Very confused about the varioius definitions of Q - Page 1
Forum section: electronics / rf-microwave

author: jrive
author_rank: Newbie
timestamp: January 22, 2022, 07:23:47 pm
content: Ok...so I have a matching circuit that is designed to match a 32+j246 load to 50 ohms at 13.56MHz. It is a simple cap divider match: Q=sqrt(50/32-1)=0.75. So, Cseries ends up being 53pF, and Cshunt is 176pF.Now, does this Q mean that the bandwidth of this matching network is 18MHz, as given by BW=f0/Q? I can see that it is not--but I"m also not sure what bandwidth it is referring to. Is it the bandwidth the transfer function at the 50 ohm match (the node connecting Cseries and Cshunt)?Where is this Q "visible" in the final circuit?Thanks!

author: bob91343
author_rank: Super Contributor
timestamp: January 23, 2022, 12:13:40 am
content: It's not that simple. A good matching network will use high Q parts but then it matches only over a narrow range of frequency. Lower the Q and the match gets poorer but less frequency sensitive. That compromise is hard to quantify. See a text on filter design.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 23, 2022, 01:56:32 am
content: There are as many ways to measure Q, as there are ways to put together reactance and resistance.When we talk about the Q of a component, we mean the impedance of the component by itself. Q of a resonator, usually its unloaded Q as measured by bandwidth, or resistance at resonance (when sqrt(L/C) is known).Q of a filter, or of components or stages in a filter, is in relation to system resistance. For practical filters, this Q will be much lower than component Q (the difference* gives insertion loss due to that respective element/section), and can be no higher than that (at full insertion loss).*Or, difference of reduced reciprocals, something like that. Anyway, consider the element's resistance in relation to system resistance, and overall loss.Note that we can go a slightly useless direction and ask the Q of a resonator as a component in isolation. That is, what is the Q of its terminal impedance? This is necessarily 0 at resonance (reactance goes to zero, impedance is all resistive), and increasing away from there, towards a peak or asymptotic value given by the respective elements making up the resonator (so, you get more inductor Q below Fo, and more capacitor Q above; the distinction may not be meaningful for a distributed element, but this is obviously the case for a lumped equivalent). So there are some questions we need to ask carefully about, say if we're taking measurements with just an impedance meter (measuring a 1-port (s1), or s11, s22, etc. on a VNA), we need to solve for the useful combination, not necessarily the apparent one.So for your case, we have two important questions:1. What is the Q of your load?2. What is the Q of Cshunt to your load?3. What is Q of the overall system? (In whatever applicable way: if as a resonator say, then we would want to use the bandwidth measure, not the terminal impedance at resonance!)I'm going to guess that your j246 is pure inductance, given the popularity of 13.56MHz for near-field applications -- implying 2.887uH. It's not obvious if this is stable -- there may be stray capacitance leading to a resonance not far above this, for example. The point being twofold: your antenna has some Q at this frequency, but also this Q might not be representative over a wide range anyway due to other effects pulling in first (like the SRF). There are also higher-order (more subtle) effects, like skin effect in the wire, or absolutely anything nearby -- if this is on top of a chassis for example (hopefully with a ferrite shield plate inbetween?), those will all have some frequency-dependent impedance reflected in the winding impedance. Ultimately, we can draw an equivalent circuit of this inductor, approximating all of these properties to arbitrary precision, and use that to further design the system. (Or just use tabulated impedances or scattering parameters, without drawing the lumped-element approximation -- that's fine too!)So, point just being, if we only ever care about exactly one frequency, sure, you'll have a Q at that frequency, given by the ratio of elements. But that won't necessarily mean anything in relation to other kinds of Q, particularly ones that are dependent on frequency -- like bandwidth, especially if that bandwidth should be over a wide range!Anyways, Cshunt is 66.7 ohms at resonance, which... let's see, that doesn't cancel out the inductive reactance in parallel, it'll leave a capacitive phase; and then Cseries acts to divide down into that, presumably presenting 50 ohms at the input port, but...Wait, is that Cseries in series with the inductor first, then Cshunt across the source? Cseries is 221 ohms, leaving a slightly inductive impedance that Cshunt can then correct, while transforming |Z| a bit. That probably makes more sense...The Q figure for your cap divider, is just that, and equals the maximum Q for resistive source/load. Obviously that's not the case here, and will be limited further by the Q of the inductor (~7.7). Also, note that's the unloaded Q, for which coupling is perfect into the winding resistance (mostly; among other things, as hinted above). The impedance will change when coupled to something else, and ideally one has a model of that, and tunes for it.Final comments -- what kind of RX/TX is this connecting to?If RX, it probably doesn't care much, so this is fine.If TX, it probably requires an inductive load impedance. (The equivalent circuit is about the same as any other inverter: a totem pole of transistors driving a load between rails. They're probably not too different from your average logic-level CMOS output pins, actually. Anyway, the problem is node capacitance, which causes additional switching loss for a capacitive or resistive load, but is negated for sufficient inductive load. Put another way: the output is square wave, and those harmonics need to bounce off an inductor, recycling that power through the supply, rather than burning it in Rds(on) and switching losses.)So a cap divider isn't usually a good idea for TX, and an inductive filter network is required. The simplest solution is probably putting Cshunt across the inductor, and changing Cseries into Lseries. So you get a traditional L-match network, with a reactive load and resistive to slightly inductive input.Many chips also have complementary (balanced) outputs (aka H-bridge). You can mate this to an unbalanced load, using a balun. A balun component (transformer or tapped inductor) is simplest, or can be avoided by using a 180° hybrid on one side, or +/-90s on each, or by splitting the circuit so that it's balanced. The latter is probably a good idea anyway, so that the winding can have on average zero voltage applied to it -- reducing the emitted electric field, and thus sensitivity to proximity of nonmagnetic / poor conducting materials. Instead of one Cshunt to GND (in parallel with the coil), split it into two in series of double the value each, and ground the middle; then the two ends are driven by opposite phases, so, the same networks going up to the respective TX pins.All of this should be covered (albeit somewhat more basic, and more directly dealing with a few recommended topologies) in the part appnotes, so give that a look as well.Tim

