Measurement of an inductor not accurate at all





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$begingroup$


I'm trying out a method of measuring the inductance of inductors with a following circuit:



enter image description here



I'm using a signal generator to create the input voltage, and an oscilloscope to read the output across both the resistor and the inductor, as well as across the inductor only. I've found from multiple sources that when you tweak the input frequency such that the voltage across the inductor is one half of the input voltage (across both components), the inductance is given by the formula:



$$L=sqrt{3}frac{R}{2pi f}$$



Here is a picture of my circuit:



enter image description here



The voltage across the inductor is half of the input when the frequency is around 90.9kHz:



enter image description here



As you can see on the right, the peak to peak voltage of the yellow channel is 5.12, while the voltage of the blue channel is 2.56. The frequency fluctuated a bit, but it was around 90.90kHz. The series resistor has a value of 98Ohms as measured by my multimeter.



Using the formula for inductance, I get an inductance of roughly 300uH. However, the actual value of the inductor should be 100uH! So the measurement is completely wrong. What did I so wrong here?



I've read that the method only really works when the series resistance of the inductor is low. I measured it using my multimeter:



enter image description here



It seems to be around 3.3 Ohms. The reactance of the inductor (100uH) at 90.0kHz should be around 57 Ohms, so the series resistance should not cause this much error. I also thought that the output impedance of the generator would make a difference, but I don't see how as the measurement is taken outside of the generator. So what is wrong here?










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$endgroup$












  • $begingroup$
    Are you sure your resistor is non-inductive? A lot of them are made coating a non-conductive core with a conductive coating and cutting a spiral in them.
    $endgroup$
    – ratchet freak
    Dec 29 '18 at 5:13


















5












$begingroup$


I'm trying out a method of measuring the inductance of inductors with a following circuit:



enter image description here



I'm using a signal generator to create the input voltage, and an oscilloscope to read the output across both the resistor and the inductor, as well as across the inductor only. I've found from multiple sources that when you tweak the input frequency such that the voltage across the inductor is one half of the input voltage (across both components), the inductance is given by the formula:



$$L=sqrt{3}frac{R}{2pi f}$$



Here is a picture of my circuit:



enter image description here



The voltage across the inductor is half of the input when the frequency is around 90.9kHz:



enter image description here



As you can see on the right, the peak to peak voltage of the yellow channel is 5.12, while the voltage of the blue channel is 2.56. The frequency fluctuated a bit, but it was around 90.90kHz. The series resistor has a value of 98Ohms as measured by my multimeter.



Using the formula for inductance, I get an inductance of roughly 300uH. However, the actual value of the inductor should be 100uH! So the measurement is completely wrong. What did I so wrong here?



I've read that the method only really works when the series resistance of the inductor is low. I measured it using my multimeter:



enter image description here



It seems to be around 3.3 Ohms. The reactance of the inductor (100uH) at 90.0kHz should be around 57 Ohms, so the series resistance should not cause this much error. I also thought that the output impedance of the generator would make a difference, but I don't see how as the measurement is taken outside of the generator. So what is wrong here?










share|improve this question









$endgroup$












  • $begingroup$
    Are you sure your resistor is non-inductive? A lot of them are made coating a non-conductive core with a conductive coating and cutting a spiral in them.
    $endgroup$
    – ratchet freak
    Dec 29 '18 at 5:13














5












5








5





$begingroup$


I'm trying out a method of measuring the inductance of inductors with a following circuit:



enter image description here



I'm using a signal generator to create the input voltage, and an oscilloscope to read the output across both the resistor and the inductor, as well as across the inductor only. I've found from multiple sources that when you tweak the input frequency such that the voltage across the inductor is one half of the input voltage (across both components), the inductance is given by the formula:



$$L=sqrt{3}frac{R}{2pi f}$$



Here is a picture of my circuit:



enter image description here



The voltage across the inductor is half of the input when the frequency is around 90.9kHz:



enter image description here



As you can see on the right, the peak to peak voltage of the yellow channel is 5.12, while the voltage of the blue channel is 2.56. The frequency fluctuated a bit, but it was around 90.90kHz. The series resistor has a value of 98Ohms as measured by my multimeter.



Using the formula for inductance, I get an inductance of roughly 300uH. However, the actual value of the inductor should be 100uH! So the measurement is completely wrong. What did I so wrong here?



