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

[Lamont Cranston]

On Wednesday, November 16, 2022 at 4:00:06 PM UTC-6, Ricky wrote:

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?

It\'s in the Boonton manual and the HP 4342A manual.

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 almost didn\'t put that in because I thought it confusing.
It takes a 8,310,000Ω in parallel with 1,071,389Ω to get 949,053Ω
What is in parallel with my 10MΩ to make it 8,310,000Ω?
If it is the 0.12pf self capacitance, I can\'t make the numbers work.

I\'ve lost track of your goal.
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.

Hope I explained that.


The question, why does a 10MΩ measure 8MΩ on the Q Meter?
The additional Xp formula in the manual, makes me think the I was wrong that resonating away the self capacitance,
made it of no consequence re: its Xc. Since they put the formula in the manual in the section about Large Resistors, I would think they would want
it included in a calculation of any resistor that is reactive. (All! a some frequency it matters!)
So, if we include the Xc of the self capacitance in parallel with the 10MΩ it will lower the measurement.
I think we have Rp of the original LC, the 10MΩ and the Xc of the self capacitance all in parallel. But, again I don\'t know how to calculate the imaginary number.

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.

Yes, but any math I have tried doesn\'t give me an answer that works for the self capacitance I measure.
See pg 22 second column for the formulas. http://hparchive.com/Boonton/Boonton-Manual-260A.pdf
Mikek

 

[Ricky]

On Wednesday, November 16, 2022 at 5:42:53 PM UTC-5, Lamont Cranston wrote:

On Wednesday, November 16, 2022 at 4:00:06 PM UTC-6, Ricky wrote:
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?
It\'s in the Boonton manual and the HP 4342A manual.
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 almost didn\'t put that in because I thought it confusing.
It takes a 8,310,000Ω in parallel with 1,071,389Ω to get 949,053Ω
What is in parallel with my 10MΩ to make it 8,310,000Ω?

Sorry, I don\'t know where these numbers are coming from. You have measurements of the unit with no added load capacitor. That gives some amount of dissipation which is equivalent to some value resistor. Which of these numbers is that?

Then you get a measurement of the same apparatus with a 10 Mohm load resistor added. What measurement is that?


> If it is the 0.12pf self capacitance, I can\'t make the numbers work.

The 0.12 parasitic capacitance is measured by retuning the rig, right? So that\'s just capacitance. It has nothing to do with the resistor and dissipative capacitance.

I\'ve lost track of your goal.
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.
Hope I explained that.


The question, why does a 10MΩ measure 8MΩ on the Q Meter?

What does the Q meter measure with no resistor?

The additional Xp formula in the manual, makes me think the I was wrong that resonating away the self capacitance,
made it of no consequence re: its Xc. Since they put the formula in the manual in the section about Large Resistors, I would think they would want
it included in a calculation of any resistor that is reactive. (All! a some frequency it matters!)
So, if we include the Xc of the self capacitance in parallel with the 10MΩ it will lower the measurement.

I don\'t think so. Adding a small capacitor across the resistor, does not materially affect your measurement, only the tuning, and even then only slightly. The value calculated is the total dissipative impact of all parasitic and load resistance.


> I think we have Rp of the original LC, the 10MΩ and the Xc of the self capacitance all in parallel. But, again I don\'t know how to calculate the imaginary number.

When you calculate Xl or Xc, you are calculating the scalar part of an imaginary number.

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.
Yes, but any math I have tried doesn\'t give me an answer that works for the self capacitance I measure.
See pg 22 second column for the formulas. http://hparchive.com/Boonton/Boonton-Manual-260A.pdf

I think you are doing the math wrong. We need to clarify the meaning of the terms like Rp.

--

Rick C.

--+ Get 1,000 miles of free Supercharging
--+ Tesla referral code - https://ts.la/richard11209

 

[Lamont Cranston]

On Wednesday, November 16, 2022 at 5:17:46 PM UTC-6, Ricky wrote:

On Wednesday, November 16, 2022 at 5:42:53 PM UTC-5, Lamont Cranston wrote:
On Wednesday, November 16, 2022 at 4:00:06 PM UTC-6, Ricky wrote:
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?
It\'s in the Boonton manual and the HP 4342A manual.
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 almost didn\'t put that in because I thought it confusing.
It takes a 8,310,000Ω in parallel with 1,071,389Ω to get 949,053Ω
What is in parallel with my 10MΩ to make it 8,310,000Ω?
Sorry, I don\'t know where these numbers are coming from. You have measurements of the unit with no added load capacitor. That gives some amount >of dissipation which is equivalent to some value resistor. Which of these numbers is that?
Then you get a measurement of the same apparatus with a 10 Mohm load resistor added. What measurement is that?

I\'m not sure if this is a good lead to chase, but,
Rp is the pure peak resistance that results at resonance. i.e. 1,071,389Ω
Then there is another Rp that results after the 10MΩ is installed. 949,053Ω
If you calculate, you find that 10MΩ parallel 1,071,389 is NOT 949,053Ω,
8,310,000Ω parallel 1,071,389Ω = 949,051Ω.
So there is something else involved that makes the 10MΩ look like 8,310,000.

If it is the 0.12pf self capacitance, I can\'t make the numbers work.
The 0.12 parasitic capacitance is measured by retuning the rig, right? So that\'s just capacitance. It has nothing to do with the resistor and dissipative capacitance.
I\'ve lost track of your goal.
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.
Hope I explained that.


The question, why does a 10MΩ measure 8MΩ on the Q Meter?
What does the Q meter measure with no resistor?
The additional Xp formula in the manual, makes me think the I was wrong that resonating away the self capacitance,
made it of no consequence re: its Xc. Since they put the formula in the manual in the section about Large Resistors, I would think they would want
it included in a calculation of any resistor that is reactive. (All! a some frequency it matters!)
So, if we include the Xc of the self capacitance in parallel with the 10MΩ it will lower the measurement.
I don\'t think so. Adding a small capacitor across the resistor, does not materially affect your measurement, only the tuning, and even then only slightly. The value calculated is the total dissipative impact of all parasitic and load resistance.
I think we have Rp of the original LC, the 10MΩ and the Xc of the self capacitance all in parallel. But, again I don\'t know how to calculate the imaginary number.
When you calculate Xl or Xc, you are calculating the scalar part of an imaginary number.
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.
Yes, but any math I have tried doesn\'t give me an answer that works for the self capacitance I measure.
See pg 22 second column for the formulas. http://hparchive.com/Boonton/Boonton-Manual-260A.pdf

I think you are doing the math wrong. We need to clarify the meaning of the terms like Rp.