author: Joel_Dunsmore
author_rank: Regular Contributor
timestamp: January 23, 2022, 07:49:27 am
content: There's a lot of discussion about definition of Q, but one thing to point out is it is very unlikely that your load is really 32+jWhatever at all frequencies. It is likely it is that at only one frequency and has either inductive or capacitive response beyond that (the + sign indicates inductive but if it is at one frequency only it could be moving in a number of directions or be rotated by feed length). So depending on how fast it transitions away from it's reactive element you can either have very wide matching or very narrow matching. thus width of your matched response depends very strongly on the nature of the reactive load.In my experience, as was indicated in other posts, you care about the Q of the matching network in terms of it's additional loss it applies when it is ideally matched. that is, if you ideally match your load to a 50 ohm source you will get the maximum power transferred. If you have reactive mismatch, some power will be reflected to the source. But if you have low-Q elements, then you can still perfectly match to the load but not all the power will be absorbed by the load as some is dissipated in the lossy nature of the matching network. Usually, I think this is the measure of Q that is of concern.

author: jrive
author_rank: Newbie
timestamp: January 23, 2022, 06:52:58 pm
content: Thank you for your response. The load is , like I said , 32+j246, so yes, an inductor with its series resistor. The cap divider match is just an LC L-section, where the series cap and the load inductance are “swapped”, and the load inductance is used as the series element of the L-section. The series component of the L-section is Xs=Xtot= Xinductive_from_ant - Xseries cap. The Q of the L-section match is given by sqrt(50/32-1) which is 0.75. Since I am using only one stage, the Q is what it is; I don’t have the degree of freedom to set it. Given this Q, then Xtot=32*0.75 =24. Substituting back into the equation for Xtot, Xseries_cap is then 24-246=-222. This results in Cseries=1/(2*pi*13.56e6*222)= 52.8, or 53pF. The shunt component of the L-section (which must be opposite to the inductor , hence a capacitor) is then obtained from Xp=Rp/Q. Substituting Q=0.75 and Rp=50, Xcshunt=66.67, which is Cshunt=176pF at 13.56MHz.Thus, my question is about how this Q is manifested in the final circuit. The circuit is a generator, with a 50ohm source resistance, feeding this cap divider match, shunt component facing the generator, connected to the 2 .89uH inductor in series with 32 ohms. I’m not concerned about component Qs or dissipation factor at this time, assume they’re ideal for the purposes of discussion.How does the Q used for the calculation of the comp reactances and values translate to the bandwidth of the match? Is this the network Q.- what does that mean?Does this mean it is the Q of the Transfer function Vo/Vi at the load ? It is a filter, for all intents and purposes.or is it the bandwidth of the matched load impedance as seen from the source (ZIn)?What I’m the circuit response can i measure to see this “Q” manifested?

author: jrive
author_rank: Newbie
timestamp: January 23, 2022, 07:21:22 pm
content: Thank you for your response.I understand that the load is not j246 at all frequencies. It is j246 at the 13.56MHz,though, and that is why it is matched at that value at 13.56Mhz. Although your statement about matching bandwidth is dependent on the nature of the reactive load is correct, there are matching techniques that trade off return loss for bandwidth, thus the matching solution can significantly influence your bandwidth as well.You refer to the Q of the components like the other responses, but that is not what I'm asking about. I get that. The component Q's are not (is not) the Q that is used for designing the matching network to achieve the best match at the desired frequency. It is assumed these are much greater than the matching network's Q. The network Qiswhat is used to design the matching network. In my case, Q was fixed by the ratio of source to load resistances, or 0.75, since I am only using one stage. If I used multiple stages, I add another degree of freedom and I can select this Q arbitrarily. But again, this is not what I'm trying to ask.I want to understand what this Q=0.75 translates to in the circuit's response. Assuming ideal components, in simulation, how does this Q=0.75 manifest itself?What can I look at in the circuit response to see the effect of this Q? an AC sweep of the transfer function? The input impedance seen by the load --what, in the end, is the meaning of this Q ?Thanks.

