RLC, but it might be more complex, too), you can attempt to separate the components, and attribute Q's to them -- now you're not so much looking at the Q of the component (as a whole), but the Qs of not-necessarily-physical elements thereof.Tim

author: Pulsepowerguy
author_rank: Regular Contributor
timestamp: December 09, 2020, 07:33:42 pm
content: Quote from: bob91343 on December 09, 2020, 06:55:35 pmThe Q at resonance is not necessarily unity. In fact it can be any number above unity (infinite is not likely).

author: Pulsepowerguy
author_rank: Regular Contributor
timestamp: December 09, 2020, 07:41:58 pm
content: Quote from: T3sl4co1l on December 09, 2020, 07:29:33 pmThis is not correct. Q is the ratio of energy stored and energy lost per cycle. A Q of zero is nonsensical. A bell has a high Q - well above 1. Any oscillator has a high Q or it would not work.

author: TimFox
author_rank: Super Contributor
timestamp: December 09, 2020, 07:59:23 pm
content: There are two common definitions of Q:1. For a resonant circuit, Q = (resonant frequency) / (bandwidth at -3 dB).2. For a component (inductance or capacitance) Q = (series reactance magnitude) / (series resistance) = (parallel resistance) / (parallel reactance magnitude). This is what a traditional Q meter measures for an actual inductor at a specified frequency.

author: mag_therm
author_rank: Frequent Contributor
timestamp: December 09, 2020, 08:09:35 pm
content: Yes, As I learned and still use it, Q factor is defined only as wL/Rs for a series loss resistance and Rp/wL for a parallel resistance.For a circuit consisting of LCR, Q is still defined only as above. It does not include capacitive reactance.Q factor came from old radio days where they often neglected capacitor dielectric loss.The bandwidth of a resonant circuit is Q = Fres/Delta_F .With some manipulation, that comes back to Q = wL/RsRs may include any resistance in the resonant loop, but Q is still = wL/Rs

author: langwadt
author_rank: Super Contributor
timestamp: December 09, 2020, 08:15:13 pm
content: I'd guess that 11MHz is the resonant frequency of the coil and its stray capacitance.and the is Q graph is measured with a ..... ~160nF cap in parallel to make it resonate at 125kHz

author: ricko_uk
author_rank: Super Contributor
timestamp: December 09, 2020, 08:33:34 pm
content: Thank you all!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 10, 2020, 01:20:17 am
content: Quote from: Pulsepowerguy on December 09, 2020, 07:41:58 pmRight, that's what I said. The network has no reactance at resonance, therefore all energy input during a cycle is dissipated at precisely the same rate.As I noted in the other paragraph of my comment, you can define Q in other ways, which may or may not be reasonable or physical, but that doesn't mean they are all unreasonable or unphysical. A typical case is to specifycomponent Qfor isolated components in a network, andsystem Qfor components, or sets of components, in a network.Example: say we have an LC lowpass filter, Butterworth, 1 ohm source and load, 1 Hz cutoff. L = C = 0.225 (in H and F respectively) (exactly, sqrt(2) / (2 pi)). At 1Hz, these have ~1.4 ohm reactance. Say the inductor has 14mohm DCR: then it has a component Q of 1.4 / 0.014 = 100. Whereas it has a system Q of 1.4: it doesn't matter how damn good the inductor is, as long as it's strung between the 1 ohm source and capacitor and load, it's only ever going to have this amount of Q (and the filter, its maximally-flat Butterworth transfer curve).Say we used a really crappy inductor, whose SRF is 1Hz. Maybe its nominal (asymptotic LF, excluding DCR) inductance is still 0.225H, but it just so happens its impedance flattens out around 1Hz. What sort of filter do we get, now? Well, at 1Hz, we just get a resistor divider into a capacitor. It has 1.4 ohms resistance and no reactance. The filter certainly isn't Butterworth anymore. Indeed, it doesn't even have asymptotic attenuation anymore, at high frequencies it's a capacitive divider (or something else more complicated). Clearly, this disaster has occurred because the Q is terrible, much less than 1.4.Again, indeed the "inductor"'s Q is 0 at this point, because it has no reactance.And yeah, we can model the inductor itself as a network, and assign Qs to those elements -- we just did the same thing to the filter network. In that case, the filter network was built of explicit physical components, so it's quite meaningful, and useful, to consider their Q factors individually. If we encounter a black-box network, some component that we don't know the structure of, it may not be all that easy (or even meaningful) to approach it in the same way -- certainly if nothing else, we have to perform many measurements to even begin piecing together a model of it.I'm discussing an abstraction -- sure, a self-resonating inductor might have all manner of fields around it, but when it comes down totwo terminals in a circuit, we don't, and can't, care about that -- it simply behaves however it behaves through that port (or two*). And how do we express the behavior of a 1-port? By impedance of course!*Since, to be fair, an inductor of this size (in the OP) will have a lot of common mode response -- it isn't likely to behave as a pure 1-port. But anyway, there are abstractions that we can use for that case as well.Tim