I think we have Rp of the original LC, the 10MΩ and the Xc of the self capacitance all in parallel.
Rp, the resistance of the LC at resonance.
10MΩ is 10MΩ
Xc, at this point you have convinced it is gone which I thought it, before the manual wording had me thinking it had to be there.
So, you brought up Dissipative capacitance, that very well could be it, because 40MΩ in parallel with 10MΩ = 8MΩ.
However, I just made a fairly measure, 9,985,727Ω, I did that by using a lower Q inductor and a small capacitance. (outlined in the manual)
It was only one measurement and I have other things I need to do, I\'ll do more measurements tomorrow to verify.
Mikek

 

[John S]

On 11/16/2022 4:18 PM, 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

--> The \"Formula used\" and the \"Using the formula\" do not correlate.
Following the \"Formula used\" gives
Rp = 1323 x 1130 /(2 x pi x 151.1 x 10^-12 x 193) = 8.159 x 10^12
(Note the missing frequency)

When you applied the formula, you put in the frequency but not the delta
Q. If you include the delta Q in your calculation, your answer should be
8.109 x 10^6/193 = 42 x 10^3

(Although I\'m not saying your formula/s is/are correct).

What is the formula you are using?

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

--
Dogs make me happy. Humans make my head hurt.

 

[Ricky]

On Wednesday, November 16, 2022 at 9:04:19 PM UTC-5, Lamont Cranston wrote:

On Wednesday, November 16, 2022 at 5:17:46 PM UTC-6, Ricky wrote:
On Wednesday, November 16, 2022 at 5:42:53 PM UTC-5, Lamont Cranston wrote:
On Wednesday, November 16, 2022 at 4:00:06 PM UTC-6, Ricky wrote:
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?
It\'s in the Boonton manual and the HP 4342A manual.
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 almost didn\'t put that in because I thought it confusing.
It takes a 8,310,000Ω in parallel with 1,071,389Ω to get 949,053Ω
What is in parallel with my 10MΩ to make it 8,310,000Ω?
Sorry, I don\'t know where these numbers are coming from. You have measurements of the unit with no added load capacitor. That gives some amount >of dissipation which is equivalent to some value resistor. Which of these numbers is that?
Then you get a measurement of the same apparatus with a 10 Mohm load resistor added. What measurement is that?
I\'m not sure if this is a good lead to chase, but,
Rp is the pure peak resistance that results at resonance. i.e. 1,071,389Ω
Then there is another Rp that results after the 10MΩ is installed. 949,053Ω
If you calculate, you find that 10MΩ parallel 1,071,389 is NOT 949,053Ω,
8,310,000Ω parallel 1,071,389Ω = 949,051Ω.
So there is something else involved that makes the 10MΩ look like 8,310,000.

I\'ve never looked at the equations involved. When a dissipative element is added to the circuit, it can change the nature of the circuit. Without the 10 Mohm resistor there are (assuming you have a parallel circuit) you have equal and opposite currents in the C and L legs, with dissipation depending on their parasitic resistances. I would guess the parasitic resistance of the coil to be the higher of the two. When you add another resistor, the Q is lowered, meaning the currents in the other two legs are lower (from the lower voltage). So the effect of the parasitic resistance is lowered, impacting the Q less. However, this is opposite what you seem to be measuring, the impact on the Q being more than expected when the 10 Mohm resistor is added.

If it is the 0.12pf self capacitance, I can\'t make the numbers work.
The 0.12 parasitic capacitance is measured by retuning the rig, right? So that\'s just capacitance. It has nothing to do with the resistor and dissipative capacitance.
I\'ve lost track of your goal.
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.
Hope I explained that.


The question, why does a 10MΩ measure 8MΩ on the Q Meter?
What does the Q meter measure with no resistor?
The additional Xp formula in the manual, makes me think the I was wrong that resonating away the self capacitance,
made it of no consequence re: its Xc. Since they put the formula in the manual in the section about Large Resistors, I would think they would want
it included in a calculation of any resistor that is reactive. (All! a some frequency it matters!)
So, if we include the Xc of the self capacitance in parallel with the 10MΩ it will lower the measurement.
I don\'t think so. Adding a small capacitor across the resistor, does not materially affect your measurement, only the tuning, and even then only slightly. The value calculated is the total dissipative impact of all parasitic and load resistance.
I think we have Rp of the original LC, the 10MΩ and the Xc of the self capacitance all in parallel. But, again I don\'t know how to calculate the imaginary number.
When you calculate Xl or Xc, you are calculating the scalar part of an imaginary number.
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.
Yes, but any math I have tried doesn\'t give me an answer that works for the self capacitance I measure.
See pg 22 second column for the formulas. http://hparchive.com/Boonton/Boonton-Manual-260A.pdf



I think you are doing the math wrong. We need to clarify the meaning of the terms like Rp.


I think we have Rp of the original LC, the 10MΩ and the Xc of the self capacitance all in parallel.
Rp, the resistance of the LC at resonance.
10MΩ is 10MΩ
Xc, at this point you have convinced it is gone which I thought it, before the manual wording had me thinking it had to be there.

I don\'t know what you mean by this. Xc is the impedance of the resistor\'s parasitic resistance? I probably thought you were talking about the capacitor in the LC circuit. That\'s the impedance to be used in calculating Q. I believe you can use either Xc or Xl since they should have the same value at resonance. However, remember that the peak voltage will be different because of the parasitic resistances in the L and the C. But in a high Q circuit, this should be a very small deviation, so probably can be ignored.


> So, you brought up Dissipative capacitance, that very well could be it, because 40MΩ in parallel with 10MΩ = 8MΩ.

I don\'t think there is even 40 Mohm in the resistor\'s parasitic capacitance\'s parasitic resistance. It\'s essentially an air capacitor and has to amount to a very, very tiny impact. But, it is possible that the body of the resistor (which can be part of the dielectric) could be the dissipative influence on the resistor. If you really think this might be an impact, you can try using multiple resistors of lower value in series. The parasitic capacitance would divide, while the resistances add up, reducing the impact of the parasitic capacitance on your circuit.


> However, I just made a fairly measure, 9,985,727Ω, I did that by using a lower Q inductor and a small capacitance. (outlined in the manual)

This is not clear at all. Is this with the 10 Mohm resistor?


> It was only one measurement and I have other things I need to do, I\'ll do more measurements tomorrow to verify.

Ok

--

Rick C.

-+- Get 1,000 miles of free Supercharging
-+- Tesla referral code - https://ts.la/richard11209

 

[Lamont Cranston]

On Wednesday, November 16, 2022 at 10:14:41 PM UTC-6, John S wrote:

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
--> The \"Formula used\" and the \"Using the formula\" do not correlate.
Following the \"Formula used\" gives
Rp = 1323 x 1130 /(2 x pi x 151.1 x 10^-12 x 193) = 8.159 x 10^12
(Note the missing frequency)

When you applied the formula, you put in the frequency but not the delta
Q. If you include the delta Q in your calculation, your answer should be
8.109 x 10^6/193 = 42 x 10^3

(Although I\'m not saying your formula/s is/are correct).

What is the formula you are using?

Oh good catch, I did miss the Delta Q.
(But only in showing the formulas, I did use it when I calculated, but, I still made a math error, can\'t figure where , it made only a small difference)
The formula is used to find the resistance (Rp) when the parallel LC is at resonance.
Rp = Q1 x Q2 / 2 x pi x Freq x capacitance x Delta Q
Rp = 1323 x 1130 / 6.28 x 10^6 x 151.1 x 10^-12 x 193
Rp = 1494,990 / 0.183139
Rp = 8,163,133Ω

Here is the formula as written in both the HP4342A and the Boonton 260A Q meter manuals.
https://www.dropbox.com/s/d45sm1vs732hyef/Rp%20Formula.jpg?dl=0

Thanks for checking, Mikek

 

[Lamont Cranston]

On Wednesday, November 16, 2022 at 11:24:45 PM UTC-6, Ricky wrote:

So there is something else involved that makes the 10MΩ look like 8,310,000.

