# measuring inductance brain teaser

Discussion in 'Electronics Homework Help' started by maark6000, Sep 20, 2012.

1. ### maark6000

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Feb 8, 2012
So I'm going thru my labs on reactance, and my measurements are coming close, but not really close enough for my tastes. I realized that I'm just assuming the inductance value from the code on the side of the inductor, and not actually measuring it. So, out of the many ways to measure inductance, I found this very simple one by Tim Daycounter that allows me to use the stuff I have, namely a scope and a DMM.

The theory is simple: with a known resistor value, build a series LR circuit, apply a small AC voltage, and adjust the frequency until voltage across just the resistor is half that of the whole. then use this formula: L= R*sqrt(3)/(2*pi*f) to get your L value.

So the brainteaser is this: Is that voltage measurement across the R peak, or rms? I get a significantly different result depending on which I measure. If the answer is peak, then that correlates pretty nicely with all the labs I've been doing. I had assumed a 100mH inductor, but with the above method (measuring peak) it looks like the dang thing is more like 83 mH. (And at rms, my value drops further to 75 mH).

Thanks.

2. ### Harald KappModeratorModerator

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Nov 17, 2011
Why the sqrt? If I haven't completely forgotten the basics, then:

Xl = 2*pi*f*L

If the voltages across R and L are the same, then Xl=R=2*pi*f*L and therefore
L=R/(2*pi*f)

The answer should not depend on using the peak or rms measurement, since for sinusoidal voltages (I assume that's what you use - if not, you should do so) the relation is Vpeak=sqrt(2)*Vrms, regardless of whether you measure across a resistor or across an inductor.

3. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Yeah, half is half, and should be independent of the voltage.

An issue might be the frequency you're using. You need to choose R and f such that you're not either saturating the inductor, nor letting the capacitance between windings have a significant effect, nor operating the inductor at a frequency where the core is excessively lossy.

Perhaps try a range of series resistors and see if the calculated value of L changes?

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Feb 8, 2012
5. ### Harald KappModeratorModerator

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Nov 17, 2011
Frankly, I'm not so self assured any longer.
The math on the page you linked to looks solid. The difference between that page and my simple explanation is that the reference takes into account the complex part of the voltage. However, your voltmeter doesn't measure the complex relationship between the input voltage and the voltage across the resistor (or inductor).
Let's see whether Laplace can help us shed some light on this.

Last edited: Sep 21, 2012
6. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
I haven't completely digested the stuff on that page, however there are a number of things I'd point out.

1) These sorts of calculations always refer to peak voltage, not RMS or peak to peak. That's because the voltage is typically V.cos(theta) or V.f(t) or similar.

2) There appear to be current terms in there. If you have to specify (or measure) both a current and a voltage then both need to be in the same type of units (be they RMS or peak)

3) in this case we're actually talking about a pulse. RMS values need to be used with extreme care (especially where the pulse frequency in not specified), and I'd suggest that working with peak values is a lot safer.

My statement about half being half (above) presupposes a sine wave and simple voltage measurements.

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Feb 8, 2012
8. ### Harald KappModeratorModerator

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Nov 17, 2011
I think I have to stand corrected. The second reference gives the same result as that "daycounter" article, only the math is a bit clearer. Setting r->0 and x=1/2 gives the same result. So my simple approximation clearly and obviously seems to be wrong.

Still: for sinusoidal voltages it should be irrelevant whether you measure rms or peak, since the ratio of Vout/Vin is the same.  