RL circuit practice

[Ratchit]

the correct calculation to factor out DCR was not considered , explained until my posts and thus low values of load R negatively affect error, significantly.

Wrong. Of course it was. I said that if the DV across the resistor and the DV across the coil were equal, then the magnitudes of each of the impedances are equal. It is obvious from then on that it is a trivial matter to find the inductance from the frequency and DC resistance. No further explanation is necessary to anyone who knows what impedance is. Your expanded explanation, while correct but unnecessary, came later. By the way, the DC resistance is subtracted out, not factored out in the solution.

If you wish to be hopelessly pedantic, then be explicit about the sources of error
; scrupulous, precise, exact, perfectionist, punctilious, meticulous, fussy, fastidious, finicky;

Before I can explain the error, there has to be an error. I know of no error in the DV method I proposed.

Your advice to equate voltage on each part was good as a one method for a solution.

Thank you.

But it lacked the reasons for details on source impedance, and more importantly DCR was essential for this method to work.

The details I did give were sufficient for anyone familiar with AC circuits to solve.

Whereas if one discovers all the sources of R and included in the geometric equation, it can be done at any practical frequency or voltage ratio. Using equal DM voltage allows you to ignore the source impedance. Anyway you look at it, the math I pointed out is essential and was overlooked.

All the OP wanted to do was to find out the inductance of the coil. I proposed a good method to do that in post #29. You went further and displayed the calculations, but it was not really necessary to go into all that detail.

Ratch

 

[Tony Stewart]

Ratchit, you did not correct the error, which started in post #10 when Elrion stated after showing correct DM readings
I'm using a pure LC, no resistor.

Then you added to the error in analysis by stating in post #11
If they add up correctly, then you have swamped out the source resistance and the reactance should be equal to the test resistor.

not true.

... The resistance comes from capacitor leakage and wire resistance of the coil.

not true.

Also latter is only SRF resonance well above 1MHz for this part and has nothing to do with wire resistance.

I could say more, but at this point it seems pointless to convince you to acknowledge the correct method is not what you stated.

 

[Ratchit]

Ratchit, you did not correct the error, which started in post #10 when Elrion stated after showing correct DM readings
I'm using a pure LC, no resistor.

I certainly did. In post #11 I stated that "no LC circuit that is completely free of resistance."

Then you added to the error in analysis by stating in post #11
If they add up correctly, then you have swamped out the source resistance and the reactance should be equal to the test resistor.

not true.

Yes, I did carelessly miswrite reactance when I knew it to be impedance. That same basic statement was correct in post #29, however.

... The resistance comes from capacitor leakage and wire resistance of the coil.

not true.

I never said that all the resistance comes from the two sources I named. But it was enough to show that no LC circuit is completely resistance free.

Also latter is only SRF resonance well above 1MHz for this part and has nothing to do with wire resistance.

Correct. Wire resistance is whatever it is. I never said that SRF has anything to do with coil resistance. Where do you find that I did?

I could say more, but at this point it seems pointless to convince you to acknowledge the correct method is not what you stated.

Please do. You have not given any reason why the method I stated would not give correct results.

Ratch

 

[Tony Stewart]

Because your math is missing.

In any case if the user has a generator, the preferred more accurate method uses a known accurate Cap to resonate and compute L from this independent of any loss resistors in source choke or load.

I modeled a TDK part exactly which is also the general model.
https://en.tdk.eu/inf/30/db/ind_2008/b78108_148s.pdf


Note LowFreq loss is shown in graph as 10Hz=-7.6dB instead of -6dB because of DCR = 20 which accounts for additional 1.6dB loss
Note High Freq loss is shown well above resonance with ideal 50 Ohm resistors and measurement methods.
Thus Simulation shows -6dB as leakage capacitance which creates the SRF of 2MHz on its own has no effect with 1nF added to the part.

 

[Ratchit]

Because your math is missing.

Detailed calculation not needed to be shown. Written description suffices for such a simple setup. The reason you gave is wrong.

In any case if the user has a generator, the preferred more accurate method uses a known accurate Cap to resonate and compute L from this independent of any loss resistors in source choke or load.