I\'ve never looked at the equations involved. When a dissipative element is added to the circuit, it can change the nature of the circuit. Without the 10 Mohm resistor there are (assuming you have a parallel circuit) you have equal and opposite currents in the C and L legs, with dissipation depending on their parasitic resistances. I would guess the parasitic resistance of the coil to be the higher of the two.

Yes, parallel circuit. Yes, even poor capacitors have less losses than good inductors. very good capacitor, Q <10,000, very good inductor Q < 1,400.
When Q is measured it is a reflection of the combined losses in the cap and inductor. As in Rloss= X/Q, although normally the capacitor losses are ignored and the formula is Rloss = XL / Q. I once read the Q of the capacitor in the 260A reached 20,000, but no data on freq or capacitance setting..

>When you add another resistor, the Q is lowered, meaning the currents in the other two legs are lower (from the lower voltage).

Any, voltage drop is very minimal as the voltage has a 0.02Ω source resistance. The HP4342A is even lower at 0.001Ω

>So the effect of the parasitic resistance is lowered, impacting the Q less..

See above.

>However, this is opposite what you seem to be measuring, the impact on the Q being more than expected when the 10 Mohm resistor is added.

Yes, additional losses in the 10MΩ resistor, but still trying to nail down what they are.
Ideally, I would have a large inductor that I could resonate at 25kHz with about 40pf.
This would eliminate a lot of the strays and go \'by the book\' with low capacitance used for measuring a large resistors.
But, that is a 1.1 Henry inductor and hard to get a reasonable Q.

If it is the 0.12pf self capacitance, I can\'t make the numbers work..
The 0.12 parasitic capacitance is measured by retuning the rig, right? So that\'s just capacitance. It has nothing to do with the resistor and dissipative capacitance.
I\'ve lost track of your goal.
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.
Hope I explained that.


The question, why does a 10MΩ measure 8MΩ on the Q Meter?
What does the Q meter measure with no resistor?
The additional Xp formula in the manual, makes me think the I was wrong that resonating away the self capacitance,
made it of no consequence re: its Xc. Since they put the formula in the manual in the section about Large Resistors, I would think they would want
it included in a calculation of any resistor that is reactive. (All! a some frequency it matters!)
So, if we include the Xc of the self capacitance in parallel with the 10MΩ it will lower the measurement.
I don\'t think so. Adding a small capacitor across the resistor, does not materially affect your measurement, only the tuning, and even then only slightly. The value calculated is the total dissipative impact of all parasitic and load resistance.
I think we have Rp of the original LC, the 10MΩ and the Xc of the self capacitance all in parallel. But, again I don\'t know how to calculate the imaginary number.
When you calculate Xl or Xc, you are calculating the scalar part of an imaginary number.
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.
Yes, but any math I have tried doesn\'t give me an answer that works for the self capacitance I measure.
See pg 22 second column for the formulas. http://hparchive.com/Boonton/Boonton-Manual-260A.pdf



I think you are doing the math wrong. We need to clarify the meaning of the terms like Rp.


I think we have Rp of the original LC, the 10MΩ and the Xc of the self capacitance all in parallel.
Rp, the resistance of the LC at resonance.
10MΩ is 10MΩ
Xc, at this point you have convinced it is gone which I thought it, before the manual wording had me thinking it had to be there.

I don\'t know what you mean by this. Xc is the impedance of the resistor\'s parasitic resistance?

Yes.

So, you brought up Dissipative capacitance, that very well could be it, because 40MΩ in parallel with 10MΩ = 8MΩ.
I don\'t think there is even 40 Mohm in the resistor\'s parasitic capacitance\'s parasitic resistance. It\'s essentially an air capacitor and has to amount to a very, very tiny impact. But, it is possible that the body of the resistor (which can be part of the dielectric) could be the dissipative influence on the resistor. If you really think this might be an impact, you can try using multiple resistors of lower value in series. The parasitic capacitance would divide, while the resistances add up, reducing the impact of the parasitic capacitance on your circuit.

Yes, I will try the Series resistors, early on I did try five 2MΩ in series but the measurement was worse, now that I have a bit more refined measurement method, it is worth trying again.
Thanks, Mikek

 

[Ricky]

On Thursday, November 17, 2022 at 8:18:26 AM UTC-5, Lamont Cranston wrote:

On Wednesday, November 16, 2022 at 11:24:45 PM UTC-6, Ricky wrote:

So there is something else involved that makes the 10MΩ look like 8,310,000.

I\'ve never looked at the equations involved. When a dissipative element is added to the circuit, it can change the nature of the circuit. Without the 10 Mohm resistor there are (assuming you have a parallel circuit) you have equal and opposite currents in the C and L legs, with dissipation depending on their parasitic resistances. I would guess the parasitic resistance of the coil to be the higher of the two.
Yes, parallel circuit. Yes, even poor capacitors have less losses than good inductors. very good capacitor, Q <10,000, very good inductor Q < 1,400.
When Q is measured it is a reflection of the combined losses in the cap and inductor. As in Rloss= X/Q, although normally the capacitor losses are ignored and the formula is Rloss = XL / Q. I once read the Q of the capacitor in the 260A reached 20,000, but no data on freq or capacitance setting.

Just to be clear, at resonance, XL = XC, so it doesn\'t matte which you use in the equation. X/Q is not a thing. It has to be XC or XL and since they are the same, it doesn\'t matter which. You can\'t separate the parasitic resistance of the cap and inductor (not without writing out the full equations for the currents and solving for resonance). As I\'ve said before, these parasitic values are different from your added resistance and cause the peak response to show at a slightly different frequency. But you can\'t use the boiler plate equations that were written ignoring them.

When you add another resistor, the Q is lowered, meaning the currents in the other two legs are lower (from the lower voltage).
Any, voltage drop is very minimal as the voltage has a 0.02Ω source resistance. The HP4342A is even lower at 0.001Ω

Then what are you measuring to see resonance? Are you measuring current using an inductive pickup? Where, at what point in the circuit? The resonance current from the source will be very low because at resonance the current in the L and C are equal and opposite. The impedance peaks.

So the effect of the parasitic resistance is lowered, impacting the Q less.
See above.

You need to share all the details.

However, this is opposite what you seem to be measuring, the impact on the Q being more than expected when the 10 Mohm resistor is added.
Yes, additional losses in the 10MΩ resistor, but still trying to nail down what they are.

I need to understand your circuit. I\'ve never used equipment like this and am trying to understand it from first principles.

Ideally, I would have a large inductor that I could resonate at 25kHz with about 40pf.
This would eliminate a lot of the strays and go \'by the book\' with low capacitance used for measuring a large resistors.
But, that is a 1.1 Henry inductor and hard to get a reasonable Q.
If it is the 0.12pf self capacitance, I can\'t make the numbers work.
The 0.12 parasitic capacitance is measured by retuning the rig, right? So that\'s just capacitance. It has nothing to do with the resistor and dissipative capacitance.
I\'ve lost track of your goal.
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.
Hope I explained that.


