If I am right, that may not be an inductor intended for low distorion, accurate, filters, tank circuits, etc. It is probably a lossy inductor, intended for supply line filtering and the like. In that application the 'worse' the inductor is the better. As you probably know, inductors (and capacitors) have a Q (Quality) factor. which = XL/ESR. (XL= 2 * π * F). A Q of 100 would be typical for a precision inductor at its specified frequency. The best inductors are air cored, but they get physicallly large at high values, so a high permeability core is used. Air (vaccuum) is the reference with a permeability of 1.
...Electrical 101
The lossy inductor would have an AC resistance dependent on the frequency of the circuit. That resistance would not show up on a simple resistance measurement with a VOM. Perhaps the measurement could be made with a 10 ohm resistor at around 1.6 kHz. Skin effect would be negligible and the energy losses would be significantly less. That should measure mostly inductance. Ratch
Using RL low pass filter when |X(f)| = R ,
You expect voltage to be 0.707 of input voltage or -3dB, which is also half power point at 16.2kHz.
Half voltage would be the -6dB point.which is ~ 27.5kHz where XL(f) = j 173 Ohms and thus the Zsum= 100 + j 173 (vector) = 200 Ohm (scaler)
Go back to basic math until clear.
-Electrical 101
University courses for 1st yr were 101Now you have used that term. I have been wondering what it means exactly. What is the derivation? Is it a Can/Us term? The 'xxx for Dummies' books even have a '101' chapter.
JimB stated it correctly, that when the real and reactive impedance magnitudes are equal, the vector impedance is not 200 but rather 141.4 (√{1+1}) so the voltage is not 50%, rather 70.7 % or -3dB. So without a basic understanding of vector math, the Op will not get the correct answer.No need to worry about input voltage, 0.707, 6 db, or basic math. As long as the differential voltage across the resistor and coil are the same, the test resistance value and coil impedance will equal each other, even if the voltages phases are different. If the frequency is low enough, I don't think the DC resistance and AC loss resistance of the coil will be enough to affect the measurements.
Ratch
JimB stated it correctly, that when the real and reactive impedance magnitudes are equal, the vector impedance is not 200 but rather 141.4 (√{1+1}) so the voltage is not 50%, rather 70.7 % or -3dB. So without a basic understanding of vector math, the Op will not get the correct answer.
I think the lesson to be learned for those who could not answer his dilemma , is beware of your false assumptions and correct them.
In no thread is the impedance of the generator or DCR of the inductor stated.
Generators are usually 50Ω.
For reference these Thru Hole 1mH chokes are 78Ω DCR.
https://www.digikey.com/product-detail/en/8230-92/M8049-ND/19693
Never assume DCR or ESR = 0, without good reason.
This also explains why error reduced with higher series R used.
Forgive my oversight.
Then if the DCR measurements are accurate, then L may be easily calculated and the simulation should always match the real world when all assumptions and measurements are correct.
So far they do not match.
BELOW each element has equal magnitude at 2.62/5= 52.4% not 40% @f = 15.8kHz not 9.6kHz
The probe in turquoise is the lowest trace.
Reported the original results were;
View attachment 963831.- voltage on each element match at a much lower frequency (9.6k vs 16k).
2.- peak-peak voltage is not source/sqrt(2), but much lower (40% vs 70%).
EDIT
In order for 9.6kHz to be true equal voltage point with L and DCR=Rs vector impedance = 100 Ω then √{X(f)²+Rs²}=100Ω for X(f) =ωL and f = 9.6 kHz then you can solve for L or Rs.
If Rs=13Ω then L=1.6mH
and if Rs=78Ω then L=1.0mH
so recheck DCR (Rs), f etc.
Hey Ratchit,I suggested that the measurement be performed at a lower frequency, such as a test resistor at around 10 ohms. That would cause an equality to be at around 1.6 kH and eliminate the effects of rf loses and skin effects.
Ratch
Hey Ratchit,
Read my datasheet, please. These are typical SRF >>1MHz and DCR ~78Ω.
The correct method was used by the OP with an audio signal range but R=100 neglected the DCR near this value.
Why is it necessary to neglect the DC resistance of the coil? If we know what it is, we can take it into consideration when calculating.A test R >> DCR allows DCR to be neglected.
A smaller variable load like 10~100 Ω makes sense to make an AC divider at 60Hz where X(f)~0 to measure DCR, but using an Ohm-meter makes more sense and SRF is far , far away in the MHz here.
Certainly. A simple VOM differential voltage measurement across each component will give the vector voltage sum of each component. That is what voltmeters do. Of course, the voltage across the resistor will consist almost entirely of voltage in phase with the current.In every, case , vector math gives the correct answer with a DCR measurement and f for equal voltage drop.
Vector Algebra 101.
I understand you perfectly, even though I may not agree with you about everything.Ratchit,
I am failing to make myself clear.
More plausible yes, but not certain. I wish the OP would measure the resistance of the coil again.If Rs=13Ω then L=1.6mH {one value is in error, which we do not know )
and if Rs=78Ω then L=1.0mH ( seems more plausible given the "reality check of a typical choke of that construction, value , size and tolerance. )
Since you mentioned "Skin effect", I though I would mention Self Resonant Frequency SRF which is more relevant but albeit both insignificant in audio range at very low current.
You are preaching to the choir. Actually, it is phasors, not vectors. Voltages have a magnitude, but no direction. They do have a phase relationship with each other, however. Some properties are vector-like as in addition and subtraction. But what does the vector dot and cross product of a voltage mean?Vector Math solves all these problems. Z= √{ R^2+X(f)^2} when DCR is taken into account with Load R.
I believe he tried that before I suggested differential voltage measurements. With differential measurements, there is no need to use one high component value to swamp out another component value.Or as the OP suggested using a load R >> DCR then simple model ( with no DCR) matches test results closer.
I was going to ask you the same thing.Any questions?
I believe the OP, and certainly I realize the importance of taking the resistance of the coil into consideration. I am trying to understand what new viewpoint or concept you have shined on this method.My whole point of measuring L with a simple load resistor must include the loss of DCR which is significant when R=100, contrary to your suggestion.
RMS=root mean square? Now you have me confused. I look forward to enlightenment as to how RMS fits into this problem.If you dont use the simple RMS math to exclude the DCR loss then your results have significant error.
Be optimistic. He might have figured it out and has gone onto different things.Given the lack of response of the OP, he may be confused by now.
Do you disagree with my calculations? using Z= √{ R^2+X(f)^2} and solving for L using Rs(DCR) ?Forgive my oversight.
Then if the DCR measurements are accurate, then L may be easily calculated and the simulation should always match the real world when all assumptions and measurements are correct.
So far they do not match.
BELOW each element has equal magnitude at 2.62/5= 52.4% not 40% @f = 15.8kHz not 9.6kHz
The probe in turquoise is the lowest trace.
Reported the original results were;
View attachment 963831.- voltage on each element match at a much lower frequency (9.6k vs 16k).
2.- peak-peak voltage is not source/sqrt(2), but much lower (40% vs 70%).
EDIT
In order for 9.6kHz to be true equal voltage point with L and DCR=Rs vector impedance = 100 Ω then √{X(f)²+Rs²}=100Ω for X(f) =ωL and f = 9.6 kHz then you can solve for L or Rs.
If Rs=13Ω then L=1.6mH
and if Rs=78Ω then L=1.0mH
so recheck DCR (Rs), f etc.
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