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

Jeroen Belleman <jeroen@nospam.please> 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 Bellema

Presumably, the capacitor is in parallel with the inductor. So any resistance 
is automatically in parallel with both.

100 pf is a lot larger than the capacitance of a 10 M resistor. Calcuating 
Q's in the millions is simply idiotic.






-- 
MRM

On Sunday, November 13, 2022 at 3:53:24 PM UTC-4, Mike Monett VE3BTI wrote:
> legg <le...@nospam.magma.ca> wrote: 
> 
> > On Fri, 11 Nov 2022 04:30:40 -0800 (PST), Lamont Cranston 
> > <amd...@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. 

Mike is not correct in a technical way, but he is probably right in a practical way.  Capacitors have dissipation, some more, some less.  I didn't think about it, but the parasitic capacitance of a resistor, is essentially an air capacitor with an extremely low amount of dissipation.  Combine that with the very low value of the capacitance and you get, what is likely an immeasurable amount of extra dissipation.  

-- 

Rick C.

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On Sunday, November 13, 2022 at 6:49:38 PM UTC-4, Lamont Cranston wrote:
> 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

When you talk about "resonating out" the added capacitance.  I seem to recall that when you have damping in a resonant system, that alone changes the frequency of peak response.  Since that is likely what you are tuning to, it may not be the same thing as equal and opposite XL and XC.  So by changing the tuning, you might be mucking with your own measurement.  Can you try adding a very low loss 0.12 capacitance (twisted wire, for example) and tuning,  then replacing the wire with the resistor and measuring without retuning? 

-- 

Rick C.

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On Sunday, November 13, 2022 at 7:04:31 PM UTC-4, Lamont Cranston wrote:
> 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

When I do the math, the Xc of 0.12 pF @ 1 MHz comes out to 1.3 Mohm.  So your 10 Mohm resistor is actually affecting the circuit as a lossy capacitor. 

-- 

Rick C.

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On Sunday, November 13, 2022 at 6:09:20 PM UTC-6, Ricky wrote:

> When I do the math, the Xc of 0.12 pF @ 1 MHz comes out to 1.3 Mohm. So your 10 Mohm resistor is actually affecting the circuit as a lossy capacitor. 

 The only way I can see my theory working now is if the Q of the 0.12pf is 5.299. That would mean the series resistance is 250.3kΩ.
A series to parallel conversion give me 7.278MΩ. Which is about what I’m getting. But, if say the 0.12pf has a Q of 529.9,
Then the Parallel R is 100 times to high.
   I don’t have any idea about the Q of the self capacitance of a resistor and tiny wire leads that I used, but 5.299 seems very low.
 I’m using this series to parallel conversion calculator. 
https://daycounter.com/Calculators/Parallel-Series-Impedance-Conversion-Calculator.phtml

As you suggested, I added the gimmick wire. The connectors on the Q meter are 1” apart, I put the gimmick wire on and started snipping, until they were no longer wrapped,
The ends overlapped 1/8” and, they were about 1/8” apart. I got close at 0.13pf. I didn’t readjust the tuning cap and the calculations found 7.91MΩ of parallel resistance.
Very similar to 7 other measurements I made of high 7MΩ to low 9MΩ range.
  Just for fun I put the gimmick back on, this time I got it right to 0.12pf, I did tune it out, and got Q with and without, it acts like a 110MΩ resistor.
Then I removed the insulation from the wire wrap wire, I could no longer see any difference in Q, with or without the gimmick, it is just a very good air cap.
i.e. as good as the cap in the Boonton 260A.

At this point, I think the self capacitance of the resistor is lossy, but not lossy enough to explain the measurement error I’m having with the 10MΩ resistor.
Again, reading the manual, there is a very wide frequency range that a 10MΩ can be measured at, it also says Delta Q should NOT be less than 5, I missed the not before, but, that’s OK, I was much higher than 5 so that is not the source of error.
  The next step is to do an alignment of the Boonton 260A and retest. I know it measures low on high Q inductors but it measures only measures 2% high on the (235Q)  Q standard I have.

 The reason this came up. I want to measure the input impedance of high input impedance amps. I thought the Q meter was a good fit, I can tune out the capacitance, so, I know the input capacitance, 
And I can use the drop in Q to calculate the input resistance. Sounds perfect, but I can’t even get a 10MΩ right so, I wonder.
 I have measured 2 high input impedance amps, off the top of my head one had an input capacitance of 0.5pf and resistance of 180,000,000Ω. The other had 1.1pf input capacitance and 200,000,000Ω input resistance.
However, those numbers are somewhat questionable, capacitance is probably good but I don’t yet know about the resistance.  (all measurements were at 1MHz.)
 I have beat this dead horse enough for today.