The question, why does a 10MΩ measure 8MΩ on the Q Meter?
What does the Q meter measure with no resistor?
The additional Xp formula in the manual, makes me think the I was wrong that resonating away the self capacitance,
made it of no consequence re: its Xc. Since they put the formula in the manual in the section about Large Resistors, I would think they would want
it included in a calculation of any resistor that is reactive. (All! a some frequency it matters!)
So, if we include the Xc of the self capacitance in parallel with the 10MΩ it will lower the measurement.
I don\'t think so. Adding a small capacitor across the resistor, does not materially affect your measurement, only the tuning, and even then only slightly. The value calculated is the total dissipative impact of all parasitic and load resistance.
I think we have Rp of the original LC, the 10MΩ and the Xc of the self capacitance all in parallel. But, again I don\'t know how to calculate the imaginary number.
When you calculate Xl or Xc, you are calculating the scalar part of an imaginary number.
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.
Yes, but any math I have tried doesn\'t give me an answer that works for the self capacitance I measure.
See pg 22 second column for the formulas. http://hparchive.com/Boonton/Boonton-Manual-260A.pdf



I think you are doing the math wrong. We need to clarify the meaning of the terms like Rp.


I think we have Rp of the original LC, the 10MΩ and the Xc of the self capacitance all in parallel.
Rp, the resistance of the LC at resonance.
10MΩ is 10MΩ
Xc, at this point you have convinced it is gone which I thought it, before the manual wording had me thinking it had to be there.

I don\'t know what you mean by this. Xc is the impedance of the resistor\'s parasitic resistance?
Yes.
So, you brought up Dissipative capacitance, that very well could be it, because 40MΩ in parallel with 10MΩ = 8MΩ.
I don\'t think there is even 40 Mohm in the resistor\'s parasitic capacitance\'s parasitic resistance. It\'s essentially an air capacitor and has to amount to a very, very tiny impact. But, it is possible that the body of the resistor (which can be part of the dielectric) could be the dissipative influence on the resistor. If you really think this might be an impact, you can try using multiple resistors of lower value in series. The parasitic capacitance would divide, while the resistances add up, reducing the impact of the parasitic capacitance on your circuit.
Yes, I will try the Series resistors, early on I did try five 2MΩ in series but the measurement was worse, now that I have a bit more refined measurement method, it is worth trying again.

Ok

--

Rick C.

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[Lamont Cranston]

On Thursday, November 17, 2022 at 9:55:59 AM UTC-6, Ricky wrote:

On Thursday, November 17, 2022 at 8:18:26 AM UTC-5, Lamont Cranston wrote:
On Wednesday, November 16, 2022 at 11:24:45 PM UTC-6, Ricky wrote:

So there is something else involved that makes the 10MΩ look like 8,310,000.

I\'ve never looked at the equations involved. When a dissipative element is added to the circuit, it can change the nature of the circuit. Without the 10 Mohm resistor there are (assuming you have a parallel circuit) you have equal and opposite currents in the C and L legs, with dissipation depending on their parasitic resistances. I would guess the parasitic resistance of the coil to be the higher of the two.
Yes, parallel circuit. Yes, even poor capacitors have less losses than good inductors. very good capacitor, Q <10,000, very good inductor Q < 1,400.
When Q is measured it is a reflection of the combined losses in the cap and inductor. As in Rloss= X/Q, although normally the capacitor losses are ignored and the formula is Rloss = XL / Q. I once read the Q of the capacitor in the 260A reached 20,000, but no data on freq or capacitance setting.

Just to be clear, at resonance, XL = XC, so it doesn\'t matte which you use in the equation. X/Q is not a thing. It has to be XC or XL and since they are the same, it doesn\'t matter which. You can\'t separate the parasitic resistance of the cap and inductor (not without writing out the full equations for the currents and solving for resonance). As I\'ve said before, these parasitic values are different from your added resistance and cause the peak response to show at a slightly different frequency. But you can\'t use the boiler plate equations that were written ignoring them.

Ya, I was trying to denote that the Rloss is in both Xc and Xl, but it didn\'t get by you .

I do know that a shift of Rloss in either Xc or Xl will slightly shift the frequency, but it\'s been a long time since I visited that.

When you add another resistor, the Q is lowered, meaning the currents in the other two legs are lower (from the lower voltage).
Any, voltage drop is very minimal as the voltage has a 0.02Ω source resistance. The HP4342A is even lower at 0.001Ω
Then what are you measuring to see resonance? Are you measuring current using an inductive pickup? Where, at what point in the circuit? The resonance current from the source will be very low because at resonance the current in the L and C are equal and opposite. The impedance peaks.
So the effect of the parasitic resistance is lowered, impacting the Q less.
See above.

You need to share all the details.

The details of the low source resistance?

However, this is opposite what you seem to be measuring, the impact on the Q being more than expected when the 10 Mohm resistor is added.
Yes, additional losses in the 10MΩ resistor, but still trying to nail down what they are.

I need to understand your circuit. I\'ve never used equipment like this and am trying to understand it from first principles.

Not sure what you need, but I\'ll list some items, first see the schematic of the source, L and its Rloss, the tuning cap and where the 10MΩ attaches.
https://www.dropbox.com/s/0wuezz8tlro03mw/Q%20meter%20HP%20injection%20transformer.jpg?dl=0
For first principles I don\'t expect you need values, so ignore if you want.. I\'m using 253uh inductor with a measured Q of 245, tuning cap was set to 40pf.
The voltage is adjustable, but adding the 10MΩ doesn\'t change the drive voltage of 20mV.
( note, previously I was using a high Q inductor, but changed to the Q Standard sold by Boonton and recommended for this resistor measurement.)
Mikek

 

[Lamont Cranston]

You may want to look at Pages 25, 26, and 27 for,
\"Correction of Errors and Formulas for Calculating Qs and Impedance Parameters of Parallel and Series Measurements\".
http://hparchive.com/Boonton/Boonton-Manual-260A.pdf]
Mikek

 

[Ricky]

On Thursday, November 17, 2022 at 11:35:34 AM UTC-5, Lamont Cranston wrote:

On Thursday, November 17, 2022 at 9:55:59 AM UTC-6, Ricky wrote:
On Thursday, November 17, 2022 at 8:18:26 AM UTC-5, Lamont Cranston wrote:
On Wednesday, November 16, 2022 at 11:24:45 PM UTC-6, Ricky wrote:

So there is something else involved that makes the 10MΩ look like 8,310,000.

I\'ve never looked at the equations involved. When a dissipative element is added to the circuit, it can change the nature of the circuit. Without the 10 Mohm resistor there are (assuming you have a parallel circuit) you have equal and opposite currents in the C and L legs, with dissipation depending on their parasitic resistances. I would guess the parasitic resistance of the coil to be the higher of the two.
Yes, parallel circuit. Yes, even poor capacitors have less losses than good inductors. very good capacitor, Q <10,000, very good inductor Q < 1,400.
When Q is measured it is a reflection of the combined losses in the cap and inductor. As in Rloss= X/Q, although normally the capacitor losses are ignored and the formula is Rloss = XL / Q. I once read the Q of the capacitor in the 260A reached 20,000, but no data on freq or capacitance setting.