author: emece67
author_rank: Frequent Contributor
timestamp: December 11, 2020, 01:06:34 am
content: .

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 11, 2020, 01:28:57 am
content: But that's a transient response -- you can't very well say you're testing a networkat a frequencyif you're starting and stopping it arbitrarily (generating sidebands at whatever rate)!AC steady state analysis really is steady state -- it may not always make intuitive sense, basically because we can't experience true sine waves, that exist for all time, backwards and forwards.So, if we're employing a different kind of analysis, like an impulse response, or wavelet or tone burst or such, or a non-LTI source (like a transistor alternating between hard saturation and cutoff), we need to say so. And we need to define how we're measuring impedance, or what it even means -- because indeed, the instantaneous voltage and current need not follow from the steady-state solution in these cases!(Which, if I haven't made it explicit yet -- I am assuming AC steady state is the default, assumed analysis used for networks. If this is not your default, that explains the confusion!)For a very simple case, consider varying the frequency around the resonance of this 1-port network. What does the impedance do? Well, presumably it's reactive above and below (of complementary signs), and more or less resistive on/off resonance. If we find the resonance is simple (reasonably approximated by 2 poles, say?), we might define a Q based on the width from R = +/-X to R = -/+X, relative to Fres, that is, the relative bandwidth. But what are we really doing here -- why does it matter what it approximates as? We're assuming some internal structure about the network, and assigning values to those parameters. Whether we mean to or not, we're solving for a model! And indeed, the network could be some odd lumpy case, like a ferrite bead, or bias tee: it need not have a single clear symmetrical peak, it might rise and fall irregularly, it might have peaks, notches, who knows.So, this is why, in the most general case, an AC impedance meter can only tell you an RL or RC equivalent circuit -- at a given frequency, it measures some amount of resistance and reactance, period. Whether those elements have any sort of underlying relationship, or meaningful values (is a parallel or series equivalent more meaningful, who knows?), isn't for the meter to decide.So it is in that sense, that a network at resonance (given a suitable definition of such), has an externally measured Q of 0. That's all.Tim

author: ElectronRob
author_rank: Contributor
timestamp: December 11, 2020, 01:56:18 am
content: I feel like your back tracking on a technicality here TimFor us mortals I think we could all agree that we would prefer our crystal oscillators had a high Q rather than a Q of zero. Regardless of which relative dimension you are measuring it from - I do get your point but suggest its more academic than practical.

author: bob91343
author_rank: Super Contributor
timestamp: December 11, 2020, 02:19:41 am
content: I made an error. If there is resonance, there is Q. A pure resistor is the same at all frequencies and so has a Q of zero. Meaning no stored energy.A real resistor has distributed capacitance and series inductance and so has a Q greater than zero. At some frequency these two parasitic components will self resonate and thus have a resistance overall that in general is different from its low frequency value. But that says nothing about its Q. If the Q is high, the resonance is sharp and vice versa.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 11, 2020, 02:59:17 am
content: Right, the real point of all these paragraphs is: we need to be careful what Q we're talking about.For the crystal, we have a generally accepted model, and values fitted to it. (But also, it may not be as often appreciated what this model lacks -- real crystals have a plurality of such resonances, that can be expressed as a parallel stack of series-RLC tanks. The rated fundamental just happens to be the most prominent, well behaved, and characterized one.) This is a good example, because when the model is discussed, it's usually noted how absurd the values are (fF and fractional H) -- after all, they are electrical equivalents of a mechanical resonance, so they can be forgiven for having improbable electrical values!Quote from: bob91343 on December 11, 2020, 02:19:41 amYep-- again, fitting a 1st order model to it, we can observe ESL at low frequencies, and EPC at high frequencies; together, we get a resonant circuit, if, hopefully, a shitty one.And now we have an intuitive foothold for considering the "bandwidth" of a resistor -- where the ESL and EPC arise from, and how they relate to resistance value, and application. Namely, that modest values, near sqrt(ESL / EPC), have the widest bandwidth, where "bandwidth" is defined as resistive-dominant; and that values far away from this case, tend to dominate in one or the other property; indeed this can sometimes be quite extreme, given the extreme range of resistance available.Tim

author: emece67
author_rank: Frequent Contributor
timestamp: December 11, 2020, 12:34:44 pm
content: .