The OP probably does not have access to a standard capacitor or a cap tester that can be traced back to a standard. Also his generator does not have the range you specified. The circuit you specified is too elaborate and expensive for most hobbyists.

I modeled a TDK part exactly which is also the general model.
https://en.tdk.eu/inf/30/db/ind_2008/b78108_148s.pdf

Note LowFreq loss is shown in graph as 10Hz=-7.6dB instead of -6dB because of DCR = 20 which accounts for additional 1.6dB loss
Note High Freq loss is shown well above resonance with ideal 50 Ohm resistors and measurement methods.
Thus Simulation shows -6dB as leakage capacitance which creates the SRF of 2MHz on its own has no effect with 1nF added to the part.
View attachment 96401

A real setup like that will not be cheap.

Ratch

 

[Tony Stewart]

I am still waiting for you to explain his errors or yours rather than deflect the issues.

I can prove that it doesn't take much cost or practical thinking to make this cheap , even free using a PC audio generator like Audacity (free) to generate the signal and record the response in a scope waveform or spectral FFT view ( all free software) using the audio ports.

But I dont need to do that as I have already shown the root cause of original errors with sufficient proof and examples.

 

[Ratchit]

I am still waiting for you to explain his errors or yours rather than deflect the issues.

Until OP confirms the DC resistance of the coil, I have no way to explain anything. What deflection are you talking about?

Edit: You edited your post, so I will edit my reply to your edits.

I can prove that it doesn't take much cost or practical thinking to make this cheap , even free using a PC audio generator like Audacity (free) to generate the signal and record the response in a scope waveform or spectral FFT view ( all free software) using the audio ports.

That is a lot of preparation, setup, debugging, tinkering, etc. Best to just buy the coils and be done with it.

But I dont need to do that as I have already shown the root cause of original errors with sufficient proof and examples.

With respect to what I suggested, I have shown your statements to be invalid, not applicable, and mistaken. Still waiting to receive an explanation of deflection.

Ratch

 

[Elerion]

Given the lack of response of the OP, he may be confused by now.

Definitely, I'm confused!
Need more time to read and understand than you guys to post !

 

[MrAl]

Definitely, I'm confused!
Need more time to read and understand than you guys to post !

See post #60.

 

[Elerion]

Until OP confirms the DC resistance of the coil

Please, tell me what measurement you want me to do.
I didn't see your reply to my question on #19

Now my question is, was or wan't I already making differential measurement? If not , what did I miss?

 

[Ratchit]

Please, tell me what measurement you want me to do.
I didn't see your reply to my question on #19

OK, first of all, measure the DC resistance of the coil with your VOM again, and tell us what it is. Second, yes, you were making differential voltage (DV) measurements if you measured the voltage across each component. You don't need a scope to do that. Next, do your comparative measurement again using a 10 ohm test resistor. Make sure the DV voltages across each component are equal by adjusting the frequency, and don't worry about their phase difference. The frequency will probably be less than 2 kHz. Let us know what the exact test resistor value is, and the frequency where their DV voltages equalize. Hope to hear from you.

Ratch

 

[Tony Stewart]

Please, tell me what measurement you want me to do.
I didn't see your reply to my question on #19

YES Elerion, If probes are matched, ( so that if applied to same signal would result in flat line(0V) it appears correct.
Recheck DCR
Recheck f for equal component voltage. "L+DCR" & Rload
then solve for L

from my post #40 you have 2 choices.

1) Subtraction by solving for L
√{X(f)²+DCR²}=100Ω at equal voltage point for OP's original question.
or L² = (100² - DCR²) /2πf

2) ignoring DCR by make R load bigger like 10k instead of 100 instead of smaller ( recall he could not generate higher f.)
using same L² = (10k² - DCR²) /2πf then DCR is negligible for f the point where amplitude is equal.

When you use scope probe, you are adding capacitance so 10~30pf, always use 10:1 probe.
Otherwise cable capacitance becomes significant if you go to 100k load and high frequency.
10:1 probe internally is corrected. if calibrated with square wave wave test signal.

3) Third method uses large cap of known precise value e.g. 10nF 100 nF 1uF to resonate with L to verify above

This has advantage of neglecting all R values and Q factor depends on impedance ratio X(f)/Rs where Q= resonant gain.