Just to be clear, at resonance, XL = XC, so it doesn\'t matte which you use in the equation. X/Q is not a thing. It has to be XC or XL and since they are the same, it doesn\'t matter which. You can\'t separate the parasitic resistance of the cap and inductor (not without writing out the full equations for the currents and solving for resonance). As I\'ve said before, these parasitic values are different from your added resistance and cause the peak response to show at a slightly different frequency. But you can\'t use the boiler plate equations that were written ignoring them.
Ya, I was trying to denote that the Rloss is in both Xc and Xl, but it didn\'t get by you .

I do know that a shift of Rloss in either Xc or Xl will slightly shift the frequency, but it\'s been a long time since I visited that.
When you add another resistor, the Q is lowered, meaning the currents in the other two legs are lower (from the lower voltage).
Any, voltage drop is very minimal as the voltage has a 0.02Ω source resistance. The HP4342A is even lower at 0.001Ω
Then what are you measuring to see resonance? Are you measuring current using an inductive pickup? Where, at what point in the circuit? The resonance current from the source will be very low because at resonance the current in the L and C are equal and opposite. The impedance peaks.
So the effect of the parasitic resistance is lowered, impacting the Q less.
See above.

You need to share all the details.
The details of the low source resistance?
However, this is opposite what you seem to be measuring, the impact on the Q being more than expected when the 10 Mohm resistor is added.
Yes, additional losses in the 10MΩ resistor, but still trying to nail down what they are.

I need to understand your circuit. I\'ve never used equipment like this and am trying to understand it from first principles.
Not sure what you need, but I\'ll list some items, first see the schematic of the source, L and its Rloss, the tuning cap and where the 10MΩ attaches.
https://www.dropbox.com/s/0wuezz8tlro03mw/Q%20meter%20HP%20injection%20transformer.jpg?dl=0
For first principles I don\'t expect you need values, so ignore if you want. I\'m using 253uh inductor with a measured Q of 245, tuning cap was set to 40pf.
The voltage is adjustable, but adding the 10MΩ doesn\'t change the drive voltage of 20mV.
( note, previously I was using a high Q inductor, but changed to the Q Standard sold by Boonton and recommended for this resistor measurement.)

I\'m asking what you are measuring. Is this a measurement built into the Boonton? What exactly is it measuring? If the voltage is constant, the only thing left to measure is a current, but which current?

--

Rick C.

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[Lamont Cranston]

On Thursday, November 17, 2022 at 12:51:02 PM UTC-6, Ricky wrote:

> I\'m asking what you are measuring. Is this a measurement built into the Boonton? What exactly is it measuring? If the voltage is constant, the only thing >left to measure is a current, but which current?

It seems the more you ask the less I know.
All the measuring is built into a Q meter. It reads out Q directly on a meter. You can set two things, the drive voltage and the tuning capacitance.
After 3 hrs of reading, my understanding has changed. My understanding had been it the L and C were in parallel, now I see it is a series connection.
So tuning to resonance, as the reactance diminishes, the current rises, at resonance the current is limited by the series resistance* in the circuit, causing a higher voltage across the tuning capacitor, this voltage is measured and displayed as Q.

* it seems the series 0.02Ω drive source and the capacitors Rloss is ignored and just incorporated as losses in the inductor.
I find no reference to this in the error correction section.
There is an awful lot of correction that can be done.

My quote is, \"Q is Elusive\" I\'m getting some feeling for why.
Here\'s a graph of the same inductor on 3 different Q meters.
https://www.dropbox.com/s/au2sp79rgta1d80/Elusive%20Q.jpg?dl=0

I tried to copy and paste the section from the pdf, but it doesn\'t come out right.
See section 4-1 of the HP 4342A manual.
file:///C:/Users/Lamon/Downloads/hp-4342a_q-meter-op-sm.pdf
It describes: i=e/Rs x Xc and E/e = Xc/Rs = XL/Rs = Q. Where i is current, Rs is the series resistance of the inductor, e is the excitation voltage, E is the voltage across the tuning capacitor.

The Socratic Method is working.

Thanks, Mikek

 

[Ricky]

On Thursday, November 17, 2022 at 3:31:00 PM UTC-5, Lamont Cranston wrote:

On Thursday, November 17, 2022 at 12:51:02 PM UTC-6, Ricky wrote:

I\'m asking what you are measuring. Is this a measurement built into the Boonton? What exactly is it measuring? If the voltage is constant, the only thing >left to measure is a current, but which current?
It seems the more you ask the less I know.
All the measuring is built into a Q meter. It reads out Q directly on a meter. You can set two things, the drive voltage and the tuning capacitance.
After 3 hrs of reading, my understanding has changed. My understanding had been it the L and C were in parallel, now I see it is a series connection..
So tuning to resonance, as the reactance diminishes, the current rises, at resonance the current is limited by the series resistance* in the circuit, causing a higher voltage across the tuning capacitor, this voltage is measured and displayed as Q.

* it seems the series 0.02Ω drive source and the capacitors Rloss is ignored and just incorporated as losses in the inductor.
I find no reference to this in the error correction section.
There is an awful lot of correction that can be done.

My quote is, \"Q is Elusive\" I\'m getting some feeling for why.
Here\'s a graph of the same inductor on 3 different Q meters.
https://www.dropbox.com/s/au2sp79rgta1d80/Elusive%20Q.jpg?dl=0

I tried to copy and paste the section from the pdf, but it doesn\'t come out right.
See section 4-1 of the HP 4342A manual.
file:///C:/Users/Lamon/Downloads/hp-4342a_q-meter-op-sm.pdf
It describes: i=e/Rs x Xc and E/e = Xc/Rs = XL/Rs = Q. Where i is current, Rs is the series resistance of the inductor, e is the excitation voltage, E is the voltage across the tuning capacitor.

The Socratic Method is working.

So if the resonance is series, as you say, the voltage goes toward zero and the current gets large. They must be measuring this current. If you place the 10 Mohm resistor across the series L and C, it is an entirely different analysis than in the fully parallel arrangement. In fact, with the voltage going toward zero, the added parallel resistor will have almost no current and have very little impact on the measurement.

Sorry, but I\'m very lost at this point. With a very low impedance power source, and a very large parallel resistor, I don\'t see how you can be getting the results you claim.

Wait, you said the tuning capacitor is inside the unit, but you are attaching a coil? So the resistor is only across the coil? That\'s not the same thing and the resistor will impact the coil Q factor. Not sure how to write the equation, since the coil parasitic resistor is in series with the pure inductance and your added resistor is in parallel. The parasitic resistor capacitance is different as well, since it in parallel with the coil.

In the application circuits, will you be using a parallel or a series tuning method?

--

Rick C.

+-+ Get 1,000 miles of free Supercharging
+-+ Tesla referral code - https://ts.la/richard11209

 

[Lamont Cranston]

> So if the resonance is series, as you say, the voltage goes toward zero and the current gets large.