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 11, 2020, 01:04:52 pm
content: Quote from: emece67 on December 11, 2020, 12:34:44 pmAnother point that's equivalent, but worth making explicit: the RL or RC equivalent of some impedance, in general will vary with frequency. That is, R and L/C (indeed whether it's L or C) are functions of frequency. They aren't constant elements, and it's not a model at all, just an equivalent. This is again more explicit when we are discussing specific use cases, like capacitor ESR. We can model a capacitor's impedance over a wide range of frequency, but we'll need much more than three (fixed value components) to do so.Tim

author: mag_therm
author_rank: Frequent Contributor
timestamp: December 11, 2020, 03:49:22 pm
content: It is interesting to see that the Q_factor as a figure of merit is still popular in electronics.I used it beccause of familiarity with it, coming from I suppose radio biased early interest.Working with electric power oriented engineers, younger ones, sometimes they'd comment that Q factor was outdated.Where an Inductor of jwL = 2 Ohm and Rs = 1 ohm, a Q person would say "2 ohm of reactance with a Q of 2."The power person would want to hear " Z = 2.236 Ohm with Cos(Phi) = 0.447 "In some ways the Cos(phi) expression is more accurate.When Q is high, we just use inductive reactance and tend to ignore R.I have seen it here on forum quite a lot. Inductors are just shown as ideal.But when Q starts to get below 10, It is no longer accurate to just consider jwL, as above example shows.Also the simulation apps tend to use inputs and results using the Cos(phi) format, with angles in degrees.

author: bob91343
author_rank: Super Contributor
timestamp: December 11, 2020, 06:31:36 pm
content: I chuckle at all this simulation talk. It's trying to rationalize something somewhat irrational. As mentioned above, there is no fixed number that can be assigned to parasitic parameters to allow for modeling. It's different at different frequencies. So the modeling experiment fails.Modeling is a curse put upon engineers by mathematicians. It's far better to understand what's going on rather than attempting to make reality conform to formulas.It reminds me of music theory, an attempt to make an art form into a science. While those with minimal talent work with fancy chord names and mode definitions, the real innovators are busy creating music.

author: Pulsepowerguy
author_rank: Regular Contributor
timestamp: December 11, 2020, 08:10:03 pm
content: Q factor is used when describing laser cavities, and acoustical systems so I wouldn't say it is some outdated term from radio.I have never in my life heard anyone say that Q = 0 at resonance - certainly not for the resonant system in question. This and other statements on this thread don't make sense to me.

author: bob91343
author_rank: Super Contributor
timestamp: December 12, 2020, 12:22:50 am
content: I retracted that Q=0 comment. I must have been typing late at night.Additionally, Q is defined for inductors, not capacitors. Capacitors have their own terminology. But that is splitting hairs.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 12, 2020, 03:02:08 am
content: Q is defined for anything you can give a definition for.Tim

author: emece67
author_rank: Frequent Contributor
timestamp: December 12, 2020, 08:12:53 am
content: .

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 12, 2020, 02:12:37 pm
content: It's funny, isn't it? You'll get the same sort of "nonsense" from this tool, too:https://hamwaves.com/inductance/en/index.html#input(models solenoid coils quite accurately, including resonances)Tim

author: langwadt
author_rank: Super Contributor
timestamp: December 12, 2020, 02:41:21 pm
content: Quote from: emece67 on December 12, 2020, 08:12:53 amthe Q vs. f curve is surely with a load capacitor to bring the resonance to the frequency that'll be used in a charger

author: emece67
author_rank: Frequent Contributor
timestamp: December 12, 2020, 03:35:10 pm
content: .

