Al Klappenberger Posted January 31, 2006 Author Share Posted January 31, 2006 All the inductors have the same inductance, just different series resistance, correct? All the inductors I tested were neary the same, yes. Have you simulated cross overs with inductors with different series resistance? Yes, I always assume a reasonable Q value when I do a computer simulation. What if you make the ESR a function of frequency? As I explained before, how do you correlate a frequency point on a curve with the same frequency point you are analyzing with the computer? You would need to make a curve-fit equation for each seperate inductor. It could be done but it would be a real pain in the posterior! Do you have data of ESR vs Frequency of the various inductors? Yes, that data IS what the Q curves are! The Q=X/ESR where X=2*Pi*Freq*L That way you can compare the Cross over response with different inductors and see what it does to frequency response, phase, group delay etc. Or just measure the crossovers built with the different inductors? The high series resistance of the inductors at the cross over frequencies would surely shift those crossover frequencies just like different speaker impedances. Not by much. It only would change the MEASURED crossover becasue of the definition being the frequency where the two channels have equal level. Losses change that level. Another dumb questions: When you design a cross over for a give frequency, you use the nominal impedance of the driver, say 4, or 8 ohms, whatever. Would it not be better to use the real complex impedance of the driver at the cross over frequency? Yes, I do that on woofer filters by synthesizing the filter to have an output inductor equal to the voice coil inductance of the woofer. For the K33, that's 1 mHy. All my networks are done that way. Most designers don't have the software to do that so they add a "Zobel" to absorb that inductive component then design for a flat resistive impedance of 4 or 8 or whatever. Quote Link to comment Share on other sites More sharing options...
sfogg Posted January 31, 2006 Share Posted January 31, 2006 Dana, Show me where I ever said RLC didn't count, don't go arguing with a straw man. Esp. considering I'm the one pointing out to you the difference between an inductor and the same length of wire not coiled up. If you go back and check the threads on ABX wire testing I flat pointed out that the only time people have passed ABX testing of wires is if the wires have 'wacky RLC' values or if the wire isn't just a wire in that it also has passive components installed in it like MIT cables. (Coils/caps...etc..) The point is that very small differences in the measured RLC characteristics of a 6" piece of wire (if they are even there at all) does not automatically correlate to any sort of audible difference. And in point of fact when scientifically tested for audibility people fail... consistantly... with any sort of normal wires. Even though going in they claimed differences were 'unmistakable.' Maybe you do really hear what you claim you do. So prove it and take the $1,000,000 challenge. Shawn Quote Link to comment Share on other sites More sharing options...
DrWho Posted January 31, 2006 Share Posted January 31, 2006 I have a $100... Only a $100? I'd wager a lot more. I'd put millions on that bet [] So Al, what is the end 'audible' result that makes the higher Q inductor better? Do you end up with different slopes or something? Or is it more of an electrical noise floor thing? Quote Link to comment Share on other sites More sharing options...
D-MAN Posted January 31, 2006 Share Posted January 31, 2006 Al, frequency has nothing to do with it! Electricity always takes the path of least resistance and that is on the outside first couple (thousand) of layers of molecules (i.e., the skin effect). That is an unalterable fact. The idea that audio frequency current flows through the core of the wire is mistaken. By extrapolating that premise, the higher in frequency the signal goes, the closer to the surface it conducts?! Not here on Earth, anyway... Also I think multi-strand solid silver wire is absolutely the way to go. You guys will spend a fortune on fancy capacitors and coils, but not wire?! I say WHY NOT?! Why draw an arbitrary line like that? Why argue, you've already made the point that all components matter - I can't feature leaving out the wiring, which enables everything in the first place. No, It is NOT a waste, but it is expensive. Sort of like your crossovers (at last count I bought 4 pairs)! Was that a waste, too?! NOT! DM Quote Link to comment Share on other sites More sharing options...