I suggest you resonate with suitable large Cap to verify this method.
Do you know formula for LC resonator vs f? ( although C is often only 10%)

Mr AL is also correct but shows different match approach.

I made it simple to just solve for L,

 

[Tony Stewart]

OK, first of all, measure the DC resistance of the coil with your VOM again, and tell us what it is. Second, yes, you were making differential voltage (DV) measurements if you measured the voltage across each component. You don't need a scope to do that. Next, do your comparative measurement again using a 10 ohm test resistor. Make sure the DV voltages across each component are equal by adjusting the frequency, and don't worry about their phase difference. The frequency will probably be less than 2 kHz. Let us know what the exact test resistor value is, and the frequency where their DV voltages equalize. Hope to hear from you.

Ratch

If you Test load R in series with your lossy choke is less than the DCR of the choke,
i.e if Choke is more than 10 Ohms it will never get down 50% of load and ALWAYS be greater

bad choice.

it will be impossible to achieve 1/2 Voltage.

 

[Ratchit]

If you Test load R in series with your lossy choke is less than the DCR of the choke,

I am counting on that at that low frequency, AC loss will not be a factor.

i.e if Choke is more than 10 Ohms it will never get down 50% of load and ALWAYS be greater

bad choice.

it will be impossible to achieve 1/2 Voltage.

Good point. So the test resistor should be somewhat higher, so the choke impedance can be raised to its value. Perhaps a better way would be to first set a low convenient frequency and use an adjustable pot. Then measure the test resistor's value when voltage equality is achieved. Lets see what the coil DC resistance is first of all.

Ratch

 

[Tony Stewart]

there are differences in L for a voltage source vs a current source. Why ?

same DM magnitude criteria is used for DUT and load R.

L = sqrt (R load² - DCR²) /2πf same formula as Mr Al

... Vector Algebra

 

[Ratchit]

there are differences in L for a voltage source vs a current source. Why ?

same DM magnitude criteria is used for DUT and load R.

L = sqrt (R load² - DCR²) /2πf same formula as Mr Al

View attachment 96417

... Vector Algebra

The denominator for your formula on the image should be (2*pi*f)^2 , not 2*pi*f . The calculations are correct, however. It should not matter if current driven or voltage driven as long as the DV's are equal across the components. If the AC resistance is significant and not taken into consideration, then the value of L will be calculated higher than it should be. It might explain why your L value at the higher frequency is higher. That is why I would like to keep the frequency as low as practical.

Ratch

 

[Tony Stewart]

typo corrected. However simulation including DCR still shows a discrepancy in results.

 

[Tony Stewart]

None of these methods we use above follow historical methods of a Maxwell bridge and others which are far more accurate for a test set as the signals are matched and DM is nulled, rather than compared with uncertainty of calibration.

https://www.electrical4u.com/maxwell-bridge-inductance-capacitance-bridge/

 

[Ratchit]

None of these methods we use above follow historical methods of a Maxwell bridge and others which are far more accurate for a test set as the signals are matched and DM is nulled, rather than compared with uncertainty of calibration.

https://www.electrical4u.com/maxwell-bridge-inductance-capacitance-bridge/

Yes, especially with expensive certified standard components. I have always known those bridges and variants as impedance bridges. Thanks for your link. Here is mine. Look at the last entry of the first page.

https://www.google.com/search?q=imp...0ahUKEwjbt623vP_JAhVIpx4KHTqnDIAQsAQIKg&dpr=1

Ratch

 

[Tony Stewart]

don't assume your page is the same size as mine

I prefer an RLC meter which uses a CC source to measure impedance at a chosen frequency to measure Rs, L, Rp, C, Q, d.f.

But when you have a reference capacitor to resonate in a suitable range, you can use a counter to measure the inductance regardless of DCR.

Here's is my interactive simulation where you can change L, DCR and C (7uF) ref
Try here

It is not that expensive to buy a 1% Reference Film Cap
1pc <$1
https://www.digikey.com/product-sea...t=0&page=1&quantity=0&ptm=0&fid=0&pageSize=25