I've read that the method only really works when the series resistance of the inductor is low. I measured it using my multimeter:



enter image description here



It seems to be around 3.3 Ohms. The reactance of the inductor (100uH) at 90.0kHz should be around 57 Ohms, so the series resistance should not cause this much error. I also thought that the output impedance of the generator would make a difference, but I don't see how as the measurement is taken outside of the generator. So what is wrong here?










share|improve this question









$endgroup$




I'm trying out a method of measuring the inductance of inductors with a following circuit:



enter image description here



I'm using a signal generator to create the input voltage, and an oscilloscope to read the output across both the resistor and the inductor, as well as across the inductor only. I've found from multiple sources that when you tweak the input frequency such that the voltage across the inductor is one half of the input voltage (across both components), the inductance is given by the formula:



$$L=sqrt{3}frac{R}{2pi f}$$



Here is a picture of my circuit:



enter image description here



The voltage across the inductor is half of the input when the frequency is around 90.9kHz:



enter image description here



As you can see on the right, the peak to peak voltage of the yellow channel is 5.12, while the voltage of the blue channel is 2.56. The frequency fluctuated a bit, but it was around 90.90kHz. The series resistor has a value of 98Ohms as measured by my multimeter.



Using the formula for inductance, I get an inductance of roughly 300uH. However, the actual value of the inductor should be 100uH! So the measurement is completely wrong. What did I so wrong here?



I've read that the method only really works when the series resistance of the inductor is low. I measured it using my multimeter:



enter image description here



It seems to be around 3.3 Ohms. The reactance of the inductor (100uH) at 90.0kHz should be around 57 Ohms, so the series resistance should not cause this much error. I also thought that the output impedance of the generator would make a difference, but I don't see how as the measurement is taken outside of the generator. So what is wrong here?







inductance






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asked Dec 28 '18 at 19:14









S. RotosS. Rotos

8871814




8871814












  • $begingroup$
    Are you sure your resistor is non-inductive? A lot of them are made coating a non-conductive core with a conductive coating and cutting a spiral in them.
    $endgroup$
    – ratchet freak
    Dec 29 '18 at 5:13


















  • $begingroup$
    Are you sure your resistor is non-inductive? A lot of them are made coating a non-conductive core with a conductive coating and cutting a spiral in them.
    $endgroup$
    – ratchet freak
    Dec 29 '18 at 5:13
















$begingroup$
Are you sure your resistor is non-inductive? A lot of them are made coating a non-conductive core with a conductive coating and cutting a spiral in them.
$endgroup$
– ratchet freak
Dec 29 '18 at 5:13




$begingroup$
Are you sure your resistor is non-inductive? A lot of them are made coating a non-conductive core with a conductive coating and cutting a spiral in them.
$endgroup$
– ratchet freak
Dec 29 '18 at 5:13










2 Answers
2






active

oldest

votes


















6












$begingroup$

First of all, you should be measuring the voltage across the resistance.
Because you want to know the current that is flowing in the circuit.



$ V_R = sqrt{V_S^2 - V_L^2} = sqrt{5.12^2 - 2.56^2} = 4.43V $



Hence the current is $I = frac{4.43V}{100Omega} = 44.3textrm{mA}$



The inductor reactance is $X_L = frac{2.56V}{44.3textrm{mA}} = 57.9Omega $



And finally $L =frac{X_L}{2 pi F} = frac{57.9Omega}{2 pi cdot 91textrm{kHz} } = 101mu H $



As for your equation, the correct one is:



$$L=sqrt{frac{1}{3}}frac{R}{2pi f} =sqrt{frac{1}{3}}frac{98Omega}{2pi 90 textrm{kHz}} = 100 mu H $$



And this equation is true only for a frequency at which the (Vgen/Vinductance) = 0.5






share|improve this answer











$endgroup$













  • $begingroup$
    Thank you, with the correct formula I also got it working nicely. Also your suggested alternative method worked.
    $endgroup$
    – S. Rotos
    Dec 28 '18 at 22:42



















1












$begingroup$

A similar way





  • $L=sqrt{3}frac{R}{2pi f}$> is incorrect and even if L was $sqrt{3}frac{98}{2pi 90k}=14.1~ it ~is ~not~ 300$


since $Z_L={2pi f}*L$ thus when $ |Z_L|=R,~ $



$L=frac{R}{2pi f}$



The equation works out OK if you choose an f << SRF so that R is not too high yet >> Rs of L.