Al Klappenberger Posted January 31, 2006 Author Share Posted January 31, 2006 Dman, However, would you agree that an arbitrary 16ft chunk would have a measurable amount of inductance (however small) and resistance (and capacitance)? It stands to reason that it would, and therefore, it is not simply a conductor with no electrical properties. Now what do these properties (however small) do to the signal being passed? Seems to me that for anyone to simply discount or ignore the properties of wires is discounting WAY too much. If your are illuding to speaker cables, NO! Remember Lt=L1+L2+2*Lm. There is a hitch to that. It is really L1+L2 +-2*Lm. The mutual inductance my be in phase or out of phase. If you reverse the polarity of one of the winding the two will add to ZERO! In speaker cable the currents are flowing in opposite directions in each side of the "pair". The inductance cancells out! Also, since Q is realted (In part) to DCR of the wire, winding the wire into a coil adds mutual inductance between the turns. This makes the inductance higher for any give length of wire. So.. the same inductance can be had with less wire and therefore less DCR. The trick is to fiund a ratio of length to diameter and layers that maximize the mutual inductance. That is to get as many turns as possible to be as close as possible to all the others. That research has yielded the shape you see coils made. Other factors come into it when frequency goes up. Maximum mutual inductance isn't always best for distributed capacity between windings. At microwave frequency, a long skiney single layer solonoid has the highest Q with the lowest idstributed capacity. Al K. Quote Link to comment Share on other sites More sharing options...
sfogg Posted January 31, 2006 Share Posted January 31, 2006 "frequency has nothing to do with it!" Of course it does. Read up on skin effect more, frequency is an integral part of it. Like you said... in a radar array frequency it matters. At audio frequencies...... http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/audio/skineffect/page1.html http://www.audioholics.com/techtips/audioprinciples/interconnects/SkinEffect_Cables.htm Shawn Quote Link to comment Share on other sites More sharing options...
DrWho Posted January 31, 2006 Share Posted January 31, 2006 Al, frequency has nothing to do with it! Electricity always takes the path of least resistance and that is on the outside first couple (thousand) of layers of molecules (i.e., the skin effect). That is an unalterable fact. Well that's not what they're teaching in the Electrical Engineering classes at UIUC... nor the materials science classes either.... Sure, electrons follow the path of least resistance and sure the skin effect exists. But the driving forces behind these things aren't really related at all.... Quote Link to comment Share on other sites More sharing options...
sfogg Posted January 31, 2006 Share Posted January 31, 2006 " Only a $100? I'd wager a lot more. I'd put millions on that bet" Me too... but to win a bet you gotta get the other side to accept the bet first. Shawn Quote Link to comment Share on other sites More sharing options...
Al Klappenberger Posted January 31, 2006 Author Share Posted January 31, 2006 Al, frequency has nothing to do with it! Electricity always takes the path of least resistance and that is on the outside first couple (thousand) of layers of molecules (i.e., the skin effect). That is an unalterable fact. NO..Electricity takes all paths at all times each according to ohms law! The idea that audio frequency current flows through the core of the wire is mistaken. By extrapolating that premise, the higher in frequency the signal goes, the closer to the surface it conducts?! Not here on Earth, anyway... Skin Effect at audio is virtually nonexistant. Also I think multi-strand solid silver wire is absolutely the way to go. You guys will spend a fortune on fancy capacitors and coils, but not wire?! I say WHY NOT?! Why draw an arbitrary line like that? Why argue, you've already made the point that all components matter - I can't feature leaving out the wiring, which enables everything in the first place. Silver has a DC resistance .95 times that of copper but costs many, many times more than that ratio. Going to soild silver is a wast of big bucks for very little return at audio frequency! I have had may microwave filters silver plated internally where skin effect IS a factor. The reduction in loss is measureable but modest. That is why silver plating is used ONLY at microwave frequencies. No, It is NOT a waste, but it is expensive. Sort of like your crossovers (at last count I bought 4 pairs)! Was that a waste, too?! NOT! You are comparing apples to oranges! Have you ever seen or even heard of silver litz wire? The Litz is what's important not the material the wire is made of! Al K. BTW: I once built one of my universal networks for a customer using silver wire and solder he supplied. It was a pain in the butt to do and the difference could not even be measured! I did it for the experience. I would never do it again! Quote Link to comment Share on other sites More sharing options...
sfogg Posted January 31, 2006 Share Posted January 31, 2006 The short answer on skin effect is that it starts to be more of an issue when the wavelength of the signal in the wire gets to be shorter then the wire itself. Pop Quiz.... what is the wavelength of a 20hz wave in a piece of cable.... Shawn Quote Link to comment Share on other sites More sharing options...