What voltage? I don\'t believe the drive voltage changes, in fact they have a level control circuit to keep it constant in the HP4342A.
( I think they put it in the wrong place, but they now more than me)

>They must be measuring this current.
Indirectly, they measure the voltage drop across the tuning capacitor.
You need to look at the schematic, it consists of a 0.02Ω resistor in series with the inductor under test (plus stray series resistances) in series with the tuning capacitor. The Drive Oscillator is across (in parallel with) the 0.02Ω resistor.

>If you place the 10 Mohm resistor across the series L and C, it is an entirely different analysis than in the fully parallel arrangement.

The 10MΩ is only across the tuning Capacitor.

> In fact, with the voltage going toward zero, the added parallel resistor will have almost no current and have very little impact on the measurement.

The drive voltage is constant, it does not go to toward zero.

Sorry, but I\'m very lost at this point. With a very low impedance power source, and a very large parallel resistor, I don\'t see how you can be getting the >results you claim.

I\'m not the only one, I have at least one on another group that has done several tests and has the same low readings. He has the newer HP4342A.

Wait, you said the tuning capacitor is inside the unit,

Yes.

but you are attaching a coil?

Yes.

So the resistor is only across the coil?

Nope, it is across the capacitor.

That\'s not the same thing and the resistor will impact the coil Q factor. Not sure how to write the equation, since the coil parasitic resistor is in series with the pure inductance and your added resistor is in parallel. The parasitic resistor capacitance is different as well, since it in parallel with the coil.

It is across the capacitor.

In the application circuits, will you be using a parallel or a series tuning method?

It is a series tuned circuit, But, the 10MΩ is called a parallel connection to the measuring circuit.
Please look at the dropbox picture, The parts in red, I added, to show where the inductor and resistor are placed.
https://www.dropbox.com/s/bjcjzdxtyy63jan/HP%20Injection%20circuit%20with%2010M%CE%A9.jpg?dl=0

Mikek

 

[Ricky]

On Thursday, November 17, 2022 at 4:56:02 PM UTC-5, Lamont Cranston wrote:

So if the resonance is series, as you say, the voltage goes toward zero and the current gets large.
What voltage? I don\'t believe the drive voltage changes, in fact they have a level control circuit to keep it constant in the HP4342A.
( I think they put it in the wrong place, but they now more than me)

I don\'t know what your drive circuit is doing, so I\'ll talk in terms of the impedance. At resonance, the voltages of the L and C are equal and opposite (both 90° out of phase from the drive voltage, one positive and the other negative phase). So the impedance will appear to approach zero. If your drive voltage is truly like a constant voltage, the current will ramp up hugely. Do you know what the drive voltage it?

They must be measuring this current.
Indirectly, they measure the voltage drop across the tuning capacitor.

Ok, that makes some sense. The current will be max, so the voltage on the capacitor and inductor will be max, just not if you add them together. It\'s hard to imagine a constant voltage drive given the nature of the circuit, but if it is low enough, that\'s fine. Remember, the drive voltage will appear across the cap, multiplied by the Q. So 1VAC drive will show up as 1,000VAC if the Q is 1,000.


> You need to look at the schematic, it consists of a 0.02Ω resistor in series with the inductor under test (plus stray series resistances) in series with the tuning capacitor. The Drive Oscillator is across (in parallel with) the 0.02Ω resistor.

??? Is the cap connected to the inductor and this 0.02 ohm resistor, making it a three element loop? Does the drive oscillator have feedback to keep the voltage constant?

If you place the 10 Mohm resistor across the series L and C, it is an entirely different analysis than in the fully parallel arrangement.
The 10MΩ is only across the tuning Capacitor.

Yeah, this makes the equations a bit more complicated. It provides more current to the coil than the capacitor sees. It distorts the phase angles.

In fact, with the voltage going toward zero, the added parallel resistor will have almost no current and have very little impact on the measurement.
The drive voltage is constant, it does not go to toward zero.

Sorry, but I\'m very lost at this point. With a very low impedance power source, and a very large parallel resistor, I don\'t see how you can be getting the >results you claim.
I\'m not the only one, I have at least one on another group that has done several tests and has the same low readings. He has the newer HP4342A.

Now that I realize the nature of the full circuit, the resistor will show up in the circuit equations in a way, that will modify the phase angle and alter the frequency of peak response, so that it is no longer at the resonance of the LC. This is not because of a change in the C, I don\'t think. You would have to write out the equations which is easiest using the Laplace transform. I\'ve done it before using simple algebra, but the equations get messy, making it not so \"simple\".

Essentially, what happens is the voltage on the cap and inductor are no longer 180 out of phase with each other, because each one sees a different drive current.

Wait, you said the tuning capacitor is inside the unit,
Yes.
but you are attaching a coil?
Yes.
So the resistor is only across the coil?
Nope, it is across the capacitor.

Ok, so they give you points to add the resistor, got it!

That\'s not the same thing and the resistor will impact the coil Q factor. Not sure how to write the equation, since the coil parasitic resistor is in series with the pure inductance and your added resistor is in parallel. The parasitic resistor capacitance is different as well, since it in parallel with the coil.
It is across the capacitor.

It should be pretty negligible anyway.

In the application circuits, will you be using a parallel or a series tuning method?
It is a series tuned circuit, But, the 10MΩ is called a parallel connection to the measuring circuit.
Please look at the dropbox picture, The parts in red, I added, to show where the inductor and resistor are placed.
https://www.dropbox.com/s/bjcjzdxtyy63jan/HP%20Injection%20circuit%20with%2010M%CE%A9.jpg?dl=0

I\'m not asking about the Boonton, I\'m asking about where you will use the coil. I\'m wondering how similar to your application this test circuit is.

--

Rick C.

++- Get 1,000 miles of free Supercharging
++- Tesla referral code - https://ts.la/richard11209

 

[Lamont Cranston]

On Thursday, November 17, 2022 at 6:42:12 PM UTC-6, Ricky wrote:

On Thursday, November 17, 2022 at 4:56:02 PM UTC-5, Lamont Cranston wrote:
So if the resonance is series, as you say, the voltage goes toward zero and the current gets large.
What voltage? I don\'t believe the drive voltage changes, in fact they have a level control circuit to keep it constant in the HP4342A.
( I think they put it in the wrong place, but they know more than me)

I don\'t know what your drive circuit is doing,

As I said, it is a constant voltage a 0.02Ω source is hard to swing.
The voltage can be adjusted, on the 260A the meter is marked with a max 250Q, then another meter monitors the drive voltage. You can set the voltage at 1X and the drive voltage is 20mV, set at 2 and the drive voltage is 10mV and you multiply the meter reading by 2. I externally monitor the drive voltage and set drive voltage at 3.33mV as 6X multiplier and measure Qs up to 1500.

>so I\'ll talk in terms of the impedance. At resonance, the voltages of the L and C are equal and opposite (both >90° out of phase from the drive voltage, >one positive and the other negative phase). So the impedance will appear to approach zero. If your drive voltage >is truly like a constant voltage, the >current will ramp up hugely.

YES! That is what happens.

