The Xenover

[GRollins]

So, what is an active crossover? The short answer is that it’s an electronic filter designed to divide frequencies into ranges suitable for tweeters, midranges, woofers, and subwoofers. Yes, and supertweeters, too…I know you’re out there.

In an ideal world we wouldn’t need filters. You’d only need one driver. Its frequency response would go from DC to light and it would get earsplittingly loud from a single watt. In the here and now, we are faced with drivers that only cover a few octaves well and need copious amounts of power to achieve reasonable volumes. Some way to combine drivers would be nice—using the extended top of one with the deep bass of another. It can be done, of course. All you have to do is assign the deep-ranging frequencies to the woofer, and the high frequencies to the tweeter. Why not let each driver carry all the music; just operate the woofer and the tweeter together in parallel or series, each playing the entire musical spectrum? Because it wouldn’t sound good, and because tweeters can’t take the sort of abuse they would suffer if asked to play organ pedal notes. So we need to separate the high frequencies from the low frequencies and send them to their respective drivers.

Most speakers use passive crossovers. They are comprised of resistors, capacitors, and inductors, usually mounted within the speakers themselves. They contain no active devices, require no external power, and, barring the occasional tweeter level control, don’t need adjustment.

An active crossover performs the same function, but does it further upstream. You’ll need at least two amplifiers, which leads us to the only real downside of multi-amping—cost. Amplifiers cost money. However, at this point enough people have built Alephs, X-variants, and whatnot that there are spare amplifiers sitting on the shelf gathering dust. An active crossover would give folks an excuse to use all that excess hardware.

Once you get past the price, it’s pretty much uphill from there. Multi-amp systems have numerous benefits. Passive crossovers are inefficient. Not all the power that goes in gets to the driver. A passive crossover burns off valuable wattage as heat, gaining you nothing in return. An active crossover loses signal too, but it does so earlier in the amplification process, when gain is cheap. Another benefit is increased damping factor. The damping factor is a ratio; divide the impedance of the driver by the output impedance of the amplifier. All things being equal, a bigger number is better. Basic Newtonian physics: An object in motion tends to remain in motion unless acted upon by an outside force. Start a woofer cone moving forward and it will tend to keep going, even though it’s time to turn around and head the other way. The driver’s suspension and some cabinet designs help oppose the wayward driver’s flight, but they could use some help. Enter another law of physics: For every action, there is an equal and opposite reaction. There is an electronic analog to this that we can use to put the brakes on. First, you have to stretch your imagination and look at the driver as a microphone. Vibration of the cone will produce electric current at the speaker terminals. If you short the driver’s terminals together, the cone suddenly finds itself faced with the current it just created, and in one of those cosmically cool coincidences, the current comes back into the driver via the short and immediately tries to push the driver the opposite direction from the way it’s moving, thus bringing it to a halt much more quickly. So what’s that got to do with amplifiers? Pretend you’re a driver, looking up through the speaker cable into the back of the amplifier. There’s a resistance there—well, actually an impedance—that varies from amplifier to amplifier. It’s generally pretty low, and that’s good. A passive crossover generally has a number of components in it, some of which are effectively increasing the impedance seen by the driver as it looks back up towards the amplifier. This lowers the damping factor considerably. By getting rid of the passive crossover, the driver can get all the benefit of that low Z output—which then serves as the ‘short’ that allows the speaker to put on the brakes. Another benefit to multi-amping is the fact that good parts cost money, and higher voltage, higher wattage parts cost exponentially more money. For the cost of a single decent high voltage cap to go into a passive crossover, you can buy an entire active crossover’s worth of small caps.

So now you’re convinced, right? You want to do the multi-amp thing and put all those Zens and Alephs and JLHs and single-ended tube amps to work. What’s next? Well, you need a topology. There are lots of ways to build an active crossover. Two of the most common are Sallen-Key (named after the guys who developed it), and multiple feedback. So which one is best? That’s a trick question. You’re seeking an absolute answer to a relative question. If you phrase it ‘Which one is better?’ you’ll be closer to getting an answer you can live with, but that still begs the question of,"better for what?" Multiple feedback filters have a clear advantage in certain areas. For instance, they’re easy to design for high Q and they’re adaptable for balanced signals. High Q is rarely needed in audio electronics, but the balanced feature might be nice to have. So what’s Sallen-Key got going for it? Well, it’s capable of relatively easy adjustments and, under the right circumstances, it simplifies parts selection. Hmmm. What to do? We could go either way, but ease of adjustment is a powerful thing to have in our favor, so we’ll reluctantly drop the ability to filter a balanced signal and go with the Sallen-Key.

