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Inductor testing

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henrylr

New Member
Hi all,

I just joined this site. The amount of info is staggering and very useful. My question concerns audio speaker crossover air coil test frequencies. I am building outboard crossovers and needed to buy 1.0mH, .91mH, .75mH and 0.2mH coils. The manufacturer guaranties their coils to be within 1%.
I based my order on tests results of the coils I was replacing with my LCR meter, which automatically selects the test frequencies as follows: 200kH at .3mH and lower, 14.925kH for .3mH to 4mH and 1kH for 4mH to10mH.
When I tested the new coils, the 1.0 measured .896, the .91 measured .82, the .75 measured .675 and the .2 measured .145. I called the manufacturer, and was told they use 1kH for the coils I bought, and this was the reason for the discrepancy.
Is there a way to find out what mH, measured at 1kH, I need based on my measurements made at 14.925kH and 200kH?

Thanks,
henrylr
 

Grossel

Well-Known Member
Hi.
What LCR meter do you have?

At higher frequensis, a coil doesn't behave as a pure coil anymore, but can say to be equivalent with a coil with capacitors at both ends. That's because it is a little capacitance between each turn in a coil.
 

MikeMl

Well-Known Member
Most Helpful Member
Do you have an audio oscillator?
 

The Electrician

Active Member
Hi all,

I just joined this site. The amount of info is staggering and very useful. My question concerns audio speaker crossover air coil test frequencies. I am building outboard crossovers and needed to buy 1.0mH, .91mH, .75mH and 0.2mH coils. The manufacturer guaranties their coils to be within 1%.
I based my order on tests results of the coils I was replacing with my LCR meter, which automatically selects the test frequencies as follows: 200kH at .3mH and lower, 14.925kH for .3mH to 4mH and 1kH for 4mH to10mH.
When I tested the new coils, the 1.0 measured .896, the .91 measured .82, the .75 measured .675 and the .2 measured .145. I called the manufacturer, and was told they use 1kH for the coils I bought, and this was the reason for the discrepancy.
Is there a way to find out what mH, measured at 1kH, I need based on my measurements made at 14.925kH and 200kH?

Thanks,
henrylr
First, frequencies are measured in kilohertz, abbreviated kHz, not kH. The abbreviation "kH" would mean "kilohenries".

There is no way to accurately get the inductance at 1 kHz from a measurement made at 14.925 kHz or 200 kHz. The best you could do is guess.

The problem is that crossover inductors have a lot of turns of wire, and therefore a lot of distributed capacitance. Here's a measurement of a frequency sweep of an inductor from 20 Hz to 1 MHz. The graph is showing the inductance on a vertical scale from 1 mH to 5 mH. There are two cursors, at 983 Hz (I can't set to exactly 1 kHz) and at 14.7 kHz. In the upper right part of the screen you can see the measured inductance at those two frequencies. At 983 Hz, the inductance is 1.726 mH, and at 14.7 kHz the inductance is 1.637 mH.



You can see how the inductance decreases as the measurement frequency increases until it reaches a minimum at just before 200 kHz, then it increases as the self resonant frequency of the inductor is approached, which is at about 800 kHz. This is why the measured inductance depends on the frequency that your meter uses to make the measurement.

If you want to compare your inductors to the manufacturer's spec, you will need to use a meter than makes the measurement at 1 kHz. You can find low cost meters that can do this on eBay, for example:

http://www.ebay.com/itm/BK-Precisio...037?pt=LH_DefaultDomain_0&hash=item20d36dc19d
 

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MrAl

Well-Known Member
Most Helpful Member
Hi,


An equation that relates the inductor voltage to the source driving voltage in an RL circuit is:

vL=sqrt(w^2*L^2+Rdc^2)/sqrt((Rx+Rdc)^2+w^2*L^2)

where
vL is the AC voltage across the inductor,
L is the inductance of the inductor,
Rdc is the measured DC resistance of the inductor coil,
Rx is the series resistor inserted in series with the inductor for the test,
w is the angular frequency 2*pi*f of the voltage source used to drive the circuit.

Solving for L we get:
L=sqrt(((Rx^2+2*Rdc*Rx+Rdc^2)*v^2-Rdc^2)/(1-v^2))/w

[LATEX]\[L=\frac{\sqrt{{Rx}^{2}\,{v}^{2}+2\,Rdc\,Rx\,{v}^{2}+{Rdc}^{2}\,{v}^{2}-{Rdc}^{2}}}{\sqrt{1-{v}^{2}}\,w}\][/LATEX]

where v is the AC voltage measured across the inductor divided by the source voltage:
v=vL/Vs

You could try that and see if you get better results. Of course you will need a 1kHz sine source to do this test.

For an example, say you use a 1v 1kHz AC sine source and you use a 20 ohm series resistor, and you measure 1 ohm DC resistance, and you measure 0.300 volts AC across the inductor at 1kHz. Using the above formula you would come out with an inductor value L of 0.0010377, which is only about 4 percent different from 1mH.
 
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dr pepper

Well-Known Member
Most Helpful Member
Your mainly interested in the inductance at a particular freq, there are several audio test programs around these days, if your serious then maybe you could look into one of these, you can take measurements of a xover in the circuit its designed for, numbers and specs are one thing but if you see it doing what its sposed to...
Impedance equalizers are also usefull when designing a xover, depending on what drivers you have and the order of the slope.
The values you mentioned were not that far out, over a prod run there will be slight variation anyways, the same will go for the drivers themselves.
One of the most important issues with inductors is that they do not saturate while in use, when staurated they act like a short, not good for sound quality.
 