DrWho Posted January 31, 2006 Share Posted January 31, 2006 lol shawn Btw, the reason a coil behaves differently than a piece of straight wire has to do with the fact that a coil creates a magnetic field which then interacts with the current flowing through it. It's the same principle behind Faraday Rings... And feel free to take the RLC values of any wire into account. As Al mentioned these things become very important at extremely high frequencies so the measuring equipment and modelling already exists. The problem at audio frequencies is that any difference in the wire is going to be swamped by errors in the rest of the system. Like how bout the target mass for the tweeter diaphragm? Or nanometer differences in the shape of the horn, room acoustics, listening position, source material. You're talking difference on the order of 0.01dB with changes in wire of sufficient guage, yet the entire system from start to finish would be lucky to be within +-10dB! Quote Link to comment Share on other sites More sharing options...
efzauner Posted January 31, 2006 Share Posted January 31, 2006 Al From my understanding of cross over design, (it has been some time, not beyond my ability, but just not a priority right now to spend the brain cells on.) , A simple 2nd order parallel butterworth high/low pass: C = 1 / (2 * p * f * (Z * sqrt2)) and L=(Z *sqrt2) / (2 * p * f) Z is normally the nominal impedance of the drive, say 4, 8 or 16 etc. but if the caps and inductors have considerable ESR, then they should be considered in the equations. Since the Cs and Ls are in different positions in the high pass vs the low pass, the resultant total Z is different, giving different cross over freqeuncies, No? Quote Link to comment Share on other sites More sharing options...
Al Klappenberger Posted January 31, 2006 Author Share Posted January 31, 2006 From my understanding of cross over design, (it has been some time, not beyond my ability, but just not a priority right now to spend the brain cells on.) ,A simple 2nd order parallel butterworth high/low pass:C = 1 / (2 * p * f * (Z * sqrt2)) andL=(Z *sqrt2) / (2 * p * f)Z is normally the nominal impedance of the drive, say 4, 8 or 16 etc.but if the caps and inductors have considerable ESR, then they should be considered in the equations. Since the Cs and Ls are in different positions in the high pass vs the low pass, the resultant total Z is different, giving different cross over freqeuncies, No? This is getting into an area that is EXTREMELY complicated! It borders on a technique called "predistortion". The simple answer to your suggestion is NO! The effective Q is like resistors in parallel. Qt = 1/(1/Ql + 1/Qc). It's doesn't matter if the L and C is in series or parallel. You can assume all the components of a filter are equal to that Qt and it will analyze the same. I do that in bandpass filters to generalize an L and a C together as a "resonator".Al K. Quote Link to comment Share on other sites More sharing options...
Deang Posted January 31, 2006 Share Posted January 31, 2006 I'm ashamed to say that most of this is completely over my head -- stop it!! For crying out loud, what are we talking here for loss with solid air cores compared to litz -- .003 ohms? I'm still an advocate of the Gestalt Effect with this stuff. The best measuring part isn't always the "best" part. Resistance (loss) plays an important role in the voicing of the speaker, and a little in the right spot can be a very good thing. Seriously, "loss" is just a form of attenuation. It seems silly to yank as much resistance out of the circuit as possible -- only to have to add it back in with a resistor or L-pad. The copper foils absolutely rule in low pass sections, and throwing a .20mH copper foil in place of the litz in the bandpass takes a bit of bite out of midrange -- I like them -- they sound very good. Quote Link to comment Share on other sites More sharing options...
DrWho Posted February 1, 2006 Share Posted February 1, 2006 I'm still an advocate of the Gestalt Effect with this stuff. The best measuring part isn't always the "best" part. That's why I was wondering what the net effect would be with the different parts...The sections in front of and behind the crossover are certainly not perfect, so the "ideal" crossover probably isn't going to sound the best either... Quote Link to comment Share on other sites More sharing options...
