Measuring the Disspation of the self capacitance of a thru hole resistor.

I have added a 10MΩ resistor across a high Q LC  (on a Q meter) and did the math on the drop in Q to get the value, but instead of 10MΩ I got 4,280,000Ω.
I'm tuning out the self capacitance, but think maybe I have the loss resistance multiplied by the Q of the resistors self capacitor (9,342,398Ω)  in parallel with the 10MΩ giving me the low measurement of 4,280,000Ω. 
 With a 10MΩ 1/2 W thru hole resistor, I'm calculating a Q of 22 or 0.045 DF.

 Does anyone have a way to measure the Dissipation of the self capacitance of a 10MΩ resistor? 

1/4 W will be fine, I'll retest with a 1/4 W to compare numbers.
                     Mikek

On Friday, November 11, 2022 at 8:30:45 AM UTC-4, Lamont Cranston wrote:
> I have added a 10MΩ resistor across a high Q LC (on a Q meter) and did the math on the drop in Q to get the value, but instead of 10MΩ I got 4,280,000Ω. 
> I'm tuning out the self capacitance, but think maybe I have the loss resistance multiplied by the Q of the resistors self capacitor (9,342,398Ω) in parallel with the 10MΩ giving me the low measurement of 4,280,000Ω. 
> With a 10MΩ 1/2 W thru hole resistor, I'm calculating a Q of 22 or 0.045 DF. 
> 
> Does anyone have a way to measure the Dissipation of the self capacitance of a 10MΩ resistor? 
> 
> 1/4 W will be fine, I'll retest with a 1/4 W to compare numbers. 
> Mikek

If your self capacitance is in parallel with the resistor, then the dissipative resistance of that capacitance is in series with the capacitance.  Depending on the frequency, the dissipative resistance will have a different impact on your circuit. Are you taking this into account with your math?  Or are you treating the dissipative resistance as being in parallel with the actual resistance?  I've never done the math, but the two circuits may be equivalent with different values of the dissipative resistance.  

I guess my point is that it would seem that the parasitic capacitance would have to have a rather low impedance at your test frequency to have this effect on the circuit.  What frequency are you testing? 

-- 

Rick C.

- Get 1,000 miles of free Supercharging
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On Friday, November 11, 2022 at 9:21:17 AM UTC-6, Ricky wrote:
> On Friday, November 11, 2022 at 8:30:45 AM UTC-4, Lamont Cranston wrote: 
> > I have added a 10MΩ resistor across a high Q LC (on a Q meter) and did the math on the drop in Q to get the value, but instead of 10MΩ I got 4,280,000Ω. 
> > I'm tuning out the self capacitance, but think maybe I have the loss resistance multiplied by the Q of the resistors self capacitor (9,342,398Ω) in parallel with the 10MΩ giving me the low measurement of 4,280,000Ω. 
> > With a 10MΩ 1/2 W thru hole resistor, I'm calculating a Q of 22 or 0.045 DF. 
> > 
> > Does anyone have a way to measure the Dissipation of the self capacitance of a 10MΩ resistor? 
> > 
> > 1/4 W will be fine, I'll retest with a 1/4 W to compare numbers. 
> > Mikek
> If your self capacitance is in parallel with the resistor, then the dissipative resistance of that capacitance is in series with the capacitance. 

If that is true then I did the calculations wrong. Unless the series resistance is over 9.34MΩ, it doesn't explain the 10MΩ measurement error.
Rethinking, maybe the resistors self capacitance is much higher than 22 and it could work out to 9.34MΩ.

>Depending on the frequency, the dissipative resistance will have a different impact on your circuit. Are you taking this into account with your math? 

I have only tested at 1MHz. Remember, I'm trying to measure the value of a 10MΩ resistor on a Q meter and got 4,280,000Ω instead of 10MΩ, I'm trying to make sense of the answer I got.

>Or are you treating the dissipative resistance as being in parallel with the actual resistance?

Yes, that is what I'm doing

> I've never done the math, but the two circuits may be equivalent with different values of the dissipative resistance. 

Or, the answer could come out right if a higher Q is used,  22 is very low for a capacitor, but it's not a designed capacitor, it is a stray capacitor.
> 
> I guess my point is that it would seem that the parasitic capacitance would have to have a rather low impedance at your test frequency to have this effect on the circuit. What frequency are you testing? 
 