With the Sallen-Key topology in mind, we find that we need an active component of some sort. It can be as simple as a follower, or so complicated that a tribe of EE’s get lost without road maps. The follower is seductive due to its simplicity, but some gain would be nice, so I’ve drawn the schematics with an opamp. Which opamp? Any opamp. You can build yourself a discrete opamp like the ones that Nelson Pass outlined in his paper on DIY opamps on the www.passdiy.com website, or you can buy a simple chip opamp at Radio Shack if you’re desperate. Try not to get that desperate.

Next comes the pesky question of what to do about the other parts. While it’s possible to build an active filter with inductors, we won’t need them. Resistors and caps are all the Sallen-Key topology requires. Parts quality? How good do you want it to sound? I use the Vishay/Dale 1% resistors from Mouser. Do you need to buy 1% resistors? Not really, but they’ll give you finer control over your crossover points. For caps, I lean towards polystyrene. Again, Mouser has them in a decent selection of values. If you want to use polypropylene or silver/mica or whatever, that’s fine. The circuit won’t care. Just try to have a reasonable match between the two channels, as odd things can happen to the image if one channel’s crossover point is higher than the other’s.

I’m using a slightly strange system of notation on the schematic. The reason being that the parts nomenclature goes along with what the part does, not its position on the schematic or due to any relation to the math behind the scenes. The first set of resistors and caps the signal sees on the way in define the second pole. The second set define the first pole. What’s a pole? A pole is a multiple of 6 dB/oct. A one pole filter is 6 dB/oct. A two pole filter is 12 dB/oct. Three poles equals 18 dB/oct., and so forth. Sorry, they come in sets of six. That’s just the way the universe works. If you want to go for an in-between number, you’re shading off into shelving networks and such and that’s a whole ‘nother ball of wax. Maybe some other day. In any event, C1 will be the cap for pole 1, C2 for pole 2, R1 for pole 1, and R2 for pole 2. I don’t show it on the schematics, but you can use jumpers to cut out pole 2 entirely and make a two pole filter a single pole filter. Purely selfish of me to name things in such an odd way, but it’ll save me from having to carry around a schematic all the time. If we’re talking about R2, I’ll immediately know that it’s the resistor that helps set the frequency for the second pole of the filter.

While there’s nothing patented about this circuit, I’d like to thank Nelson Pass for two ideas which I lifted wholesale from his commercial crossover, the XVR1. I’ve been building crossovers for quite some time, and though some of them had selectable frequencies, I always hard-wired the Q. The idea of selectable Q tickled me so much that I dropped what I was doing and redid everything so as to have the ability to set the Q. Neat. The other thing is a purely practical matter. Headers. You know…those little jumper things they use in computers to set configurations. Cheap, zillions of positions, gold-plated. Love it. So when’s the last time you priced a ten-position switch of decent quality? I betcha it cost more than 39 cents. Mind you, if you want to use Elma or Electroswitch or Alps or whatever’s popular this month, they’ll do just fine. Better than fine. But they’re likely to cost as much as all the other parts put together. For frequency selection you don’t need 100,000 cycle switches. You’ll be working things hard if you change frequencies as much as a hundred times during the life of the crossover. Headers are your friends. Particularly if you don’t have an infinite R&D budget.

And the name of the circuit? Well, remember that X is pronounced as Z, as in xylophone. Also, xeno comes from the Greek language, meaning strange or alien. To some people, this crossover may seem a little strange. So is it Xen-over or Xeno-ver? I’ll leave that up to you.

I’m nearing the limit for length on posts, so I’m going to break here and start another chapter.