author: Siwastaja
author_rank: Super Contributor
timestamp: December 12, 2020, 03:59:51 pm
content: Quote from: emece67 on December 12, 2020, 03:35:10 pmDepending on series/parallel viewpoint, it's either 0 or infinity.Again, it's more important to understand theconceptthan to learn one definition of the Q factor by heart then be confused about it and unable to utilize it.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 12, 2020, 06:38:52 pm
content: Quote from: emece67 on December 12, 2020, 03:35:10 pmIt's automatic on entering a full set of parameters. Enable JS and use a current browser.Tim

author: bob91343
author_rank: Super Contributor
timestamp: December 12, 2020, 06:56:14 pm
content: The plot thickens. We are confusing circuit Q with component Q. The thing that is wrong is, as mentioned above, applying formulas without understanding them.You care about component Q when using an inductor. Circuit Q varies all over the place and isn't a terribly useful parameter.

author: SiliconWizard
author_rank: Super Contributor
timestamp: December 12, 2020, 07:12:23 pm
content: Yeah. There is an obvious difference in the definition of the Q factor for an inductor (or rather, a coil, as opposed to an ideal inductor) and for a resonant circuit, and that can be confusing.https://en.wikipedia.org/wiki/Q_factor

author: KE5FX
author_rank: Super Contributor
timestamp: December 12, 2020, 08:27:49 pm
content: Another problem with Q is that it conflates energy loss/efficiency with bandwidth and stability. In precision timing work, it's almost an axiom that stability improves as Q increases and line width shrinks.Not necessarily true for pendulum clocks, though. This is an area where pendulums are more interesting than they appear at first. On page 73, Tom suggests that we should be using P for stability and Q for energy loss, at least where pendulums are concerned.What are some examples where this is true in electronics?

author: SiliconWizard
author_rank: Super Contributor
timestamp: December 12, 2020, 08:35:05 pm
content: A high Q factor for a resonant circuit is a double-edged sword: it improves efficiency at the exact resonant frequency, but degrades efficiency elsewhere much more so than a circuit with a lower Q factor. So in practical power transfer applications, it's often a trade-off.

author: TimFox
author_rank: Super Contributor
timestamp: December 12, 2020, 09:34:47 pm
content: Going back to my earlier post about the Q for a resonant circuit and the Q of a reactive component. I leave the actual algebra as an exercise for the reader.1. As pointed out by several, for resonators in general (e.g., cavities, RLC circuits, piezoelectric crystals): at the resonant frequency, Q is defined in terms of the energy dissipated per cycle compared with the energy stored in the resonator.2. From this definition, we see that Q is also the ratio of the resonant frequency to the bandwidth (at half-power, - 3 dB).Q= (resonant frequency) / (-3 dB bandwidth).3. For a component such as a practical coil or capacitor, with a reactance and resistance at a given frequency (assuming the frequency of measurement is far enough from the self-resonant frequency of the component so that we can ignore the other reactance), a Q meter will measure the component Q asQ= Abs(Xs) /Rs= Abs(Rp/Xp), where the absolute value is needed since the reactance will be negative for a capacitance and positive for an inductance. Note that the series and parallel reactances for a given component are not equal, although if Q is very high the difference is small. For a high-Q component, the series and parallel resistances differ by a large factor, approximatelyQ2.4. We now wire a practical coil and capacitor together to make a resonant circuit, with the resonant frequency well below the self-resonant frequencies of the two components for simplicity.5. A series circuit gives a low impedance at resonance, where the positive inductive series reactance and negative capacitive series reactance cancel. Since both components have positive series resistance, the total impedance does not go to zero, but the current reaches a maximum at the resonant frequency when driven by a low-impedance voltage source.6. A parallel circuit gives a high impedance at resonance, where the positive inductive parallel reactance and the negative capacitive parallel reactance cancel. What is left is the parallel combination of the two parallel resistances. If the two components were lossless, which is not physically possible, then the impedance would increase without bound (infinite impedance at resonance).7. From this elementary circuit theory, we calculate that the Q of the resonant circuit is given by 1/Q= 1/QL+ 1/QC. Usually, the capacitor's Q is higher than the inductor's Q and the inductor Q dominates the calculation.8. When the resonant circuit is used to transfer power from a source to a load at a single frequency, the efficiency is reduced by the unloaded Q of the resonant circuit, where the loaded Q includes the effect of the load resistor on the circuit (series or parallel).

author: emece67
author_rank: Frequent Contributor
timestamp: December 12, 2020, 10:07:44 pm
content: .