Do you know what the drive voltage it?
I explained that above, the max on the 260A is 20mV. I\'m not sure on the 4342A.

They must be measuring this current.
Indirectly, they measure the voltage drop across the tuning capacitor.
Ok, that makes some sense. The current will be max, so the voltage on the capacitor and inductor will be max, just not if you add them together. It\'s hard to imagine a constant voltage drive given the nature of the circuit, but if it is low enough, that\'s fine. Remember, the drive voltage will appear across the cap, multiplied by the Q. So 1VAC drive will show up as 1,000VAC if the Q is 1,000.

Yes, on the 260A to read a Q of 1000, the drive voltage will need to be reduced to 5mv.

You need to look at the schematic, it consists of a 0.02Ω resistor in series with the inductor under test (plus stray series resistances) in series with the tuning capacitor. The Drive Oscillator is across (in parallel with) the 0.02Ω resistor.

??? Is the cap connected to the inductor and this 0.02 ohm resistor, making it a three element loop?

Yes. I know, it looks strange.

>Does the drive oscillator have feedback to keep the voltage constant?

The 260 has a knob where the human adjusts the drive voltage to keep it at the at the correct level.
The 4342A has an ALC that monitors the input the the transformer. ( seems like it should monitor the output!)

If you place the 10 Mohm resistor across the series L and C, it is an entirely different analysis than in the fully parallel arrangement.
The 10MΩ is only across the tuning Capacitor.
Yeah, this makes the equations a bit more complicated. It provides more current to the coil than the capacitor sees. It distorts the phase angles.
In fact, with the voltage going toward zero, the added parallel resistor will have almost no current and have very little impact on the measurement.
The drive voltage is constant, it does not go to toward zero.

Sorry, but I\'m very lost at this point. With a very low impedance power source, and a very large parallel resistor, I don\'t see how you can be getting the >results you claim.
I\'m not the only one, I have at least one on another group that has done several tests and has the same low readings. He has the newer HP4342A.

Now that I realize the nature of the full circuit, the resistor will show up in the circuit equations in a way, that will modify the phase angle and alter the >frequency of peak response, so that it is no longer at the resonance of the LC. This is not because of a change in the C, I don\'t think. You would have to >write out the equations which is easiest using the Laplace transform. I\'ve done it before using simple algebra, but the equations get messy, making it not >so \"simple\".

Essentially, what happens is the voltage on the cap and inductor are no longer 180 out of phase with each other, because each one sees a different drive current.

Wait, you said the tuning capacitor is inside the unit,
Yes.
but you are attaching a coil?
Yes.
So the resistor is only across the coil?
Nope, it is across the capacitor.
Ok, so they give you points to add the resistor, got it!
That\'s not the same thing and the resistor will impact the coil Q factor. Not sure how to write the equation, since the coil parasitic resistor is in series with the pure inductance and your added resistor is in parallel. The parasitic resistor capacitance is different as well, since it in parallel with the coil.
It is across the capacitor.
It should be pretty negligible anyway.
In the application circuits, will you be using a parallel or a series tuning method?
It is a series tuned circuit, But, the 10MΩ is called a parallel connection to the measuring circuit.
Please look at the dropbox picture, The parts in red, I added, to show where the inductor and resistor are placed.
https://www.dropbox.com/s/bjcjzdxtyy63jan/HP%20Injection%20circuit%20with%2010M%CE%A9.jpg?dl=0

I\'m not asking about the Boonton,

That is actually the 4342A.

>I\'m asking about where you will use the coil.

The coil I\'m using is called a Q standard or work coil.
From the Radio Museum, \"
The Q-Standard Type 513-A is a shielded reference inductor which has accurately-measured and highly-stable inductance and Q characteristics. Specifically designed for use with Q-Meters Type 260-A and 160-A, the Q-Standard is particularly useful as a check on the overall operation and accuracy of these instruments, as well as for providing precisely-known supplementary Q-circuit inductance desirable for many impedance measurements by the parallel method. (Mikek here, as in measuring resistors)
The Q-Standard consists of a specially-designed, high-Q coil of Litz wire, wound on a low-loss Steatite form. The coil is hermetically sealed inside a copper shield can which is filled with inert gas under pressure. The desired Q-versus-frequency characteristics are provided by a carbon film resistor shunted across the coil. Two replaceable banana plug connectors mounted on the base serve to connect the unit to the Q-meter circuit.
The Q-Standard is supplied in a convenient wooden carrying and storage case.. Each unit is individually calibrated and marked with its true inductance (L), distributed capacity (Cd), effective Qe and indicated Qi at 0.5, 1.0 and 1.5 Mc, respectively. Tolerance: L ±1%, Cd ±2%, Qe ±3%, measured at 73°F. Any instrument deviating more than ± 7% from the marked value is not operating in accordance with original specifications.


I\'m wondering how similar to your application this test circuit is.

I\'m using it as the work coil, it is part of the LC to do other measurements, like a resistor.
I don\'t have an application for the coil, it is a tool for checking calibration and to be used
as a work coil for other parallel and series external component measurements.
You need to add the work coil to setup a resonant LC to measure resistors, capacitors,
and you can even do dielectric constant measurements. This requires a fixture.

The main purpose of a Q meter is to measure Q of inductors, however, it has found other uses.
Thanks, Mikek

 

[Anthony Stewart]

If you want to compare with a lower capacitance resistor, this depends on the surface area so 1/8th watt ought to be 0.03 pF or so. But I don\'t know the purpose of 10Mohm when that would exceed the Q=242 @ 1MHz of the Reference L. the tuning cap is almost 100 pF for the reference L = 249 uH // 7.9 pF

Practical Q\'s of 1500 are getting in the perfect helix types. Practical limits are 100 for decent sensitivity and are easiest to self-clock in Colpitts or Pierce oscillators..

This Falstad tool may assist in eyeball measurements for Q up to 100 with 40 dB gain. Or customer filter design.
There isn\'t enough resolution on the display to go sharper. But I tuned yours with a Q of 10k or 80 dB gain without a resistor.It was closer to 99 pF.

https://tinyurl.com/mrwm6uxr

 

[Ricky]

On Thursday, November 17, 2022 at 8:44:17 PM UTC-5, Lamont Cranston wrote:

On Thursday, November 17, 2022 at 6:42:12 PM UTC-6, Ricky wrote:
On Thursday, November 17, 2022 at 4:56:02 PM UTC-5, Lamont Cranston wrote:
So if the resonance is series, as you say, the voltage goes toward zero and the current gets large.
What voltage? I don\'t believe the drive voltage changes, in fact they have a level control circuit to keep it constant in the HP4342A.
( I think they put it in the wrong place, but they know more than me)
I don\'t know what your drive circuit is doing,
As I said, it is a constant voltage a 0.02Ω source is hard to swing.
The voltage can be adjusted, on the 260A the meter is marked with a max 250Q, then another meter monitors the drive voltage. You can set the voltage at 1X and the drive voltage is 20mV, set at 2 and the drive voltage is 10mV and you multiply the meter reading by 2. I externally monitor the drive voltage and set drive voltage at 3.33mV as 6X multiplier and measure Qs up to 1500.
so I\'ll talk in terms of the impedance. At resonance, the voltages of the L and C are equal and opposite (both >90° out of phase from the drive voltage, >one positive and the other negative phase). So the impedance will appear to approach zero. If your drive voltage >is truly like a constant voltage, the >current will ramp up hugely.
YES! That is what happens.
Do you know what the drive voltage it?
I explained that above, the max on the 260A is 20mV. I\'m not sure on the 4342A.