Grey

[GRollins]

Frequency selection:
The crossover frequency is selected by the values of R1, R2, C1, and C2. I’ve drawn the schematics to show a single cap value and three resistor values for each pole. Simply jumper one resistor from the R1 bank and one resistor from the R2 bank. You’re done. You can easily set it up for more than one cap and as many resistors as you can afford to buy. If you want real flexibility in frequency selection, use a pot instead of fixed resistors.

Now, let’s get down to the nitty-gritty. How do you decide what values to use for your resistors and capacitors? There’s only one formula you need to know:

F=1/(2*PI*R*C)
where
F=frequency
R=resistance
C=capacitance in Farads

Generally, you’ll want to put the same values in the R1 bank as you put in the R2 bank. For the purposes of the schematic, I arbitrarily chose a nominal crossover frequency of 500Hz, then bracketed it with values at +-10%. You can change the crossover frequency in either of two ways. You can crank the handle on the formula until the answer pops out, or you can simply scale the parts values in the schematic. It’s an inverse linear relationship. If you want a nominal crossover frequency of 50Hz, then that’s one-tenth of 500Hz. You then increase either the cap or the resistor values by a factor of ten. Since 681k is going to sound like a really high value to a solid state person (tube folks sprinkle 681k resistors on their breakfast cereal—doesn’t bother them a bit), you might be happier going with a 47000pF cap instead.

The idea of selecting the poles separately is a little scary. Most people, most of the time, are going to want to switch them together. That is, if you select resistor R1b, then you’re going to want to select resistor R2b at the same time. This will give you a nice, conventional crossover slope. But if you like to walk on the wild side, you can choose different values. That will give you a compound slope. A compound slope is one that starts at one rate, but changes to another, steeper slope later on. If the values you select are near each other, you won’t notice much difference, but if you select frequencies two or three octaves apart, you can get some really interesting curves. Use this feature to equalize drivers that have a rising response near the end of their usable frequency range (we’ll deal with falling response when we get to Q). If you want to get a two pole rotary switch and switch both poles together, that’ll be just great. The separate pole adjustment thing just sort of falls out when you go with headers for frequency selection. There’s no law that says you have to use it.

Good caps are generally more expensive than resistors, so I set the circuit up with a minimum number of caps and used resistors for frequency selection. If you happen to have a box full of caps you’re itching to use, feel free to reconfigure the circuit to have switchable cap values. For more versatility, use lots of each. If you really want to go to the limit, jumper more than one value at a time. This will place the selections in parallel, and may take a little calculation to get the desired result, but jumpers are only a few cents apiece, and you can fiddle to your heart’s content.

How do you know which slope to choose? A good place to start is to duplicate what the manufacturer was using in their passive crossover. Same frequency, same slope. It’s often in the specifications. Once you get used to that, you can try varying things a bit. If you’re starting from scratch, building a DIY speaker, you’re going to have to wing it. There are too many driver choices to try to give guidelines here, although you generally won’t want to go much below about 2kHz with most tweeters; 3kHz is safer still. If you don’t have something specific in mind, stick with a 12 dB/oct slope until you get further along in the design process. Yes, there are people who are passionate about first order crossovers, and for every one of them, there’s a Linkwitz-Riley fan who’s just as rabid about 24 dB/oct slopes. 12 dB/oct is a good compromise to start with. You can always change it later.

First order slopes are easy and the Sallen-Key topology does a good job of handling second order, but what do you do if you want a third or fourth order slope? Add two stages together. A 6 dB/oct stage adds to a 12 dB/oct stage to yield 18 dB/oct, just as you might expect. Likewise, two 12’s yield a 24 dB/oct slope. Crossover slopes above fourth order are rare in audio work.