  Testing at 1MHz and the parasitic capacitance measures at 0.37pf.
                           Thanks, Mikek
> 
> -- 
> 
> Rick C. 
> 
> - Get 1,000 miles of free Supercharging 
> - Tesla referral code - https://ts.la/richard11209

> On Friday, November 11, 2022 at 9:21:17 AM UTC-6, Ricky wrote: 
> > On Friday, November 11, 2022 at 8:30:45 AM UTC-4, Lamont Cranston wrote: 

> > I guess my point is that it would seem that the parasitic capacitance would have to have a rather low impedance at your test frequency to have this effect on the circuit. What frequency are you testing?
> Testing at 1MHz and the parasitic capacitance measures at 0.37pf. 
> Thanks, Mikek

OK, time for my theory, feel free to tell me if it is not good.
Since the 0.37pf is being resonated, its R is magnified by Q of the capacitor.
 So, I get a large R in parallel with my 10MΩ, lowering the total R.
 Here is a drawing showing the details as I see them.
https://www.dropbox.com/s/yjxq29f23zhft6t/Cp%20Magnified%282%29.jpg?dl=0
                             Thanks,   Mikek

On Fri, 11 Nov 2022 04:30:40 -0800 (PST), Lamont Cranston
<amdx62@gmail.com> wrote:

>I have added a 10M? resistor across a high Q LC  (on a Q meter) and did the math on the drop in Q to get the value, but instead of 10M? I got 4,280,000?.
>I'm tuning out the self capacitance, but think maybe I have the loss resistance multiplied by the Q of the resistors self capacitor (9,342,398?)  in parallel with the 10M? giving me the low measurement of 4,280,000?. 
> With a 10M? 1/2 W thru hole resistor, I'm calculating a Q of 22 or 0.045 DF.
>
> Does anyone have a way to measure the Dissipation of the self capacitance of a 10M? resistor? 
>
>1/4 W will be fine, I'll retest with a 1/4 W to compare numbers.
>                     Mikek

It sounds like you're operating outside of your 
test equipment's rated tolerances.

RL

On Sunday, November 13, 2022 at 1:17:08 PM UTC-6, legg wrote:

> It sounds like you're operating outside of your 
> test equipment's rated tolerances. 
> 
> RL

 Looking in the manual it shows 1000MΩ can be measured at 50kHz and over 900MΩ at 1MHz. I'm measuring at 1MHz.
 I can clearly define the change in capacitance when I add the resistor. So I don't see a problem there.
 I will note,  since the previous 1/2W 10MΩ, I have measured a 10MΩ 0805 smd resistor and I have closer answers of 7MΩ to 9.2MΩ, dependent on how the resistor is mounted. 
The surface mount resistor measures 0.12pf to 0.14pf, well, 0.3pf with the apparently bad mounting that I thought would be good. But it still gave me an almost 8MΩ measurement.
 I am out of the range the injection voltage the internal meter will read, so I have an external meter measuring the lower injection voltage, (a well known practice.)

  I hate to harp, but I'm looking to understand, does resonance of the resistors self capacitance result in a large parallel R?
Here is a drawing showing the details as I see them.
https://www.dropbox.com/s/yjxq29f23zhft6t/Cp%20Magnified%282%29.jpg?dl=0 
                      Thanks, Mikek

legg <legg@nospam.magma.ca> wrote:

> On Fri, 11 Nov 2022 04:30:40 -0800 (PST), Lamont Cranston
> <amdx62@gmail.com> wrote:
> 
>>I have added a 10M? resistor across a high Q LC  (on a Q meter) and did
>>the math on the drop in Q to get the value, but instead of 10M? I got
>>4,280,000?. I'm tuning out the self capacitance, but think maybe I have
>>the loss resistance multiplied by the Q of the resistors self capacitor
>>(9,342,398?)  in parallel with the 10M? giving me the low measurement of
>>4,280,000?. 
>> With a 10M? 1/2 W thru hole resistor, I'm calculating a Q of 22 or
>> 0.045 DF. 
>>
>> Does anyone have a way to measure the Dissipation of the self
>> capacitance of a 10M? resistor? 
>>
>>1/4 W will be fine, I'll retest with a 1/4 W to compare numbers.
>>                     Mikek
> 
> It sounds like you're operating outside of your 
> test equipment's rated tolerances.
> 
> RL

Also operating beyond the limits of comprehension. The Q of an LC circuit 
is mainly determined by the losses in the inductor. This results in a Q of 
20 to 40. 

There is no point in adding a 10M resistor across the inductor.

The "Dissipation of the self capacitance of a 10M resistor?" is an utterly 
meaningless question.