$L=frac{V_L}{V_R}*frac{1}{2pi f R} ~~= frac{2.56V}{~ 4.43V*~6.28~ *~91kHz~*~98Ω}=103 ~mu H$




  • computing errors from DCR=3.3 Ω may be done or corrected in the formula


When using a digital scope, it is not necessary to find 50% V point but close is ok , as long as phase shift is 90 deg in the inductor ( no SRC capacitance nor DCR effects visible) but you must use voltage drop across R and L separately ( differential scope mode A-B)






share|improve this answer











$endgroup$













  • $begingroup$
    Thanks! So basically my error was that I used an incorrect formula? And just a couple of questions: Why is it needed that the phase shift across inductor is 90 degrees? Isn't current always 90 apart from the voltage across the inductor?
    $endgroup$
    – S. Rotos
    Dec 28 '18 at 22:41










  • $begingroup$
    wrong formula and wrong measurement, there will be a few % error.using R=100 Ohms, next time measure VL and I.
    $endgroup$
    – Sunnyskyguy EE75
    Dec 28 '18 at 23:12














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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

First of all, you should be measuring the voltage across the resistance.
Because you want to know the current that is flowing in the circuit.



$ V_R = sqrt{V_S^2 - V_L^2} = sqrt{5.12^2 - 2.56^2} = 4.43V $



Hence the current is $I = frac{4.43V}{100Omega} = 44.3textrm{mA}$



The inductor reactance is $X_L = frac{2.56V}{44.3textrm{mA}} = 57.9Omega $



And finally $L =frac{X_L}{2 pi F} = frac{57.9Omega}{2 pi cdot 91textrm{kHz} } = 101mu H $



As for your equation, the correct one is:



$$L=sqrt{frac{1}{3}}frac{R}{2pi f} =sqrt{frac{1}{3}}frac{98Omega}{2pi 90 textrm{kHz}} = 100 mu H $$



And this equation is true only for a frequency at which the (Vgen/Vinductance) = 0.5






share|improve this answer











$endgroup$













  • $begingroup$
    Thank you, with the correct formula I also got it working nicely. Also your suggested alternative method worked.
    $endgroup$
    – S. Rotos
    Dec 28 '18 at 22:42
















6












$begingroup$

First of all, you should be measuring the voltage across the resistance.
Because you want to know the current that is flowing in the circuit.



$ V_R = sqrt{V_S^2 - V_L^2} = sqrt{5.12^2 - 2.56^2} = 4.43V $



Hence the current is $I = frac{4.43V}{100Omega} = 44.3textrm{mA}$



The inductor reactance is $X_L = frac{2.56V}{44.3textrm{mA}} = 57.9Omega $



And finally $L =frac{X_L}{2 pi F} = frac{57.9Omega}{2 pi cdot 91textrm{kHz} } = 101mu H $



As for your equation, the correct one is:



$$L=sqrt{frac{1}{3}}frac{R}{2pi f} =sqrt{frac{1}{3}}frac{98Omega}{2pi 90 textrm{kHz}} = 100 mu H $$



And this equation is true only for a frequency at which the (Vgen/Vinductance) = 0.5






share|improve this answer











$endgroup$













  • $begingroup$
    Thank you, with the correct formula I also got it working nicely. Also your suggested alternative method worked.
    $endgroup$
    – S. Rotos
    Dec 28 '18 at 22:42














6












6








6





$begingroup$

First of all, you should be measuring the voltage across the resistance.
Because you want to know the current that is flowing in the circuit.



$ V_R = sqrt{V_S^2 - V_L^2} = sqrt{5.12^2 - 2.56^2} = 4.43V $



Hence the current is $I = frac{4.43V}{100Omega} = 44.3textrm{mA}$



The inductor reactance is $X_L = frac{2.56V}{44.3textrm{mA}} = 57.9Omega $



And finally $L =frac{X_L}{2 pi F} = frac{57.9Omega}{2 pi cdot 91textrm{kHz} } = 101mu H $



As for your equation, the correct one is:



$$L=sqrt{frac{1}{3}}frac{R}{2pi f} =sqrt{frac{1}{3}}frac{98Omega}{2pi 90 textrm{kHz}} = 100 mu H $$



And this equation is true only for a frequency at which the (Vgen/Vinductance) = 0.5






share|improve this answer











$endgroup$



First of all, you should be measuring the voltage across the resistance.
Because you want to know the current that is flowing in the circuit.