Al Klappenberger Posted February 1, 2006 Author Share Posted February 1, 2006 " Resistance (loss) plays an important role in the voicing of the speaker, and a little in the right spot can be a very good thing. Seriously, "loss" is just a form of attenuation. It seems silly to yank as much resistance out of the circuit as possible -- only to have to add it back in with a resistor or L-pad. " Dean! You, of all people, should know better! The difference between distributing the loss through a filter and "lumping" it into an Lpad are many! 1- Losses in a filter casue the response to round off. Attenuation in an L-pad are constant with a flat respones. 2- Losses in a filter are STUCK THERE. With the pad you have the option to adjust the level to your liking. 3- Losses in general divorce your speaker driver from the amplifiers damping factor. It isolates the drivers back EMF. This is the same theory as using a transformer to attenuate the squawker RATHER than an L-Pad. Then there will be virtually no resistive losses at all! As to foil inductors, they have their point of least loss too high and distributed capacity is not factor. Al K. Quote Link to comment Share on other sites More sharing options...
Al Klappenberger Posted February 1, 2006 Author Share Posted February 1, 2006 Who, You ask how the different inductors will "sound". The answer is the same as how the different capacitor brands "sound". The answer is different for every person you ask and every speaker they go into and every room they are located in! You can't pin it down. Measured Q however is very easy to pin down. High Q makes a better filter and a better filter gives you a better network. What is a "better network", you ask. The answer again is who do you ask! The fact is that people hear what they want to hear. As only one person, my opinions count no more or less than anyone else. I restrict my comments only to what I can measure. This is called Science and can't be easily disputed. Al K. Quote Link to comment Share on other sites More sharing options...
DrWho Posted February 1, 2006 Share Posted February 1, 2006 Aye, but engineering is taking that science and doing something with it...in this case turning it into something that sounds good [] What you said in your previous post was more of what I was looking for. I just can think of scenarios where all of those "flaws" would be beneficial to the real world application. But to use these to your advantage would require insane amounts of testing and as you said - it's like shooting in the dark because the losses are set in stone. Quote Link to comment Share on other sites More sharing options...
efzauner Posted February 1, 2006 Share Posted February 1, 2006 Having just done some simulations (I am lazy, I have $100K of simulation software on my laptop-thats my job) of simple 12db HP and LP filters, I see that the cap and inductor Q just increases the overall loss just like you said. From your Q graphs I was able to calculate the following (approximately):for 200uH inductorat 2KHz Inductor Q=15 ESR=170mOhm for Solid WIre Q=20 ESR=120mOhm for Litz Wire at 10KHz Inductor Q=10 ESR=1.2 Ohm for Solid WIre Q=65 ESR=200mOhm for Litz WireThe added resistance of the solid wire at 10KHz results in about a 1db loss with an 8ohm speaker load. This is basically the loss of a 1 ohm resistor in series with the 8 ohm speaker. Are you suggesting that the difference in sound perceived with the Litz inductors is a result of the 1 dB increase vs the Solid Wire inductor? I cannot believe that this 1 db makes all the difference. Changing the sqwaker tap would do the same. Or is there something else going on that the ESR does not measure? In other words, something is happening, but ESR is not the right measurement. Send me the ESR vs Freq and I would be happy to setup a simulation of your networks with ESR modeled vs freqeuncy. But I don't think we will see much more. The above simulations are the min/max already. Also dont forget that for a constant ESR, Q will increase linearly with frequency anyway because the reactive component 2piFL increases linearly with frequency. Quote Link to comment Share on other sites More sharing options...
popbumper Posted February 1, 2006 Share Posted February 1, 2006 Al: Please expound on your comment "This is the same theory as using a transformer to attenuate the squawker RATHER than an L-Pad. Then there will be virtually no resistive losses at all!". I would like to understand (and maybe you have answered already) how and autoformer and L-pad or fixed resistance differ performance-wise. I realize both achieve the same end, but why would you choose one over the other; what is the rationale? Again, I think you answered, but I am hoping for a bit deeper explanation. Thanks! Chris Quote Link to comment Share on other sites More sharing options...
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