MrAl

Well-Known Member
Most Helpful Member
Hello dr pepper,

Yes maybe using the sound card and a small driver circuit would produce the 1kHz tone required. I was also thinking of fooling the tester he has into thinking there was a larger inductor at the test terminals so it would put out a 1kHz tone, but that might introduce other problems such as distortion or the signal might be too low.
That tester would have been better if it allowed switching the output test frequency manually.
 

henrylr

New Member
LCR meter

Hi The Electricain
Thanks for the tip about the BK LCR on ebay. There are many others there with low prices.

Thanks again,
henrylr
 

henrylr

New Member
I bought a BK 878 and measured coils I want to replace. It shows L and Q or D values. I went shopping for coils on the web and it seems all are listed by L and ohm values. Is there a way to relate or convert Q or D to ohms?

Thanks,
henrylr
 

The Electrician

Active Member
It would be good if you would post a link to a description of an inductor in terms of inductance and resistance.

They may just be referring to the DC resistance of the coil, or they may be referring to the AC resistance of the coil at the measurement frequency.

Your new meter shows the Q of the coil at the measurement frequency, which could be 120 Hz or 1000 Hz. You can calculate the AC resistance with the formula R[sub]AC[/sub] = (2*pi*f*L)/Q, where f is the measurement frequency.

The AC resistance of inductors varies with frequency. Here's another sweep of the crossover inductor I measured in post #5. This sweep shows the AC resistance vs. frequency. You see at low frequencies the AC resistance is fairly constant and at really low frequencies is essentially equal to the DC resistance, but as the frequency increases, the AC resistance increases. It reaches about 10 ohms at 10 kHz, and 161 ohms at 100 kHz.

 

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henrylr

New Member
I have tested the inductor at 1kHz and it measured .952mH with a Q of 5.6. The BK inductance mode only gives L and the option to get the value for Q or D. If I swith the meter to Resistance mode will it give a meaningfull ohm value of the .952mH inductor. As you can probably tell I am a real novice in electronics. Sorry for asking what are probably, very boring, basic questions.

Thanks,
henrylr
 

The Electrician

Active Member
Using the formula, R[sub]AC[/sub] should be (2*pi*1000*.000952)/5.6 = 1.068 ohms. If you switch the meter to resistance mode, does it give a value near 1.068 (make sure you're still on 1 kHz frequency)? If it does, then you're golden.

Please post a link to the inductors you found on the web where they give resistance. I'd like to see what they're specifying.
 
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henrylr

New Member
I need to buy 8R6 and 1R0 10W resistors for audio crossovers I have designed. I found Vishay Dale 10W 1% wire wound Power Resistors encased in finned aluminum which is supposed to act as a heat sink. The words "Current Shunt" are also in the description and I don't know what that means. Can these be used as audio resistors? Do I need to mount these on metal heat sink plates?

Thanks,
henrylr
 

dr pepper

Well-Known Member
Most Helpful Member
Mr Al,
Back in the day I used a software package called speaker workshop, it did exactly as you mentioned, and would plot impedance, phase, group delay and a load of usefull info.
I built a test box to use with it, which contained switches, a little power amp and reference resistors/inductors/caps, all in all an excellent package.
Dont know if its still incirculation, if so it might not be free anymore.
 

MikeMl

Well-Known Member
Most Helpful Member
If you will be dissipating 10W, then yes, they will have to be bolted to a heatsink to get the rated dissipation.

They should work fine for audio frequencies.

"Current Shunts" are usually sub-1Ω (0.001Ω is typical), but they are just precise (1% or better) power resistors
 

henrylr

New Member
What size heat sink plates should I use ( LxWxT).? What is the best material for the plates? Can both resistors be mounted on the same plate? And lastly, do I need heat sink grease between the resistors and plates?

Thanks,
henrylr
 

MikeMl

Well-Known Member
Most Helpful Member
What size heat sink plates should I use ( LxWxT).?
What is the best material for the plates?
Can both resistors be mounted on the same plate?
And lastly, do I need heat sink grease between the resistors and plates?
20W of dissipation with 85degC rise (very hot, will burn you) means a thermal resistance of 85/20 = 4.25 degC/W, which requires a vertically-mounted plate of Al of about 140 sq. cm. total surface area (count both sides).

Use thermal compound between the body of the resistors and the plate.

Best way is to mount the resistors on the plate in-situ, hook them to a suitable DC power supply, crank the supply so the resistors are dissipating 10W each, and measure the final temperature reached after a few minutes. Check the maximum temperature rating for the resistors...
 

henrylr

New Member
Great info and fast reply. I will machine a 4"X 7" equaling 28 sq. in. (71.12 sq. cm for each side) = >140 sq cm plate. Two final questions are, what thickness should the plate be and, can two 10w resistors be mounted on this size plate?

Thanks again,
henrylr
 
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MikeMl

Well-Known Member
Most Helpful Member
I used 20W (10W per resistor) in the calc. I would use no thinner than 1/8", 3/16" would be better. You will likely not be dissipating 20W all the time, because of the peak to average ratio of music...
 
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