Capacitors have a Q, what would the Q be of the self capacitance of a resistor? - Page 1
Forum section: electronics / rf-microwave

author: Qmavam
author_rank: Regular Contributor
timestamp: November 10, 2022, 02:57:46 am
content: Long story, but I'll make it short, if I get an answer that supports my thinking I'll describe how I got to the question.I have a Q meter, I want to measure a high value resistor by noting the Q drop when I put it across the tuning cap (and do the math). I resonate an LC at 1MHz, find the Q, thenadd a parallel 10 MΩ 1/2W MF resistor, when I add the resistor to get back to resonance, I have to reduce the tuning capacitor by 0.37pf.So my resistor has 0.37pf of parallel capacitance. Capacitors have a Q, what would the Q be of the self capacitance of a resistor? 10, 100, 1000?Thanks, Mikek

author: CatalinaWOW
author_rank: Super Contributor
timestamp: November 10, 2022, 04:11:20 am
content: I don't think there is a simple answer to this question. Q has meaning with respect to a resonant circuit, which requires at least two energy storage elements. Usually an inductance and a capacitance, though strain from magnetostriction and other things can provide a second storage element. A resistor can have several resonances. A series resonance from the lead inductance and parasitic capacitance. A in traditional wire wound and spiral cut film resistors a parallel resonance from the inductance of the coiled conducting element with the capacitance between coils. There can be others. You may also be observing effects from other elements of your test circuit. Since the values of the inductances and capacitances are usually difficult to obtain, and because the resistance of lossy elements in the resonant circuit can also be tough it is common just to do a frequency sweep of the part and determine Q by the relative bandwidth (as defined by 3dB or 6 dB drop from the amplitude) of the resonant peaks. Often some care is required to separate the resonances that are internal to the resistor from resonances that are partly or completely external to the resistor.

author: bdunham7
author_rank: Super Contributor
timestamp: November 10, 2022, 04:19:35 am
content: Quote from: CatalinaWOW on November 10, 2022, 04:11:20 amWell, there is but it is practically meaningless. For a parallel RC circuit Q = ωRC, so his answer is about 23 (@ 1MHz). The impedance of the capacitor is 430K, which overwhelms the 10M resistor--which is why you don't see ridiculous values like this in RF circuits. But without considering the parasitic inductance of the resistor I don't think that gets you anywhere in this case.

author: Qmavam
author_rank: Regular Contributor
timestamp: November 10, 2022, 10:15:49 am
content: OK, seems like enough understand to do the long story.Here’s the scenario:I have a 167uH capacitor resonating with a 151pf cap on my Q meter, Q = 1092.Now I put a resistor across the tuning capacitor, I retune* for resonance and measure the lowered Q. Because of the resistors capacitance I must reduce the capacitance to get back to resonance.The idea is to measure the resistor I put in parallel with the capacitor by the drop in Q.I’m having trouble getting the Q meter to measure the resistor as the correct value.First my Q meter measures LC in parallel resonance. And I think it might be becauseI'm resonating the self capacitance of the resistor.The detailsI set up an inductor and measured Q at 1MHz, the Q is 1092, the reactance is 1,053Ω.So, Rp = Q times X =1092 x 1053 = 1,149,876Ω =Rp (Rp = Parallel resistance at resonance)Then I put a 10MΩ Metal Film ½ W resistor in parallel with the tuning capacitor.I had to reduce* the tuning cap by 0.37pf and the Q is reduced to 882. Making Rp = 928,746ΩQ times X = 882 x 1053 = 928,746Ω*Meaning the resistor has 0.37pf of capacitance across it.The parallel combination of a 1,149,876Ω and 10MΩ is 1,031,290Ω.But, my Q meter gives an Rp of 928,746Ω saying my 10MΩ has a value of,4,820,000 Ω. i.e. 1,149,876Ω//4,820,000Ω = 928,746ΩBut wait, my resistor has 0.37pf of capacitance. At 1MHz that is 430,128ΩSo, I have 430,128Ω of capacitive reactance in parallel with 10MΩ or 429,751Ω.That doesn’t jive with my measurement.What I don’t have a handle on is this, the capacitance of the resistor is tuned out by the inductor, so I would think it is not seen and we have just Rp. (a pure parallel resistance)But, the self capacitance of the resistor has a Q, if the R loss of the capacitor is magnified by the Q of the capacitor, Then we have a pure resistance across the 10MΩ.For an example, let's say the Q of the “self capacitance” of the resistor is 21.72, and if Rp=Q times Xc, (that is a question, I’m not sure)Then 21.72 x 430,128Ω = 9,342,398Ω and 10MΩ//9,342,398Ω = 4,820,000 Ω . This the answer my Q meter gives.If any of this makes sense, is it possible the “self capacitance” of the resistor has a Q of 21.72?The reason I'm doing this is, I want to measure the input impedance of high input impedance amplifiers, maybe as high as 100MΩ and a few pf.Mikek