Not all 12 dB/oct slopes are created equal, nor are 18’s or 24’s or any of the higher order slopes. Only 6 dB/oct slopes sidestep the question of Q. Q is a measure of how "peaky" or "droopy" a crossover slope is at its cutoff frequency. Q is best illustrated with graphs, which I can’t easily generate and post. Look at the final page of the owner’s manual for the Pass Labs XVR1 (available online in PDF format at www.passlabs.com) and you’ll find graphs showing the effect of Q on the slope of the crossover. Nelson has three settings for Q on the XVR1: Low, Medium, and High. If my scribblings are correct, these correspond to Qs of .5, .63, and 1. Note that Q doesn’t have a named unit, like Ohms or Farads, it’s just a number. The schematic shows three crossover selections, .577, .707, and 1. The lowest setting, .577, has a name: Bessel. Likewise, the middle selection, .707, is called Butterworth. These are merely convenient labels for specific points on a continuous line. There’s no reason you can’t design a crossover with a Q of .8153 or any other figure that you might need, although it’s unusual to need anything below .5 or above 2. The Bessel configuration’s claim to fame is that it has the least phase shift. Butterworth has the flattest passband response. The .63 figure that Nelson appears to use for the XVR1 is a good compromise between the two, allowing for flat response with little phase shift. A Q of 1 is useful for bumping up a sagging frequency response curve right before the rolloff. It will provide a boost of 1 dB. Higher boosts are possible; a Q of 2, for instance, will provide 6 dB of boost, but this sort of trick is best used in moderation as it can take a considerable amount of amplifier power and be hard on the driver. Approach equalization with caution. Like fire, it is a good servant, but a bad master.

Suppose you want a proper Butterworth 18 dB/oct filter. How would you go about putting one together? Choose a crossover frequency. Then calculate convenient resistor and capacitor values for that frequency. You’ll need a 6 dB/oct filter and a 12 dB/oct filter. In principle, you can put them in either order, but if you put the 6 first, you’ll get a slight advantage in noise reduction. The last thing to do is to set the 12 dB/oct filter section for a Q of 1. Why 1, if Butterworth is supposed to be .707? Because the 6 dB/oct filter is a bit droopy and the slight boost from the 12’s Q of 1 offsets this and the two work hand in hand to create a good rolloff characteristic. There are extensive tables of ratios to be used for setting filters to various desired Qs, and eventually this thread will probably be filled with scads of suggestions for various filters. It’s a tedious job to transcribe all that information, so I’ll leave it for later, as I’m running short on time.

I’ll close with one final note on how to set Q should you want something other than the ratios that I put in the example schematic. Q is determined entirely by the ratio of Rfb and RQ. That’s one of the benefits of using this filter configuration. It doesn’t really matter whether you change Rfb, RQ, or both. In the final analysis, only the ratio between them matters. In my case, both the schematic and my PCB layout worked best when I left Rfb constant and varied RQ. The formula for determining Q is:

Q=1/(3-(1+(Rfb/RQ)))

The main caveat here is that you’ll want to stick to low Q values. If you start pushing the circuit too hard, it gets really, really sensitive to parts values and you’re going to need to go to very tight tolerance parts in order to get predictable performance. For most purposes, audio filters don’t need to go above a Q of 2 or 3, so this shouldn’t be a problem in the real world.

Not all of this is as clear as I’d like, but one of the nice things about an interactive format like this is that we can hammer at it until (almost) everyone is satisfied. I still haven’t covered phase relationships, amp selection, and power requirements, but that’ll have to wait. I’m going to go pull some gear and go camping.

Grey

[GRollins]

Okay, the thread is open for business. Sorry about the size of the files, but I've spent waaay too much time on this and I ran out of juice trying to shrink them down while retaining enough resolution.
I'll try to get back in here later this evening.

Grey

[Nelson Pass]

I say we call it the Rollover.

[Variac]

Grey, This is just what I needed. A kindergarten chat about the subject. Lots of us know how hard it is to write clearly, so your efforts are appreciated. Maybe eventually it can go in the wiki
as part of a basic primer aas you had mentioned in the past.

I'm using a balanced preamp and amp, but the CD into the pre is unbalanced. So, do I need to put the crossover before the preamp(s) and have 2 preamps?

I say we call it the Rollover.

If you agree, it's up to Dave to change the posts till now. It wouldn't be hard at this point!

Nelson, have you surrendered to the High/Lo Pass name for your future effort?

[ergo]

Grey, thank you very much.

You have a wonderful way of explaining things. Sometimes it is needed just to have this kind of explanation to finally "get" something that you have anyway tought about and studied for quite a while.

Just for curiousity, what king of solution are you using in place of the "opamp" and if you have tried many, which seems to give the best results in your mind?

