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Dr_EM 28th April 2011 04:29 PM

Help on speaker impedance, estimating inductance
Hello all. Firstly, yes, this is "homework", however I'm not asking for anyone to solve this, only to maybe word it somehow that makes more sense. I can't speak to my lecturer right now and need to complete a report on this.

We have completed an experiment measuring a driver's impedance. Fine. But in the write-up, we need to estimate the inductance from the high frequency portion of our results. Here's how it's worded in the guide:

"For frequencies above 2kHz, plot loudspeaker impedance against frequency (done). Considering the loudspeaker as an impedance (L)and series resistor (R) then:

Z = (sqrt) R(squared) + Xl(squared)

where Xl = 2pi fL at frequency f

At high frequencies Xl is much greater than R, so Z tends to Xl hence:

Z = Xl = 2pi fL

Part that confuses me;

For a graph of impedance against frequency, the slope or gradient is

Z/f = 2pi L

Since 2pi L is a constant, then the graph of Z against frequency should tend to a straight line (it is indeed). From this portion of the graph (5-20kHz), calculate the slope of the line and hence estimate the value of the loudspeakers inductance L."

Ok, so I see the equations, and the graph with the straight line, but what data am I to put into these equations? If I select one point of data, say 10kHz, and put it into a transposed last equation I can get 0.329mH as per:

(10.33ohms/5kHz)/2pi = 0.329mH

Which sounds reasonable, but then it's different if I choose 5k or 20k. So, it's to do with finding the graphs slope, but I just can't understand exactly what or how I'm doing that?

Any help appreciacted!

DF96 28th April 2011 06:14 PM

The stuff you have been given is slightly confusing, perhaps because of an attempt to avoid calculus? Don't put any particular point in the equation, but estimate the slope of the graph from a suitable pair of points. Don't just plug numbers into formulas, but think about what the formulas mean. You are almost there, and asking the right questions.

By the way, thanks for saying that this is a homework problem. Few are as honest as you!

Dr_EM 28th April 2011 07:24 PM

Thanks for your response! Re-reading it for about the 100th time I noticed more that the gradient IS 2pi L, and as such the gradient/2pi is L.

I looked around for finding gradients and it appeared simple:

Gradient (Slope) of a Straight Line

So in this case I have an increase of 8ohms from 5k to 20k. So 8ohms/15,000Hz (got to keep to the base units right?) = 0.0005333. So this is the gradient. 0.000533/2pi = 0.00008488, which should be the inductance in Henrys, so 84.88uH.

The answer now almost seems too small though! It is a small radio type driver, 1cm coil sort of thing though. Some of the smaller Visaton drivers are 0.1mH, which is very similar, so it could be right.

DF96 28th April 2011 07:49 PM

Well done. You can now calculate the line from the formula for Z, and see how well it fits your experimental results. You may want to round your answer, as 4 sig figs is too much unless your measurements were very precise.

Dr_EM 28th April 2011 09:08 PM

Excellent, thanks for your guidance!

Dr_EM 29th April 2011 11:35 AM

Hello again. I thought I'd check this method against some commercial drivers, by applying it to thier impedance graphs and comparing to the published inductance. The results have not matched at all.

I've confused myself more now, apparently the published inductance is at 1kHz and might legitimately be different? I'm not sure what inductance value I've found now, is it the effective inductive component of the driver system as a whole, whereas the inductive value published is for a voice coil not installed in a driver system?

Do you think that the method I found yesterday is correct for the problem set?

I found this pdf whilst googling, but I can't use those formulas as we didn't measure phase nor Mms or BL!

DF96 29th April 2011 12:34 PM

There would be little point in specifying the coil inductance in isolation, as it will change considerably when placed in the magnet etc.

Checking against quoted figures is a sensible thing to do, but can produce puzzles. Bear in mind that the formula you have been given is for a resistance in series with a perfect (infinite Q) inductance. This is a good model, unless there are eddy current losses which can be roughly modelled by an extra resistor in parallel with the inductance. My guess is that there may be such losses in a speaker - does the impedance curve flatten off at higher frequencies?

Dr_EM 29th April 2011 01:05 PM

Hmm, starting to seem as though there is a lot they have neglected to explain about this. I think my best bet is to write out as I did in post 3 for my report and hope that's what they want to see!

Thanks again!

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