author: Qmavam
author_rank: Regular Contributor
timestamp: November 10, 2022, 10:32:53 am
content: >But without considering the parasitic inductance of the resistor I don't think that gets you >anywhere in this case.I'm unsure, but an inch of wire at 1MHz is 20nH.20nH at 1 MHz is 126 milliohms, so is that really a problem?At 1 GHZ yes, but 1MHz, not so sure.Thanks, Mikek

author: jjoonathan
author_rank: Frequent Contributor
timestamp: November 10, 2022, 01:52:11 pm
content: This is my favorite resource on the subject:https://www.vishay.com/docs/49427/vse-tn00.pdf

author: RoV
author_rank: Regular Contributor
timestamp: November 11, 2022, 08:50:21 am
content: Quote from: Qmavam on November 10, 2022, 10:15:49 amIt does actually make sense according to me. That parasitic capacitor is not simply due to the opposite metallic electrodes of the resistor, with their area and distance, but it is a distributed capacitor that encounters lossy parts along the resistor.I have experienced exactly what you say when I built a medium-high voltage (few kV) oscilloscope probe, see photo. It was a 1:100 to be added to the standard 1:10 to reach 1:1000. I deliberately added a small capacitance across the high valued resistors (about 25 Meg), built with a piece of coax cable, in order to compensate the load capacitance. However, a shunt load capacitor wasn't sufficient to compensate acceptably, and I traced this to the fact that there was some additional lossy capacitance across the resistors, that I compromise compensated with series RC on the load side. The final response is in the final Bode plot, where 0 dB correspond to the nominal attenuation of 60 dB. You don't see variable capacitors, because I finally replaced them with fixed SMD parts.

author: Qmavam
author_rank: Regular Contributor
timestamp: November 11, 2022, 11:53:24 am
content: Thanks for the response RoV,I have posed this question to a couple of people and a few forums or newsgroups and no one has addressed the main question as you did. Thanks!Capacitors usually have low losses, low dissipation, high Q, but that is by design, the capacitance of a standard resistor is just there as a stray with no attempt to reduce the dissipation. So, I wonder if 22 is a reasonable Q for the stray capacitance of a resistor.I have some SMD 0805 resistors they have much lower capacitance, but I don't know how that would affect the Q of that lower capacitance.Do you have any thoughts about how I could find the Q of the capacitance of a resistor?Or have a I already did that, (and just don't trust the answer) :-)All this comes about as I'm trying to build a high input impedance amplifier without the knowledge to do so!MikekP.S. years ago, we needed to measure 800v with a 400v max scope, I built a voltage divider circuit at the scope input, and had to compensate it, and shield it to make it work properly. The probe made it through the tests, but I always wondered if the probe cable was going to arc through.

author: bdunham7
author_rank: Super Contributor
timestamp: November 11, 2022, 02:50:23 pm
content: Quote from: Qmavam on November 10, 2022, 10:32:53 amI think the parasitic inductance of a resistor is typically much higher than the same length of wire, but in any case you are probably right that in such a high impedance circuit, it may be negligible. So, the Q as I calculated is about 23, your measured result is ~22. Seems legit!What are you using for a 'Q meter' ?

author: bdunham7
author_rank: Super Contributor
timestamp: November 11, 2022, 02:55:50 pm
content: Quote from: Qmavam on November 11, 2022, 11:53:24 amMy apologies in advance if my answer here is a simplistic insult, but you are aware that Q is the inverse of D, dissipation? And if you have a known capacitance in parallel with a known resistance, their ratio determines Q and D? You solved this earlier when you calculated the impedance of the capacitor at 1MHz and stated that the resistor was 10M. Q = ωRC is just all that condensed.