This is all much more clear now. It was a bit of work to get out of the parallel resonance mindset. I think you explained this gear a couple of years (or a few) ago when you were looking to improve its sensitivity or something.

They must be measuring this current.
Indirectly, they measure the voltage drop across the tuning capacitor..
Ok, that makes some sense. The current will be max, so the voltage on the capacitor and inductor will be max, just not if you add them together. It\'s hard to imagine a constant voltage drive given the nature of the circuit, but if it is low enough, that\'s fine. Remember, the drive voltage will appear across the cap, multiplied by the Q. So 1VAC drive will show up as 1,000VAC if the Q is 1,000.
Yes, on the 260A to read a Q of 1000, the drive voltage will need to be reduced to 5mv.
You need to look at the schematic, it consists of a 0.02Ω resistor in series with the inductor under test (plus stray series resistances) in series with the tuning capacitor. The Drive Oscillator is across (in parallel with) the 0.02Ω resistor.

??? Is the cap connected to the inductor and this 0.02 ohm resistor, making it a three element loop?
Yes. I know, it looks strange.
Does the drive oscillator have feedback to keep the voltage constant?
The 260 has a knob where the human adjusts the drive voltage to keep it at the at the correct level.
The 4342A has an ALC that monitors the input the the transformer. ( seems like it should monitor the output!)

I don\'t know what an ALC is.

If you place the 10 Mohm resistor across the series L and C, it is an entirely different analysis than in the fully parallel arrangement.
The 10MΩ is only across the tuning Capacitor.
Yeah, this makes the equations a bit more complicated. It provides more current to the coil than the capacitor sees. It distorts the phase angles.
In fact, with the voltage going toward zero, the added parallel resistor will have almost no current and have very little impact on the measurement.
The drive voltage is constant, it does not go to toward zero.

Sorry, but I\'m very lost at this point. With a very low impedance power source, and a very large parallel resistor, I don\'t see how you can be getting the >results you claim.
I\'m not the only one, I have at least one on another group that has done several tests and has the same low readings. He has the newer HP4342A.

Now that I realize the nature of the full circuit, the resistor will show up in the circuit equations in a way, that will modify the phase angle and alter the >frequency of peak response, so that it is no longer at the resonance of the LC. This is not because of a change in the C, I don\'t think. You would have to >write out the equations which is easiest using the Laplace transform. I\'ve done it before using simple algebra, but the equations get messy, making it not >so \"simple\".

Essentially, what happens is the voltage on the cap and inductor are no longer 180 out of phase with each other, because each one sees a different drive current.

Wait, you said the tuning capacitor is inside the unit,
Yes.
but you are attaching a coil?
Yes.
So the resistor is only across the coil?
Nope, it is across the capacitor.
Ok, so they give you points to add the resistor, got it!
That\'s not the same thing and the resistor will impact the coil Q factor. Not sure how to write the equation, since the coil parasitic resistor is in series with the pure inductance and your added resistor is in parallel. The parasitic resistor capacitance is different as well, since it in parallel with the coil.
It is across the capacitor.
It should be pretty negligible anyway.
In the application circuits, will you be using a parallel or a series tuning method?
It is a series tuned circuit, But, the 10MΩ is called a parallel connection to the measuring circuit.
Please look at the dropbox picture, The parts in red, I added, to show where the inductor and resistor are placed.
https://www.dropbox.com/s/bjcjzdxtyy63jan/HP%20Injection%20circuit%20with%2010M%CE%A9.jpg?dl=0

I\'m not asking about the Boonton,
That is actually the 4342A.
I\'m asking about where you will use the coil.
The coil I\'m using is called a Q standard or work coil.
From the Radio Museum, \"
The Q-Standard Type 513-A is a shielded reference inductor which has accurately-measured and highly-stable inductance and Q characteristics. Specifically designed for use with Q-Meters Type 260-A and 160-A, the Q-Standard is particularly useful as a check on the overall operation and accuracy of these instruments, as well as for providing precisely-known supplementary Q-circuit inductance desirable for many impedance measurements by the parallel method. (Mikek here, as in measuring resistors)
The Q-Standard consists of a specially-designed, high-Q coil of Litz wire, wound on a low-loss Steatite form. The coil is hermetically sealed inside a copper shield can which is filled with inert gas under pressure. The desired Q-versus-frequency characteristics are provided by a carbon film resistor shunted across the coil. Two replaceable banana plug connectors mounted on the base serve to connect the unit to the Q-meter circuit.
The Q-Standard is supplied in a convenient wooden carrying and storage case. Each unit is individually calibrated and marked with its true inductance (L), distributed capacity (Cd), effective Qe and indicated Qi at 0.5, 1.0 and 1.5 Mc, respectively. Tolerance: L ±1%, Cd ±2%, Qe ±3%, measured at 73°F. Any instrument deviating more than ± 7% from the marked value is not operating in accordance with original specifications..
I\'m wondering how similar to your application this test circuit is.
I\'m using it as the work coil, it is part of the LC to do other measurements, like a resistor.
I don\'t have an application for the coil, it is a tool for checking calibration and to be used
as a work coil for other parallel and series external component measurements.
You need to add the work coil to setup a resonant LC to measure resistors, capacitors,
and you can even do dielectric constant measurements. This requires a fixture.

The main purpose of a Q meter is to measure Q of inductors, however, it has found other uses.

You are using it as an ohm meter???

--

Rick C.

+++ Get 1,000 miles of free Supercharging
+++ Tesla referral code - https://ts.la/richard11209

 

[Lamont Cranston]

On Thursday, November 17, 2022 at 10:10:19 PM UTC-6, Ricky wrote:

> I don\'t know what an ALC is.

Automatic Level Controller
Keeps the voltage into the transformer at a constant voltage.

The main purpose of a Q meter is to measure Q of inductors, however, it has found other uses.
You are using it as an ohm meter???

Only because it was mentioned on another group, I tried it and had a big error,
I want to know why.

Mikek

 

[Lamont Cranston]

On Thursday, November 17, 2022 at 10:52:42 PM UTC-6, Lamont Cranston wrote:

On Thursday, November 17, 2022 at 10:10:19 PM UTC-6, Ricky wrote:

I don\'t know what an ALC is.
Automatic Level Controller
Keeps the voltage into the transformer at a constant voltage.
The main purpose of a Q meter is to measure Q of inductors, however, it has found other uses.
You are using it as an ohm meter???
Only because it was mentioned on another group, I tried it and had a big error,
I want to know why.

Mikek

I just found a reference to skin effect in resistors.
https://resources.system-analysis.cadence.com/blog/msa2021-understanding-resistor-behavior-at-high-frequencies
I wonder if that could explain all of the error I\'m seeing?
I need figure out a way to do the test at low frequency and still have reasonable Delta Q.
Mikek