                                              Thanks for participating, Have a good night, Mikek
 

On Monday, November 14, 2022 at 12:23:27 AM UTC-4, Lamont Cranston wrote:
> On Sunday, November 13, 2022 at 6:09:20 PM UTC-6, Ricky wrote: 
> 
> > When I do the math, the Xc of 0.12 pF @ 1 MHz comes out to 1.3 Mohm. So your 10 Mohm resistor is actually affecting the circuit as a lossy capacitor.
> The only way I can see my theory working now is if the Q of the 0.12pf is 5.299. That would mean the series resistance is 250.3kΩ. 
> A series to parallel conversion give me 7.278MΩ. Which is about what I’m getting. But, if say the 0.12pf has a Q of 529.9, 
> Then the Parallel R is 100 times to high. 
> I don’t have any idea about the Q of the self capacitance of a resistor and tiny wire leads that I used, but 5.299 seems very low. 

I think you are looking at it wrong.  The parasitic capacitance of the resistor has an extremely high Q, I'm sure... if you ignore the resistor itself.   My point is, that the reactance of the resistor is much lower than the resistance.  


> I’m using this series to parallel conversion calculator. 
> https://daycounter.com/Calculators/Parallel-Series-Impedance-Conversion-Calculator.phtml 
> 
> As you suggested, I added the gimmick wire. The connectors on the Q meter are 1” apart, I put the gimmick wire on and started snipping, until they were no longer wrapped, 
> The ends overlapped 1/8” and, they were about 1/8” apart. I got close at 0.13pf. I didn’t readjust the tuning cap and the calculations found 7.91MΩ of parallel resistance. 
> Very similar to 7 other measurements I made of high 7MΩ to low 9MΩ range. 
> Just for fun I put the gimmick back on, this time I got it right to 0.12pf, I did tune it out, and got Q with and without, it acts like a 110MΩ resistor. 
> Then I removed the insulation from the wire wrap wire, I could no longer see any difference in Q, with or without the gimmick, it is just a very good air cap. 
> i.e. as good as the cap in the Boonton 260A. 
> 
> At this point, I think the self capacitance of the resistor is lossy, but not lossy enough to explain the measurement error I’m having with the 10MΩ resistor. 

You can't argue with measurements.  But they have to be analyzed correctly.  


> Again, reading the manual, there is a very wide frequency range that a 10MΩ can be measured at, it also says Delta Q should NOT be less than 5, I missed the not before, but, that’s OK, I was much higher than 5 so that is not the source of error. 
> The next step is to do an alignment of the Boonton 260A and retest. I know it measures low on high Q inductors but it measures only measures 2% high on the (235Q) Q standard I have. 
> 
> The reason this came up. I want to measure the input impedance of high input impedance amps. I thought the Q meter was a good fit, I can tune out the capacitance, so, I know the input capacitance, 
> And I can use the drop in Q to calculate the input resistance. Sounds perfect, but I can’t even get a 10MΩ right so, I wonder. 
> I have measured 2 high input impedance amps, off the top of my head one had an input capacitance of 0.5pf and resistance of 180,000,000Ω. The other had 1.1pf input capacitance and 200,000,000Ω input resistance. 
> However, those numbers are somewhat questionable, capacitance is probably good but I don’t yet know about the resistance. (all measurements were at 1MHz.) 
> I have beat this dead horse enough for today. 
> 
> Thanks for participating, Have a good night, Mikek

Good luck.

-- 

Rick C.

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I just finished building Ringdown Q meter, with a PCB designed and built by a follow in China.

https://www.dropbox.com/s/31iw8mm8ctwz1em/Q%20meter%20Ming%27s%201.jpg?dl=0

 I performed the same measurements of a 10MΩ 0805 resistor at 1MHz.
I ended up with pretty much the same result, measuring slightly over 8MΩ with a 10MΩ resistor.


                             Mikek
                                  

On Wednesday, November 16, 2022 at 1:12:00 PM UTC-5, Lamont Cranston wrote:
> I just finished building Ringdown Q meter, with a PCB designed and built by a follow in China. 
> 
> https://www.dropbox.com/s/31iw8mm8ctwz1em/Q%20meter%20Ming%27s%201.jpg?dl=0 
> 
> I performed the same measurements of a 10MΩ 0805 resistor at 1MHz. 
> I ended up with pretty much the same result, measuring slightly over 8MΩ with a 10MΩ resistor. 

Show your work. 

-- 

Rick C.

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++ Tesla referral code - https://ts.la/richard11209

On Wednesday, November 16, 2022 at 12:19:37 PM UTC-6, Ricky wrote:
> On Wednesday, November 16, 2022 at 1:12:00 PM UTC-5, Lamont Cranston wrote: 
> > I just finished building Ringdown Q meter, with a PCB designed and built by a follow in China. 
> > 
> > https://www.dropbox.com/s/31iw8mm8ctwz1em/Q%20meter%20Ming%27s%201.jpg?dl=0 
> > 
> > I performed the same measurements of a 10MΩ 0805 resistor at 1MHz. 
> > I ended up with pretty much the same result, measuring slightly over 8MΩ with a 10MΩ resistor.
> Show your work. 
> 
> -- 
> 
> Rick C. 

