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Old 15th March 2003, 02:48 PM   #1
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Default How to measure inductance?

I was told by the designer of the x-over I'm using to measure inductance like this:

"Inductance is pretty easy to measure with a sound card, an amplifier and a precision resistor. Some programs, audioTester you can try for free for 45 days do an impedance plot and this can be used to easily work back to inductance. Actually for large inductors, using your sound card to generate a sign wave using an amplifier to drive a resistor and inductor in series works quite accurately if you have a good RMS voltmeter. Use a 4 or 8 ohm resistor and pick a frequency where the impedance of the inductor is similar to the resistor [so linearity errors in the multimeter are essentially cancelled out] and work back to the inductance. Remember the inductance will have a resistance portion. I have been able to measure this way to within 1 or 2% when comparing them to what I was able to measure on an HP Bridge."

What is a precision resistor and where can I get one? And are there any articles on the internet explaining this in more detail? How does one find out at what freqency the resistor and the inductor have a similar impedance? Is it done like this? You wire the resistor to one channel of your amp, with the resistor between pos. and neg. wires (called series wiring, I think) that usually go to your speakers. Then take an impedance plot and then do the same for the inductor(s). Find out at which frequency the impedance is the same and then wire the resistor and inductor in series (pos. to resistor, resistor to inductor, inductor to neg. wire). Okay then you send the sign wave at the frequency and measure the resistance with RMS Voltmeter (what's a RMS voltmeter?) on one side of the resistor to the other side of the inductor. You then get say 4.56 ohms for resistance and then subtract 4 ohm (if I use a 4 ohm resistor) and get .56 ohms resistance for the inductor and then somehow work figure out the inductance. I will figure out how to get from resistance to inductance later, I should have no problem with this part of it, but is the rest of my procedure correct? Is there an easier way to do this, besides approximating the inductance which I already know how to do? Is an HP Bridge hard to obtain?

Also, I was told by the designer of the x-over I'm using to use iron core inductors for the 2 - 12 mH inductors I'm using in order to keep the resistance low. Can I use steel instead, iron is hard to find? Or maybe a better question is where can I find the iron I need for the inductors?

Last question, if you use a wooden dowel to wind inductors you have to remove the wire from the dowel before you can measure the inductance right? Would be a lot easier to wind or add inductance after measuring the inductance if it could be left on the dowel, but I suspect this is not possible. Maybe winding more than I need and subtracting is a better idea. But I haven't seen a website for calculating turns for an iron core inductor.


-George, I mean Jimmy
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Old 15th March 2003, 03:33 PM   #2
SY is offline SY  United States
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It depends on how accurately you want to measure L. If you want a highly accurate measurement, you'll need to use an inductance bridge, a fairly specialized piece of equipment. If you want to get an approximate measurement, that's easier: Choose a precision resistor (that's just a resistor with a tightly spec'd tolerance) with a resistance that's much larger than the DC resistance of the inductor. For xover coils, something on the order of 20-30 ohms will work fine.

Place the resistor in series with the coil. Now, connect the output of your sine wave source (generator or sound card) to a power amp, and connect the series LR combo to the output of the amp. Adjust the output level so that you're impressing 4 or 5 volts acrosss the LR series combo. Alternately measuring the voltage across the resistor and the coil, adjust the sine wave frequency until the two voltages are equal. Note that frequency.

Now you can calculate the inductance, L = R/(6.26 * f), where R is the resistance of the precision resistor and f is the frequency you just noted. If you want to get a little more accurate, you can slip the DCR of the coil into the equation, but in reality, getting the coil's inductance better than a couple percent accurate isn't necessary.

And the dowel will make close to zero difference in the inductance.
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Old 15th March 2003, 04:20 PM   #3
e96mlo is offline e96mlo  Sweden
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Most electronics suppliers have iron powder cores or ferrite cores. You can not use just any piece of iron or steel.

When choosing your core you should look at the data sheet and make shure that the BH-graph is linear and that it has as small as possible hysteresis.

It is possible to calculate the numer of turns before you start to wind. The obtained value will only point in the right direction, though.

/Marcus
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Old 15th March 2003, 04:54 PM   #4
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Default Hey, thanks . . .

that seems a lot simpler than I though it would be. But why does the designer of my x-over say that I have to find a point where the impedance of the resistor is similar to the impedance of the inductor? Is this insignificant or is finding a point where the voltage of the resistor and the inductor the same, a similar technique or concept.

I guess an RMS voltmeter is similar to a multimeter. If you need to measure volts in this setup. I have one of those multimeters that have a needle that tells you the volts, but maybe a need to get a digital multimeter. Any suggestions?
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Old 18th March 2003, 07:36 AM   #5
ir33t is offline ir33t  Australia
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just use a HP impedance meter ....

/me is luck as he works in the used test and measurment equipment industry.

god i love those HP automatic impedance meters, measures C, L, R
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Old 20th March 2003, 07:39 PM   #6
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Default Thanks for the help . . .

But I don't think I have access to an HP Bridge. If I can't buy one, I don't think I can get access to one and even when I buy equipment for my speakers I usually wind up returning them after I use them cause I can't afford them. Which makes sense (using my logic) cause if I had money I could just buy inductors, although winding inductors does sound like fun in a demented sort of way. Also this HP Bridge sounds like a rather complex machine that cost hundreds or even thousands of dollars, which means I probably can't buy and therefore return one at Home Depot. Which is where I do all of my shadey dealings, for the greater good of my speakers.


Where can I get one of those precision resistors (resistors of a tolerence of 1%?). Does it have to be metal oxide? Or the other kind (wire resistors, maybe? Forgot what they're called). I was going to order my caps and resistors from solen.ca. Does anybody have any better opinions. Maybe order everything I need from one place, a place with 1% resistors and the ferrite iron cores I need also. I live in Boston, MA if anybody knows a place.
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Old 20th March 2003, 07:49 PM   #7
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Any electronic supply house will have 1% resistors. Use non-inductive ones like metal film, carbon film, metal oxide, whatever is cheap.
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Old 21st March 2003, 03:51 PM   #8
Ilianh is offline Ilianh  Canada
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What about using a transistor and a reference coil to create a simple oscilator, then by putting other coils in series, the frequency will change, and with some mathematics, you could find out the coil value.


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Old 21st March 2003, 04:31 PM   #9
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If you're going to that much trouble, you can just set up a parallel resonant circuit with a precision capacitor, sweep a sine wave, and get the frequency where Z is at a maximum. Back calculating to get L is then pretty easy.
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Old 21st March 2003, 04:45 PM   #10
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Default Jameco and others

sell an add-on unit for your DVM which does a pretty good job. It's less than $10.

an GenRad inductance bridge can be had for around $50,

if you are going to build speakers some method of measuring inducatance is a necessity.

btw, you can measure inductance implicitly if you have a triangle wave generator L = E * di/dt -- this little equation was the basis for an article in Popular Electronics in the 1970's. for a triangle wave di/dt is a constant. Similarly C = I de/dt. The article was a bit of a kluge.
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