author: G0HZU
author_rank: Super Contributor
timestamp: January 23, 2022, 07:44:24 pm
content: I get a result of loaded Q =1.919 if that helps?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 23, 2022, 08:20:50 pm
content: Looking at just the matching network itself, you can measure that Q. For example, s11 when port 2 is loaded by the matching resistance. With respect to s22, it's the minimum Q of that network, I think? That is, any other impedance gives a higher Q, because the impedance doesn't match up or the reactances aren't balanced.With the antenna, that figure won't correspond to much in the circuit itself. (Much as component Q becomes less meaningful once in a circuit.)Tim

author: G0HZU
author_rank: Super Contributor
timestamp: January 23, 2022, 08:35:23 pm
content: Note that if I measure the loaded Q for the 50R match to the real Rp part of the inductor I get a loaded Q of just over 3.8. That's how I would estimate loaded Q if this C+C+L network was part of a Colpitts oscillator design.

author: jrive
author_rank: Newbie
timestamp: January 23, 2022, 10:41:52 pm
content: yes, I understand all that....thank you.so then, what is the meaning of this 0.75 Q? Assume ideal components in the matching circuit. The source is 50+0j ohms, frequency is 13.56MHz. The load is 32+j246 (32 ohm resistor in series with 2.89uH inductor). I design the circuit using the that Q given by the ratio of the source and load resistances I"m trying to match, so 0.75. What does this ultimately mean?I think I just figured this out, as I was writing the paragraph above! This Q is the optimal value for max power transfer to the load. However, this is not the Q of the final network. The matching network is ultimately a transformation of the circuit to an equivalent RLC circuit. The current through this circuit is then max at the resonant frequency. The Q of this circuit is then the ratio of the peak current to the |Ipk|/|I_f3dB|, which is still dominated by the load reactance, and the total series  resistance, which is 2*32ohms. Thus, the final Q of the system is w0*L/Rtotal, or 246/(2*32)= 13.56MHz/64 = 3.8.I was able to confirm this in the simulation. I tried attaching it below:

author: jrive
author_rank: Newbie
timestamp: January 23, 2022, 10:43:23 pm
content: Thanks, but it doesn't help if you don't tell me how you did it. I'm looking for an explanation, not an "answer". I think I figured it out. I replied to someone's response explaining what I figured.

author: jrive
author_rank: Newbie
timestamp: January 23, 2022, 11:09:48 pm
content: thank you! yes, that is what I figured out.!! The match circuit is equivalently an RLC at the resonant frequency, where the Q is then woL/Rtot = woL/2*Rload, since the source is matched to the load. Exactly as you said, the series 50ohms from the source is 50ohms in parallel with Cshunt after an impedance transformation of the voltage source-in-series-with-a -esistance to a current-source-in-parallel-with-50 ohms.This 50 ohm Rp becomes 32 ohms Rs via the matching network. Thus Rtot = 2*32, and the Q becomes 246/64=3.8. Perfect!I guess the design Q optimizes the match, but ultimately, the network Q will never be greater than the load reactance/2*load resistance! The lossier the match Q, though, the broader the match, I think. So for instance, if I added additional stages, and lowered the Q even more (than the 0.75, I mean) , then I would end with "less" perfect match, which would translate to a larger bandwidth (at the expense of power transferred to the load).

author: G0HZU
author_rank: Super Contributor
timestamp: January 23, 2022, 11:16:53 pm
content: Quote from: jrive on January 23, 2022, 10:43:23 pmI did it by calculating the loaded Q based on the (90ns?) group delay through the resonator when matched to the resistive part of the load. This is how it is often done for oscillator design although I would normally convert to Rp for oscillator designs. QL = (w*GD)/2

author: jrive
author_rank: Newbie
timestamp: January 23, 2022, 11:33:28 pm
content: this is interesting; your Q is exactly half of the Q I calculated. How exactly did you calculate the group delay of the matching network terminated by the load (2.89uH inductor plus 32) load? I'll try it myself to see what I get and to see if I understand how you got your number.Thanks again.

author: G0HZU
author_rank: Super Contributor
timestamp: January 24, 2022, 12:01:02 am
content: Quote from: jrive on January 23, 2022, 11:33:28 pmIf you look back a few posts I think I get the same Q as you at 3.83 if I treat the network the same as the feedback network in a Colpitts oscillator. See my earlier quote below.QuoteThe group delay appears to be about 90ns so loaded Q = (w*GD)/2 = 3.83

author: G0HZU
author_rank: Super Contributor
timestamp: January 24, 2022, 12:11:05 am
content: QuoteYou can measure the rate of change of s21 phase with respect to frequency at 13.56MHz to get the group delay. I have a simulator here that plots it for me.It looks like the phase changes 6.48 degrees in 200kHz so the rate of change of phase wrt this frequency change computes to 90ns. This is the way I would measure the loaded Q but there may be other ways.