$ V_R = sqrt{V_S^2 - V_L^2} = sqrt{5.12^2 - 2.56^2} = 4.43V $



Hence the current is $I = frac{4.43V}{100Omega} = 44.3textrm{mA}$



The inductor reactance is $X_L = frac{2.56V}{44.3textrm{mA}} = 57.9Omega $



And finally $L =frac{X_L}{2 pi F} = frac{57.9Omega}{2 pi cdot 91textrm{kHz} } = 101mu H $



As for your equation, the correct one is:



$$L=sqrt{frac{1}{3}}frac{R}{2pi f} =sqrt{frac{1}{3}}frac{98Omega}{2pi 90 textrm{kHz}} = 100 mu H $$



And this equation is true only for a frequency at which the (Vgen/Vinductance) = 0.5







share|improve this answer














share|improve this answer



share|improve this answer








edited Dec 28 '18 at 20:54

























answered Dec 28 '18 at 19:39









G36G36

5,6061511




5,6061511












  • $begingroup$
    Thank you, with the correct formula I also got it working nicely. Also your suggested alternative method worked.
    $endgroup$
    – S. Rotos
    Dec 28 '18 at 22:42


















  • $begingroup$
    Thank you, with the correct formula I also got it working nicely. Also your suggested alternative method worked.
    $endgroup$
    – S. Rotos
    Dec 28 '18 at 22:42
















$begingroup$
Thank you, with the correct formula I also got it working nicely. Also your suggested alternative method worked.
$endgroup$
– S. Rotos
Dec 28 '18 at 22:42




$begingroup$
Thank you, with the correct formula I also got it working nicely. Also your suggested alternative method worked.
$endgroup$
– S. Rotos
Dec 28 '18 at 22:42













1












$begingroup$

A similar way





  • $L=sqrt{3}frac{R}{2pi f}$> is incorrect and even if L was $sqrt{3}frac{98}{2pi 90k}=14.1~ it ~is ~not~ 300$


since $Z_L={2pi f}*L$ thus when $ |Z_L|=R,~ $



$L=frac{R}{2pi f}$



The equation works out OK if you choose an f << SRF so that R is not too high yet >> Rs of L.



$L=frac{V_L}{V_R}*frac{1}{2pi f R} ~~= frac{2.56V}{~ 4.43V*~6.28~ *~91kHz~*~98Ω}=103 ~mu H$




  • computing errors from DCR=3.3 Ω may be done or corrected in the formula


When using a digital scope, it is not necessary to find 50% V point but close is ok , as long as phase shift is 90 deg in the inductor ( no SRC capacitance nor DCR effects visible) but you must use voltage drop across R and L separately ( differential scope mode A-B)






share|improve this answer











$endgroup$













  • $begingroup$
    Thanks! So basically my error was that I used an incorrect formula? And just a couple of questions: Why is it needed that the phase shift across inductor is 90 degrees? Isn't current always 90 apart from the voltage across the inductor?
    $endgroup$
    – S. Rotos
    Dec 28 '18 at 22:41










  • $begingroup$
    wrong formula and wrong measurement, there will be a few % error.using R=100 Ohms, next time measure VL and I.
    $endgroup$
    – Sunnyskyguy EE75
    Dec 28 '18 at 23:12


















1












$begingroup$

A similar way





  • $L=sqrt{3}frac{R}{2pi f}$> is incorrect and even if L was $sqrt{3}frac{98}{2pi 90k}=14.1~ it ~is ~not~ 300$


since $Z_L={2pi f}*L$ thus when $ |Z_L|=R,~ $



$L=frac{R}{2pi f}$



The equation works out OK if you choose an f << SRF so that R is not too high yet >> Rs of L.



$L=frac{V_L}{V_R}*frac{1}{2pi f R} ~~= frac{2.56V}{~ 4.43V*~6.28~ *~91kHz~*~98Ω}=103 ~mu H$




  • computing errors from DCR=3.3 Ω may be done or corrected in the formula


When using a digital scope, it is not necessary to find 50% V point but close is ok , as long as phase shift is 90 deg in the inductor ( no SRC capacitance nor DCR effects visible) but you must use voltage drop across R and L separately ( differential scope mode A-B)






share|improve this answer











$endgroup$













  • $begingroup$
    Thanks! So basically my error was that I used an incorrect formula? And just a couple of questions: Why is it needed that the phase shift across inductor is 90 degrees? Isn't current always 90 apart from the voltage across the inductor?
    $endgroup$
    – S. Rotos
    Dec 28 '18 at 22:41










  • $begingroup$
    wrong formula and wrong measurement, there will be a few % error.using R=100 Ohms, next time measure VL and I.
    $endgroup$
    – Sunnyskyguy EE75
    Dec 28 '18 at 23:12
















1












1








1





$begingroup$

A similar way





  • $L=sqrt{3}frac{R}{2pi f}$> is incorrect and even if L was $sqrt{3}frac{98}{2pi 90k}=14.1~ it ~is ~not~ 300$


since $Z_L={2pi f}*L$ thus when $ |Z_L|=R,~ $



$L=frac{R}{2pi f}$



The equation works out OK if you choose an f << SRF so that R is not too high yet >> Rs of L.



