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 = 949,053I 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
Measuring the Disspation of the self capacitance of a thru hole resistor.
Started by ●November 11, 2022
Reply by ●November 16, 20222022-11-16
Reply by ●November 16, 20222022-11-16
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.pdfI 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
Reply by ●November 16, 20222022-11-16
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
Reply by ●November 17, 20222022-11-17
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.
Reply by ●November 17, 20222022-11-17
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
Reply by ●November 17, 20222022-11-17
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
Reply by ●November 17, 20222022-11-17
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
Reply by ●November 17, 20222022-11-17
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 0.001ΩThen 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. -++ Get 1,000 miles of free Supercharging -++ Tesla referral code - https://ts.la/richard11209
Reply by ●November 17, 20222022-11-17
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
Reply by ●November 17, 20222022-11-17
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