author: Qmavam
author_rank: Regular Contributor
timestamp: November 11, 2022, 03:42:07 pm
content: Quote from: bdunham7 on November 11, 2022, 02:50:23 pm

author: Qmavam
author_rank: Regular Contributor
timestamp: November 11, 2022, 04:34:19 pm
content: Quote from: bdunham7 on November 11, 2022, 02:55:50 pmI calculated 2pifRC, 6.28 x 10^6 x 10^7 x 0.37 x 10^-12 and get 23.236, very close to 22.72.I can't help but think that is a coincidence.But as far as my thinking, that 0.37pf has been tuned out, so we are not talking about a RC to calculate DF.Not sure this matters, but since, the reactance of 0.37pf at 1MHz is 429,751Ω, the parallel impedance of that and a 10MΩ resistor at 1MHz is 429,751Ω. If that is what was going on while measuring on the Q meter my calculation is of by a factor of 10. and I don't think it is.Happy to have your response, it's only my theory about this self capacitance.But it is the only way I can make sense of the 10MΩ measurement error.Thanks, Mikek

author: Qmavam
author_rank: Regular Contributor
timestamp: November 11, 2022, 05:18:49 pm
content: I was told on another group,"If your self capacitance is in parallel with the resistor, then the dissipative resistance of that capacitance is in series with the capacitance."I don't understand, I thought you could convert series to parallel and parallel to series.I don't know how to calculate if the resistance is in series, if some one does this and finds a higher Q that makes me a little more comfortable, because I think 22 is a low for the Q of a capacitor. Please let me know if you recalculate this.Thanks, Mikek

author: TimFox
author_rank: Super Contributor
timestamp: November 12, 2022, 05:33:03 pm
content: The series R-C to parallel R-C conversion, found in textbooks, is a powerful design technique, but it is only valid at a single frequency (or reasonably useful over a narrow bandwidth, such as a resonant circuit).

author: RoV
author_rank: Regular Contributor
timestamp: November 12, 2022, 05:42:17 pm
content: Quote from: Qmavam on November 11, 2022, 05:18:49 pmSee attachment. I think it is actually correct to assume that the capacitance loss is in series with the parasitic capacitor. For such a kind of parasitic capacitance, 20 kohm in series are perfectly reasonable.

author: Qmavam
author_rank: Regular Contributor
timestamp: November 12, 2022, 07:01:25 pm
content: I was writing the below and when I tried to post and got the warning that you had posted. Thanks for that. I'll work on the series equivalent math later, re-learn how to deal with -1!> I think it is actually correct to assume that the capacitance loss is in series with the >parasitic capacitor. For such a kind of parasitic capacitance, 20 kohm in series are >perfectly reasonable.Do you still think or did you ever think, the Rloss of the self capacitance is multiplied by the self capacitance Q?If we assume 20kΩ series resistance, the Q of the self capacitance of the resistor would need to be 463, which seems quite reasonable for a poor capacitor.I followed up this morning by soldering wire wrap wire (minus the insulation) to an smd 10MΩ resistor.First answer I got was 7MΩ, higher that the 4,280,000Ω I had, still not 10MΩ though.I changed the way the resistor leads fly, then I got 9,160,000Ω.With the smd 10MΩ, the capacitance is 0.12pf to 0.14pf depending on lead dressing.Then, rather than clamping the part near the Q meter top, I moved the resistorto the top of the banana connectors, (hoping to reduce capacitance to the Q meter case). Then I got 9,210,000Ω.More and more I'm leaning towards the capacitance causing the error.MikekI thought maybe 5 series* 2MΩ smds would get even better, but it gave 8,315,000Ω.After my coffee wears off, I might try putting a 1/4" of wire between each resistor, to reduce the capacitive interaction between each resistor.* I read that up to 5 in series helps reduce capacitance for through hole resistors.Mikek

author: TimFox
author_rank: Super Contributor
timestamp: November 12, 2022, 07:57:40 pm
content: Back around 1990, when I was working on analog electronics for MRI, we encountered difficulty with resistors, for example 1 megohm, used to bias the gate of a JFET amplifier.The noise in NMR depends on theQof the resonant sensor, and a good system has an unloadedQ(without patient) that is much higher than the loadedQ(after inserting patient and obtaining useful signal), so we carefully investigated the auxiliary components (such as protection diodes and bias resistors), as well as the primary components (windingQand capacitor loss) to improve SNR.When looking at Allen-Bradley carbon composition resistors (still popular then), we looked at the equivalent parallel model (high resistance in parallel with small capacitance) for 1/4-watt leaded resistors.The capacitance was buried in the tuning of the circuit, but the parallel resistance lowered the totalQ.As we increased the frequency (say, from 1 to 20 MHz), the equivalent parallel resistance in this model dropped dramatically.Checking the literature, we found that the resistor construction for high-value CC resistors comprised a mixture of carbon and talc particles, where a higher talc fraction increased the total resistance.However, at the microscopic level, the carbon-talc-carbon interfaces inside the resistor formed internal capacitors that shorted out the carbon, thus reducing the parallel resistance at higher frequency.At the time, we replaced the A-B resistors with "carbon ink" resistors from IRC, that were similar to modern thick-film parts: these had much higher high-frequency parallel resistance at the same DC resistance.More recently, I was testing a classic Wayne-Kerr B601 admittance bridge (15 kHz to 5 MHz useful range) that measures equivalent parallel resistance and capacitance, after minor repairs.I wanted to see how small a conductance could be resolved at a few MHz: I tried both A-B carbon resistors and "flameproof" cermet resistors of roughly the same physical size and found a clear difference where the carbon composition parts showed higher conductance (lower equivalent parallel resistance) than the cermet parts.

