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#1 |
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diyAudio Member
Join Date: Sep 2011
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Does anyone know how to calculate the impedance of a given inductor? I'm building a crossover filter for a two way system and I'm learning that in order to properly design my filter I need to consider the added impedance caused by the inductor .. thanks
Daniel |
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#2 |
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diyAudio Member
Join Date: Feb 2009
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If you're asking how to calculate the inductive reactance, it's 2pi times frequency times inductance. As you can see, the inductive reactance increases proportionally with an increase in frequency. This probably won't tell you what you're really after, however.
Paul |
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#3 |
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diyAudio Member
Join Date: Jun 2002
Location: USA, MN
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As paul said, you might not get much use out of the answer to your question. Why not take a couple steps back and explain what you are doing, as your question is a little bit vague?
__________________
Our species needs, and deserves, a citizenry with minds wide awake and a basic understanding of how the world works. --Carl Sagan Science is a way of thinking much more than it is a body of knowledge. --Carl Sagan |
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#4 |
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diyAudio Member
Join Date: Sep 2011
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I want to build a crossover filter for a two way system. Both the woofer and tweeter have a nominal impedance of 8 ohm. I want the crossover frequency to be 2600 hz. Well, I recently read somewhere that I have to take into consideration the added system impedance caused by the inductors. I was planning on using air coil inductors with 18awg wire.
Is this correct?, does the nominal impedance of the drivers change due to the effect of the inductors. I hope my question doesn't sound too stupid, lol |
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#5 | |
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diyAudio Member
Join Date: Jun 2002
Location: USA, MN
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Quote:
Designing a crossover without impedance and frequency response measurements may end up as a frustrating exercise. It is probably best that the drivers have a flat frequency response an octave or two above and below crossover, and that you have made their impedance as flat as possible with zobels. Baffle step is tricky to do without measurements. One place inductor DC resistance affects speaker design is in box sizing - a high DCR inductor will raise Qts and require a bigger box (and lower tuning in vented boxes)
__________________
Our species needs, and deserves, a citizenry with minds wide awake and a basic understanding of how the world works. --Carl Sagan Science is a way of thinking much more than it is a body of knowledge. --Carl Sagan |
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#6 | |
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diyAudio Member
Join Date: Feb 2009
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The impedance of an inductor is primarily determined by its inductive reactance, but it appears what you're really asking about it how much an inductor's d.c. resistance attenuates the signal to the woofer. The smaller that resistance, the less attenuation (loss) will occur, and that means the smaller the inductor's wire gage, the higher its d.c. resistance and the greater the signal loss. It always becomes a choice, then, you have to make, compromising between resistive losses and size of the inductor, with cost usually being the final determining factor.
Paul Quote:
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#7 |
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diyAudio Member
Join Date: Oct 2008
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+1 Guessing that the way you asked the question looks like you are interested in the DC resistance.
Measure it with any multimeter that can do small resistances. Assume it will always be that value. Theory suggests you keep it within one tenth of the speaker load, i.e. 0.8 ohms. Some people value the air core and are not so worried about the higher DC resistance of those inductors. Since you can't go too far wrong with all this I would suggest you try a few. You'll probably not notice much unless you are at the latter stages of dialling in your speakers. |
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#8 |
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diyAudio Member
Join Date: Sep 2011
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Yes, I think that the DC resistance is what I was after. What I read in an article was basically what pkitt said about "the d.c resistance attenuates the signal to the woofer." What got me a bit confused was trying to understand if this ment that I needed to change the values on my inductor in order to achieve the 2600hz crossover frequency that I wanted.
The only instrument I have at this time is the wt3 woofer tester. Will that help me in any way with figuring out the crossover filter. Thanks for the help guys |
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#9 |
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diyAudio Member
Join Date: Oct 2008
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When it comes to DC resistance, simply imagine you have a perfect inductor and you connected a resistor in series with it to the value of the DCR. No crossover frequency shift. In theory it will attenuate all power to the driver equally. In practice this depends on how well behaved the impedance curve for the driver is, but it's still not huge because the resistance isn't that large. It may seem to increase the bass just a little (i.e. it will reduce the lower mids a little more than other frequencies).
I have no experience with WT3 but I think it is an impedance jig (tell me if I'm wrong). Unless you can measure the responses of the drivers you won't have a clear picture of what the crossover needs to do. You can use manufacturers specs and you can do various simulations of cabinet and baffle effects for a good starting point. |
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