-- 
MRM

On 2022-11-13 20:53, Mike Monett VE3BTI wrote:
> legg <legg@nospam.magma.ca> wrote:
>
>> On Fri, 11 Nov 2022 04:30:40 -0800 (PST), Lamont Cranston
>> <amdx62@gmail.com> wrote:
>>
>>> I have added a 10M? resistor across a high Q LC  (on a Q meter) and did
>>> the math on the drop in Q to get the value, but instead of 10M? I got
>>> 4,280,000?. I'm tuning out the self capacitance, but think maybe I have
>>> the loss resistance multiplied by the Q of the resistors self capacitor
>>> (9,342,398?)  in parallel with the 10M? giving me the low measurement of
>>> 4,280,000?.
>>> With a 10M? 1/2 W thru hole resistor, I'm calculating a Q of 22 or
>>> 0.045 DF.
>>>
>>> Does anyone have a way to measure the Dissipation of the self
>>> capacitance of a 10M? resistor?
>>>
>>> 1/4 W will be fine, I'll retest with a 1/4 W to compare numbers.
>>>                      Mikek
>>
>> It sounds like you're operating outside of your
>> test equipment's rated tolerances.
>>
>> RL
>
> Also operating beyond the limits of comprehension. The Q of an LC circuit
> is mainly determined by the losses in the inductor. This results in a Q of
> 20 to 40.
>
> There is no point in adding a 10M resistor across the inductor.

One gets surprised now and then. I was measuring capacitive plate electrodes
of a beam position pick-up for an accelerator one day. The electrodes
were supposed to all have the same capacitance in the 100pF ballpark.

I was using an old-fashioned Rohde & Schwarz 'KARU' because that instrument
is excellent for comparing capacitances. You tune it to resonance on one
of the capacitive plates and then adjust the other plates to the same
resonance.

I was surprised that one of the plates had a rather poor response, even
though a quick check with an ohmmeter showed nothing out of the ordinary.
It turned out that plate had a 30 MOhm leakage to GND, because one of the
vacuum feedthroughs had been improperly dried after degreasing. That was
beyond the range of my ohmmeter.

My point, after all this, is that even a very large resistance in parallel
with either the capacitor or the coil will reduce the Q significantly.

Jeroen Belleman

On Sunday, November 13, 2022 at 3:53:52 PM UTC-6, Jeroen Belleman wrote:

> 
> My point, after all this, is that even a very large resistance in parallel 
> with either the capacitor or the coil will reduce the Q significantly. 
> 
> Jeroen Belleman

 Hi Jeroen,
With an inductor having a Q of around 1000, I'm getting Q reductions of I'm getting Q reductions of 90 to 130 with a 10MΩ resistor, after I went to the smd 0805 resistor. The math works out to 7MΩ to 9MΩ, not the 10MΩ expected. (When I measured the 1/2 W thru hole resistor, I got 4,280,000Ω.)
 I'm sure this has something to do with the capacitance, but on the Q meter I'm resonating that out, so should have only R left.
  I saw your write up about shunt capacitance of 1206 resistors, I would have liked to see you go to 10MΩ on you graphs :-). You weren't resonating out that capacitance, so not sure that would have got me any further.

 Do you thing resonating the self Capacitance results in a large R?

My numbers are 0.12pf resistor self capacitance, with an 0805 10MΩ resistor at 1MHz put in parallel with 168uh and  151pf.
I just played with a series to parallel calculator and I can't make it work.
 Do you have number in mind for the Q or DF of the self capacitance of 0805 resistor in a only slightly restricted surrounding (meaning more lossy than if done to higher measurement standards.)
 I'm only 10% to 20% off at this point, and it may have to do with as the manual says, "keep delta Q under 5",  I changed my coil and got it under 12, but it still gave me 8MΩ instead of 10, so I'm not hopefully performing the test fully my the manual is going to help.
                                         Thanks, Mikek

On Sunday, November 13, 2022 at 1:53:24 PM UTC-6, Mike Monett VE3BTI wrote:
> legg <le...@nospam.magma.ca> wrote: 

> Also operating beyond the limits of comprehension. The Q of an LC circuit 
> is mainly determined by the losses in the inductor. This results in a Q of 
> 20 to 40.

 The Q of my 170uH inductor is around 1250 measured on an HP 4342A.

> There is no point in adding a 10M resistor across the inductor. 

Actually it is across the capacitor, and that is exactly what the manual says to do, with measurements up to 1 Gigaohm.

> The "Dissipation of the self capacitance of a 10M resistor?" is an utterly 
> meaningless question. 

It very well might be,  I'm trying to figure out what the affect is at 1MHz,
 it may be very low, but it certainly is important at higher frequencies.

  So, after a more detailed reading on measuring "Large resistors",  I changed, to using a Q standard (having a lower Q) for the measurement,
 The manual says to keep the Q change to 5 or below, I'm now at 12 to 20, so, I'll find a way to solve that
and see if that gets me closer, I appreciate the feedback.
                        Thanks, Mikek