author: joeqsmith
author_rank: Super Contributor
timestamp: November 12, 2022, 08:59:22 pm
content: I doubt it's helpful but I had attempted to repeat Jeroen Belleman's (2010 CERN) experiment where they measured the shunt capacitance of some surface mount resistors. I was curious about an odd trend I noticed in their data. I wrote Jeroen but I don't believe they ever repeated the experiment. The link to their original paper and my effort may be found here:https://www.eevblog.com/forum/rf-microwave/shunt-capacitance-of-1206-smd-resistors-jeroen-belleman-december-2010/

author: Qmavam
author_rank: Regular Contributor
timestamp: November 12, 2022, 10:13:13 pm
content: Quote from: joeqsmith on November 12, 2022, 08:59:22 pmMy first question was going to be is a 50Ω network Analyzer going to be accurate at 1MΩ.It looks like you have good fixturing, that is probably 90% of the battle!Scanning the article I see 50fF this is 2.4 times what I measure, but, at these levels, the capacitance of my leads to everything around them and each other could make up that.My question is still, does the Cp resonate and multiply the value of Rs?(Rp? I'm confused at this point)Here's my question reformulated with models and questions. Please tell me if I have something wrong or if I need to clarify something.Thanks, Mikek

author: MathWizard
author_rank: Super Contributor
timestamp: November 16, 2022, 02:53:22 am
content: What level of capacitance might be in some little air coil inductor in a radio ? Not enough for me too worry about. That must be pretty small too.I made a low RF osc. this weekend and compared to LTSpice, it really under performs on the breadboard. Just adding in some 5-10pF caps around the BJT in LTSPice, I can't get it to perform as bad as IRL (it works fine, just at one fifth the Av tho)

author: TimFox
author_rank: Super Contributor
timestamp: November 16, 2022, 03:47:12 am
content: Quote from: MathWizard on November 16, 2022, 02:53:22 amThe self-resonant frequency of any coil or inductor is an important parameter for the component.You may be surprised by the capacitance (that produces the self-resonance), and it is foolish to merely laugh it off.There are simple ways to measure it--consult the literature.

author: aweatherguy
author_rank: Contributor
timestamp: November 16, 2022, 04:28:51 am
content: You are measuring a Q of "1092". That's very high, and not easy to measure accurately. I don't see any measurement tolerances being discussed. Has the equipment been recently calibrated? What is the accuracy of that measurement? Are the measurements repeatable -- if you make them again tomorrow, how much different are the results?Answers to these quesitons may help to figure out what's going on. How far off are your measurements compared to the equipment accuracy?This might end up not being an issue, but it is important to have those ducks lined up.

author: aweatherguy
author_rank: Contributor
timestamp: November 16, 2022, 06:15:16 pm
content: Okay, I did a little digging, and a typical accuracy spec for Q-meters measuring a Q of 1,000 at 1MHz is ±15%. So that Q of 1092 is really somewhere between 928 and 1256 -- and only if the meter has been calibrated within the specified calibration interval. The "2" in "1092" is meaningless. The other value, 882 is in reality somewhere between 750 and 1014. The two ranges overlap, so based on the published specification you cannot even say the Q changed with any certainty.However, you could improve things by relying on a "differential accuracy" for the meter -- but that is not typically specified. To get a handle on that you would need to perform an analysis of the internal design of the Q-meter and come up with your own specification.The discussions about the impedance of a 10M-ohm resistor at 1MHz are all valid, but I think you have some measurement issues to resolve before you can go there.

