Capacitors have a Q, what would the Q be of the self capacitance of a resistor?

[aweatherguy]

I should add, strike what I said about the "2" in "1092" being meaningless. It may or may not be.

[vk6zgo]

  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.

I 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' ?

Many 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.

[aweatherguy]

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 there might not 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!

[TimFox]

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.

[aweatherguy]

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 is not 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.

[RoV]

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.

The model with a lossy parallel capacitor that I described in https://www.eevblog.com/forum/rf-microwave/capacitors-have-a-q-what-would-the-q-be-of-the-self-capacitance-of-a-resistor/msg4531178/#msg4531178 is perfectly consistent with that. See plot.

[RoV]

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.

Why 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.

[aweatherguy]

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.

[aweatherguy]

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.

[mawyatt]

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.

We'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,

[aweatherguy]

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...