Ergo

[macka]

Ergo,

I have possed a similar question in the Grey High Low Pass Please Thread.

macka

[amo]

Unfortunately I do not know enough about electronics to agree or disagree with Uli's comments, however I do have a few burning questions (BTW thanks for going over the basics!!):

1: The crossover looks fascinating due to its simplicity. However, lets say a person has built a dual-mono, all balanced BOSOZ, and hopes to preserve the entire signal chain as balanced. With this in mind, could that person build two identical crossovers as described above, and use both to process the two halves of the balanced signal??? Just think how cool it would be to have 4 rows of "precision ganged" jumpers!

2: (perhaps a little of subject - sorry) I have previously read that a balanced system rejects noise by eliminating elements not common to both signals, the regular and the inverse. I further read that "this finally happens when the two halves of the signal meet at the speaker terminals". Please help me comprehend how this actually happens. I come from the computer science side of things, and this would make perfect sense if there was some intelligent piece of hardware that filtered noise not common to both signals. But there seems to be no circuitry for that, and it happens completely due to the laws of physics. Some insight would be most helpful.

Thanks again for the great effort!!!

[GRollins]

amo,
1) Yes, in principle, you can handle each phase of the signal this way, but it's not a truly balanced circuit. It would be a good idea to run everything through a differential afterwords to clean up the accumulated garbage, small though it may be.
2) Look at a differential. There are two gain devices in a sort of a Y shape, with the bottom of the Y being a resistor or current source. If you put a balanced input into a differential, you'll get a balanced signal out--minus anything that was common mode. When you say that something is common mode, you're saying that it shows up the same in both the + and - phases of the signal; same amplitude, same phase, same everything. An example of a common mode signal might be hum due to the signal passing near a transformer.
What happens is this: Let's say a signal goes in the left half of the differential. It's amplified, and in the process the gain device (whether tube, bipolar, or FET) draws current from the base of the Y. At the same time, the other phase of the balanced signal is being amplified by the other half of the differential, but it's going in the opposite direction--if the left half is going positive, then the right half is going negative. This works wonderfully, because as the left half of the differential pulls more current, the right half is using less, then they reverse and as the right half takes more, the left half is releasing exactly the same amount. It's a marvelous balancing act, like well matched, experienced dancers whirling across the floor. It looks effortless.
If a common mode signal is presented to the two inputs they both try to pull or push current at the same time. It's as though two kids on a see-saw both decided to go up at the same time. It doesn't work. The way a see-saw is built means that in order for one kid to go up, the other must go down. Got to. Period. If they both try to go up, nothing happens. The same thing happens in a differential. If you put in a common mode signal, nothing comes out. Well, almost nothing. In the real world, the parts aren't perfectly matched, or the current source doesn't have infinite impedance, etc. and a tiny amount leaks through anyway. This imperfection is rated using a specification called Common Mode Rejection Ratio (aka CMRR--lots less typing). More CMRR is good. It means that the differential is functioning closer to the ideal. It won't ever quite get there, but you can sneak up on it by tweaking this and that.
To have the common mode signal 'cancel at the speaker' you're pretty much going to have to have a balanced amp, e.g. a bridged amp. Nelson's X amps are bridged. The Aleph-X is bridged, but normal Alephs are not. A bridged amp has two halves that amplify separate phases of the signal (even though the circuit may not be based on a differential), then present them to the two ends of the speaker. Pretty much the same thing happens as with the differential. If both sides go positive at the same time, the speaker sees nothing, so common mode stuff is lost in the wash.
If all else fails, you sometimes have to sit down and draw it out on a piece of paper and watch the current go this way and that. Hint: The laws of thermodynamics apply, even though they have different names in electronics. You can't get current from nowhere, and you can't just make it 'go away.'
I'm not quite clear on where Fred was going with his post. If it's meant as a reply to your questions, it seems to be pretty foggy.

Grey

[Variac]

I'm using a balanced preamp and amp, but the CD into the pre is unbalanced. So, do I need to put the crossover before the preamp(s) and have 2 preamps?

I'm still wondering about this..........
I'm using a SOZ (balanced) and a Hafler) (unbalanced)
with a BSOZ

But want to maintain the balanced connection between the BSOZ and SOZ........