author: RoV
author_rank: Regular Contributor
timestamp: November 17, 2022, 09:26:57 pm
content: Quote from: aweatherguy on November 16, 2022, 06:15:16 pmIt seems to me that you are confusing accuracy with uncertainty. The fact that accuracy is +/-15% doesn't mean that the measurement is not stable and repeatable. The OP can verify himself if the "1092" reading is stable and consistent among readings done after some time, in all cases after instrument warmup. What you call "differential accuracy" is uncertainty, normally a much smaller figure respect to accuracy.

author: aweatherguy
author_rank: Contributor
timestamp: November 17, 2022, 10:51:35 pm
content: QuoteI partly agree with that, but I'm not confusing accuracy and uncertainty. Resolution and measurement stability to be more specific. Even if the measurement is stable and has the suggested resolution, there's still the separate question of relative accuracy.For example, let's say the two Q values measure 1092 and 882, all the time, day in and day out. There's still two unknowns that are necessary to make the final estimate of added loss (the 10M-ohm resistor). The first is what is the actual Q of the first measurement, because that will determine the reference point from which the final answer is deduced. We know the tolerance is of that number is 15% -- the manufacturer of the Q-meter specifies it.The second is the linearity associated with a differential measurement. The second Q of 882 is about 81% of the first measurement, but what is the actual drop in Q? Using only the 15% spec doesn't help much, but the meter is going to have some relative accuracy or linearity and that is not specified by the manufacturer. The only way to get that is to check the meter against standards that have known, traceable values of Q.And that's not really enough in the general case because that measured linearity may be different over a differrent range of Q values. For example, a change in Q from 1230 to 1000 (also 81% of the original value) may not show the same drop on the meter. Some combination of engineering analysis and more thorough measurement against standards would be required to develop a justifiable general spec for relative accuracy.And just to be clear, I'm talking about a process that would yield a traceable specification -- something you could be assured is accurate compared to NIST or other standards organizations.

author: aweatherguy
author_rank: Contributor
timestamp: November 17, 2022, 11:49:49 pm
content: I should add, strike what I said about the "2" in "1092" being meaningless. It may or may not be.

author: vk6zgo
author_rank: Super Contributor
timestamp: November 18, 2022, 12:47:13 am
content: Quote from: bdunham7 on November 11, 2022, 02:50:23 pmMany years ago, I remember a lecturer giving us a demonstration of the inductance of different types of resistor.Wire wound of course, looked very much like inductors, & carbon composition types exhibited the lowest inductance.A surprise was the carbon film types that were being pushed by Philips at the time, which were very inductive.He had one with the paint removed, & pointed out that the resistance value was set by making a helical cut through the carbon film.I believe some metal film resistors still use this technique.I would have thought surface mount resistors would be the least reactive type these days.

author: aweatherguy
author_rank: Contributor
timestamp: November 18, 2022, 05:43:08 pm
content: So first, let me apologize for going on and on about metrology issues, but with this kind of measurement they are important.I did the math, and theremightnot be an issue with the differential measurement. With an ideal 10M resistor across the circuit, the Q will drop by roughly 10%, and a 1% linearity error would create roughly a 10% measurement error. So you would have a possible 15% error on the initial measurement plus another 10% for a total error of 25%. I don't know if a 1% linearity error is in fact reasonable, but it at least seems reasonable.Change the scenario to a 100M resistor, and the Q will only change by 1%. With such a small change in Q, the linearity error would also be less, so again perhaps adding only a 10% error to the net result again. At this point however, resolution and repeatability become more important -- the change in Q would be from say 1000 to 990, and the difference between a reading of 990 and 991 represents a 10% change in the answer.This is in addition to the issues folks have brought up with the 10M test resistor.The inductor should be looked at too. Does it have a closed magnetic circuit so that external fields are minimal? If it's an open circuit such as an air core coil, the Q can be altered significantly by nearby objects, so that would need to be managed carefully. It may not take much to alter losses in the inductor by a few percent. It would be easy to explore that issue using the Q meter.Suffice it to say this is a very challenging task!

author: TimFox
author_rank: Super Contributor
timestamp: November 18, 2022, 05:55:29 pm
content: Again, I have found experimentally that high-value resistors (say, 1 to 100 megohm) will have an equivalent parallel resistance (parallel R-C model) at a high frequency that is lower than their DC resistance value.(See my reply above.)The appropriate model is not simply the DC resistance value in parallel with a small capacitance, and the parameters change with increasing frequency.