 Formula used: 
Rp = Q1 x Q2 / 2 x pi x capacitance x Delta Q
Q1 =1323
Q2= 1130 (loaded with 10MΩ)
C =151.1pf  used to resonate the inductor
Frequency 1MHz
Delta Q= 1323-1130 =193
Using the formula,
( 1323 x 1130 )   /   (6.28 X 10^6 x 151.1 X10^-12)
1,494,990 / 0.18435 = 8,109,160Ω

2nd Test:
 Q1 =1320
Q2= 1130 (loaded with 10MΩ)
C =151.1pf  used to resonate the inductor
Frequency 1MHz
 Delta Q= 1320 - 1130 =190
( 1320 x 1130 )   /   (6.28 X 10^6 x 151.1 X 10^-12 x 190)
1,491,600 / 0.18435 = 8,091,131Ω

"The manual under measuring 'Large Resistors' says,
"If the resistor is also Reactive,
 Xp = 1 / 2 x pi x F x (C1-C2) "  
(C1-C1 =0.12pf)

 Xp = 1,326,964Ω (for the capacitor, which I thought was tuned out,  but somehow, still affects the R?)
Rp = Q x Xc  (of tuning capacitor)  -----Xc = 1  / ( 2 x pi x 1MHz x 151.39) = 1019.4
Rp = 1019.4 x 1051 = 1,071,389 
Rp with parallel 10MΩ = 949,053

I don't know how to calculate imaginary numbers, but
 I don't see how the parallel 10MΩ and Xp will lower Rp to 949,053Ω
(I think much lower)

                                Mikek

On Wednesday, November 16, 2022 at 4:19:02 PM UTC-5, Lamont Cranston wrote:
> On Wednesday, November 16, 2022 at 12:19:37 PM UTC-6, Ricky wrote: 
> > On Wednesday, November 16, 2022 at 1:12:00 PM UTC-5, Lamont Cranston wrote: 
> > > I just finished building Ringdown Q meter, with a PCB designed and built by a follow in China. 
> > > 
> > > https://www.dropbox.com/s/31iw8mm8ctwz1em/Q%20meter%20Ming%27s%201.jpg?dl=0 
> > > 
> > > I performed the same measurements of a 10MΩ 0805 resistor at 1MHz. 
> > > I ended up with pretty much the same result, measuring slightly over 8MΩ with a 10MΩ resistor. 
> > Show your work. 
> > 
> > -- 
> > 
> > Rick C.
> Formula used: 
> Rp = Q1 x Q2 / 2 x pi x capacitance x Delta Q 
> Q1 =1323 
> Q2= 1130 (loaded with 10MΩ) 
> C =151.1pf used to resonate the inductor 
> Frequency 1MHz 
> Delta Q= 1323-1130 =193 
> Using the formula, 
> ( 1323 x 1130 ) / (6.28 X 10^6 x 151.1 X10^-12) 
> 1,494,990 / 0.18435 = 8,109,160Ω 
> 
> 2nd Test: 
> Q1 =1320 
> Q2= 1130 (loaded with 10MΩ) 
> C =151.1pf used to resonate the inductor 
> Frequency 1MHz 
> Delta Q= 1320 - 1130 =190 
> ( 1320 x 1130 ) / (6.28 X 10^6 x 151.1 X 10^-12 x 190) 
> 1,491,600 / 0.18435 = 8,091,131Ω 
> 
> "The manual under measuring 'Large Resistors' says, 
> "If the resistor is also Reactive, 
> Xp = 1 / 2 x pi x F x (C1-C2) " 
> (C1-C1 =0.12pf) 

I assume your equations came from Boonton?  


> Xp = 1,326,964Ω (for the capacitor, which I thought was tuned out, but somehow, still affects the R?) 
> Rp = Q x Xc (of tuning capacitor) -----Xc = 1 / ( 2 x pi x 1MHz x 151.39) = 1019.4 
> Rp = 1019.4 x 1051 = 1,071,389 
> Rp with parallel 10MΩ = 949,053 

I think you are measuring different things.  Rp in the above is the total combined resistive component of the circuit measured, the combined impact of all dissipative elements.  Is this what you are trying to measure?  This includes the loading resistor and the parasitic resistances in the reactive components. 

I've lost track of your goal.  


> I don't know how to calculate imaginary numbers, but 
> I don't see how the parallel 10MΩ and Xp will lower Rp to 949,053Ω 
> (I think much lower) 

The reactance portion uses imaginary numbers (a scalar multiplied by i, the square root of -1).  It's easier to picture this on an XY graph.  The real numbers (resistance) on the X axis, and the imaginary part on the Y axis (inductance is typically positive and capacitive is negative).  So combining resistance and reactance is a vector sum of measurements on the two orthogonal axes.  

It's actually pretty simple.  With a parallel circuit, the voltages on each component are the same, amplitude and phase.  The currents will be 90° out of phase, the inductor current is 90° ahead of the resistive current, and the capacitor current will be 90° behind the resistor.  When the circuit is in resonance, the capacitor and inductor currents are equal and opposite, so the reactive part looks like a high impedance, limited only by the parallel resistor.  

In a series circuit, the current is the same in all elements, but the voltages vary.  When the voltage on the capacitor and inductors are equal in magnitude (with opposite signs) the circuit is in resonance.  The opposite voltages on the cap and inductor will cancel, leaving only the series resistance. 

-- 

Rick C.

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