$L=frac{V_L}{V_R}*frac{1}{2pi f R} ~~= frac{2.56V}{~ 4.43V*~6.28~ *~91kHz~*~98Ω}=103 ~mu H$




  • computing errors from DCR=3.3 Ω may be done or corrected in the formula


When using a digital scope, it is not necessary to find 50% V point but close is ok , as long as phase shift is 90 deg in the inductor ( no SRC capacitance nor DCR effects visible) but you must use voltage drop across R and L separately ( differential scope mode A-B)






share|improve this answer











$endgroup$



A similar way





  • $L=sqrt{3}frac{R}{2pi f}$> is incorrect and even if L was $sqrt{3}frac{98}{2pi 90k}=14.1~ it ~is ~not~ 300$


since $Z_L={2pi f}*L$ thus when $ |Z_L|=R,~ $



$L=frac{R}{2pi f}$



The equation works out OK if you choose an f << SRF so that R is not too high yet >> Rs of L.



$L=frac{V_L}{V_R}*frac{1}{2pi f R} ~~= frac{2.56V}{~ 4.43V*~6.28~ *~91kHz~*~98Ω}=103 ~mu H$




  • computing errors from DCR=3.3 Ω may be done or corrected in the formula


When using a digital scope, it is not necessary to find 50% V point but close is ok , as long as phase shift is 90 deg in the inductor ( no SRC capacitance nor DCR effects visible) but you must use voltage drop across R and L separately ( differential scope mode A-B)







share|improve this answer














share|improve this answer



share|improve this answer








edited Dec 28 '18 at 20:24

























answered Dec 28 '18 at 20:18









Sunnyskyguy EE75Sunnyskyguy EE75

71.9k227103




71.9k227103












  • $begingroup$
    Thanks! So basically my error was that I used an incorrect formula? And just a couple of questions: Why is it needed that the phase shift across inductor is 90 degrees? Isn't current always 90 apart from the voltage across the inductor?
    $endgroup$
    – S. Rotos
    Dec 28 '18 at 22:41










  • $begingroup$
    wrong formula and wrong measurement, there will be a few % error.using R=100 Ohms, next time measure VL and I.
    $endgroup$
    – Sunnyskyguy EE75
    Dec 28 '18 at 23:12




















  • $begingroup$
    Thanks! So basically my error was that I used an incorrect formula? And just a couple of questions: Why is it needed that the phase shift across inductor is 90 degrees? Isn't current always 90 apart from the voltage across the inductor?
    $endgroup$
    – S. Rotos
    Dec 28 '18 at 22:41










  • $begingroup$
    wrong formula and wrong measurement, there will be a few % error.using R=100 Ohms, next time measure VL and I.
    $endgroup$
    – Sunnyskyguy EE75
    Dec 28 '18 at 23:12


















$begingroup$
Thanks! So basically my error was that I used an incorrect formula? And just a couple of questions: Why is it needed that the phase shift across inductor is 90 degrees? Isn't current always 90 apart from the voltage across the inductor?
$endgroup$
– S. Rotos
Dec 28 '18 at 22:41




$begingroup$
Thanks! So basically my error was that I used an incorrect formula? And just a couple of questions: Why is it needed that the phase shift across inductor is 90 degrees? Isn't current always 90 apart from the voltage across the inductor?
$endgroup$
– S. Rotos
Dec 28 '18 at 22:41












$begingroup$
wrong formula and wrong measurement, there will be a few % error.using R=100 Ohms, next time measure VL and I.
$endgroup$
– Sunnyskyguy EE75
Dec 28 '18 at 23:12






$begingroup$
wrong formula and wrong measurement, there will be a few % error.using R=100 Ohms, next time measure VL and I.
$endgroup$
– Sunnyskyguy EE75
Dec 28 '18 at 23:12




















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