author: aweatherguy
author_rank: Contributor
timestamp: November 19, 2022, 06:50:53 pm
content: If it helps any, here's a plot of the impedance of a 100k-ohm MF resistor, measured with a HP4275A LCR meter. The meter isnot in calibration.When calibrated, the specified accuracy at 100k and 10MHz is 2% on magnitude and 0.08 degrees on the angle. I've converted that data to real (resistive) and imaginary (capacitive) parts. The measurement was capacitive at all frequencies. I don't have any larger resistors to measure, and the meter cannot go much larger anyway (1M-ohm is maximum specified value at 1MHz).Calibration issues not withstanding, you wouldn't want to use this resistor much above 2MHz, suggesting that a 10M-ohm resistor may likely be a no-go at 1MHz.

author: RoV
author_rank: Regular Contributor
timestamp: November 19, 2022, 08:47:15 pm
content: Quote from: TimFox on November 18, 2022, 05:55:29 pmThe model with a lossy parallel capacitor that I described inhttps://www.eevblog.com/forum/rf-microwave/capacitors-have-a-q-what-would-the-q-be-of-the-self-capacitance-of-a-resistor/msg4531178/#msg4531178is perfectly consistent with that. See plot.

author: RoV
author_rank: Regular Contributor
timestamp: November 19, 2022, 08:58:30 pm
content: Quote from: aweatherguy on November 19, 2022, 06:50:53 pmWhy not, if compensated? Clearly, impedance will decrease at high frequency.Tektronix sells the P6015A 1000x high voltage scope probe: BW is 75 MHz; input impedance is 100 Meg + 3 pF, with the real part of impedance going quite low at high frequency (~100 k @ 1 Meg).Tek has a patent on this probe design (actually on the previous version). I saw it once, but now I'm not able to find it.

author: aweatherguy
author_rank: Contributor
timestamp: November 19, 2022, 09:24:36 pm
content: What I meant was, you wouldn't want to use it as a "100k resistor" at those frequencies. Yes, it could be compensated, but you would be relying on unspecified parameters. That's okay for one-off prototypes, but in a production environment, a discussion w/the mfr would be necessary to get some assurances about that.On another note, I found the data I measured is very well modeled with only a 0.16pF parallel capacitance...a series loss resistance isn't required, and inductance effects aren't apparent below 10MHz. Below 100kHz the phase angle is so small that LCR meter accuracy is in play, and it's not clear what is causing the mismatch down there. I could make another plot with error bars...that would be interesting too.Of course, this applies to this particular resistor, and other resistors may not be so easy to model.

author: aweatherguy
author_rank: Contributor
timestamp: November 20, 2022, 06:50:50 pm
content: Is there a better way to accurately measure such high impedances at 1MHz? Maybe. Here's an idea.Measurements on a Bourns 100k MF resistor indicate it is a good approximation to a 100k resistor, even at 1MHz. The circuit diagram below is essentially an attenuator which has an apparent impedance of 10M between the input and output ports. The actual amount of attenuation depends on the impedance of the voltmeter on the output port, Rvm. For this example, it is 1M but much lower values (e.g. 1k) will potentially work too.With the DUT removed, measure the output voltage, Vo. Now connected the DUT as shown and Vo will rise -- in this case a factor of two (6dB) since the DUT has the same resistance as the equivalent resistance between the source and output ports. Not shown here, but you can develop equations for computing the DUT resistance as a function of the change in output voltage.Now, there's one possible issue with this scheme. The input voltage required to make this measurement will depend on the attenuator ratio, voltmeter input impedance and sensitivity. The voltage applied to the DUT will be nearly as large as the source voltage, and if that overloads the DUT then the measurement might not be valid.Anyway, it is another idea for you to try if you cannot get the Q-meter idea to work.

author: mawyatt
author_rank: Super Contributor
timestamp: November 20, 2022, 11:01:22 pm
content: Quote from: aweatherguy on November 19, 2022, 09:24:36 pmWe've seen about the same ~0.16pF for a 1/4 watt MF leaded resistor, and ~0.2pF for a 1/8 watt measured from 10K to 8MHz (HF limit) with IM3536 LCR, and same from 10K to 100KHz (HF limit) on TH2830 LCR.Best,

author: aweatherguy
author_rank: Contributor
timestamp: November 20, 2022, 11:57:01 pm
content: Rats, I think my suggestion about the alternate measurement method won't work -- the original post mentioned a few pF in parallel with the resistance to be measured. So the goal is to measure something like 10M in parallel with a few pF...and estimate the resistive part accurately. This needs some more thinking...