Introduction to designing crossovers without measurement

[AllenB]

The resistance has an effect which depends on where you use it, among other things. Used before the woofer, this resistance will waste power reducing the woofer's level slightly. Normally this effect is small and reasonably evenly spread across the woofer's range.
Any resistance in the inductor used for the tweeter will reduce the effect of the filter. Again, the effect is small.

Theory suggests it should be kept an order of magnitude (one-tenth) lower than the circuit it is a part of (as an example, one-tenth of 8 ohms). If you buy an inductor which is intended for use in a crossover, you should expect that the designer had this in mind.

[AllenB]

Some notes for a three-way midrange. There are some special considerations although it's possible to approach the two crossovers separately first treating the mid as if it were a tweeter, then as if it were a woofer. This way it gets the treatment on both ends The relevant aspects will still be the same e.g. the low end of a mid (its impedance, response etc..) looks and acts similarly to the low end of a tweeter etc..

One thing that designers sometimes choose to do is use just a capacitor to remove the bass, after all a midrange will sometimes be more rugged in its application than a tweeter. The good thing about using a second order filter is that it will be less critical of the impedance peak that is found down at the mid driver's resonance (just a part of why I chose to use one for the tweeter). Simply put, the capacitor and inductor tend to pay more attention to each other than of just the impedance of the driver. (If measurements are involved then you have more choice in the matter).

If you happen to have a mid driver that is a few dB less sensitive than the woofer, you can try crossing them at the baffle step frequency, there is a chance it will be helpful there.

If your mid needs to be brought down in level you might consider using an L-pad. (Although the resistor layout I used for the tweeter tutorial will also work, the inductor (low-pass from the woofer tutorial) will end up being a little larger as you should add the value of the series resistor to the driver impedance just before you calculate that inductor). If you decide to use an L-pad you could use an online calculator as the maths is more involved than for the other components. Put the impedance compensation (resistor and capacitor) directly before the mid driver (if the L-pad is used to bring the mid down more than a few dB, you might even be able to dispose of this compensation, but it doesn't hurt). Put the L-pad just before the impedance compensation and don't use the other (tweeter tutorial) resistors, then put the high and low-pass filters before that.

The order of placing these two filters with each other shouldn't matter. At low frequencies the low-pass filter almost acts as if it is not there, and vice-versa with the high pass. Therefore they cannot really see each other so to speak. In reality they will interact a little and may push the mid cross points out wider than you intended, or make the middle of the range a little louder. These could be tweaked by adjusting the filter components or level resistors but there is a better suggestion to begin with.....If you keep the mid upper and lower cross points a couple of octaves or so apart, you'll have less high/low interaction.

Hope this helps.

[gpapag]

I think Allen's work deserves a place in the Articles section.
Thank you Allen

George

[shmb]

Love this thread. Well done.

Just a quick general question, when tweaking the woofers crossover by increasing the capacitor value (in the impedence compensation network) will this have a negative effect on overall system impedence? ie: lowering it too much. I would like to ideally maintain roughly 6-8 ohms impedence overall. (woofer is rated 8 ohm)

Reason I ask is, this works really well with one of the systems I'm tweaking at the moment. But I had to increase the capacitor from 10uF to approx 22uF. Being a bit concerned, I then increased the resistor value from 6.8 ohm to 8.2 ohm thinking it will sort of counteract it without changing the sound too much. It sort of worked. (I think) Am I being too cautious?

I realise this is a very general question, but I'm just trying to get a grip on how the theory works.

Thanks

[AllenB]

Thank you gpapag and shmb.

This is a good question, and I thought it would help if I showed an example. I used an example similar to the tutorial with a 6.8Ω resistor, an 8.2µF capacitor, which I then increased to 40µF which seemed extreme enough to prove a point. In all cases the grey traces are the raw driver, green is the normal case, and magenta is the extreme case.

The first image shows the effect of the impedance compensation on the driver. There are six traces, the lower three are the impedance curves that are normally shown, and the upper three are the impedance phase (which is not the same as acoustic phase). Although technically both work together, it is sometimes easier to just look at the normal impedance curve, but with regards to phase, just look at whether it is close to zero degrees or far from it meaning for example is it an easy 8 ohms to drive, or a not so easy 8 ohms to drive? If that helps.

So, the impedance drops to 4 ohms around 1kHz with the large value of capacitor, and as the phase is near zero degrees for that case, it will act similarly to a simple 4 ohm resistor, a load that most amps will handle.

The second image shows the impedance once the 1mH inductor is added. This is the impedance that matters because it is what the amp will see. By adding the inductor, the effect of using the larger capacitor is partially hidden from the amp... the impedance is well on the rise at 1kHz and upward.

The lowest impedance here is around 5 ohms. If you look at around 600Hz, the phase is a little further from 0 degrees although still less than 90 degrees. You could take a non-technical guess that this makes it equivalent to maybe a 4 ohm load at that point.

Summarising the effect of the larger capacitor (looking at the first plot), there is little difference at the highest frequencies because all capacitors act like a short circuit at high frequencies. The effect is to act on the impedance at a lower point than before, encroaching on the lower midrange as the capacitor grows in value.

I've also shown the frequency response on the second plot (the upper two traces) for interests sake. The end result it is showing in this case, is that the greatest effect of that inductor will be above 1kHz, and the large capacitor is producing an effect both above and below 1kHz. The impedance reduction below 1kHz may be an unwanted side effect, but if it gets you the result you want then it is valid, and in this case relatively harmless.

[shmb]

Thanks once again, that helps a lot.

One last thing. Will increasing the resistor value from 6.8 to about 12 ohms sort of help bring it back up a bit? I'm guessing it will, but will that have any other side effects?

I've already done this as a trial and it doesn't seem to have made as much of an audible difference as increasing the capacitor did (which is good) so I've left the 12ohm in there for now.

[AllenB]

By and large, increasing the resistor will compensate like you say... but there is a catch. Impedance is complex (literally meaning it has a phase component). Generally speaking, if you overlook this you'll be OK most of the time but I thought I'd stretch the point, starting with your stated values.

If you look at the first sim attached below, yellow is 6.8Ω / 8.2µF, blue is 6.8Ω / 22µF, and red is 12Ω / 22µF (I've also changed the scales). My example will not be the same as yours but it should give a good idea of what can happen.

Both 22µF cases have a more similar frequency response than the other so that may explain why they sound more similar. The impedance is also generally more high for the 12Ω / 22µF case but this case in fact shows the lowest point of any of them.

Taking it to an extreme with the second plot, I took 6.8Ω / 8.2µF, and changed the resistor to 25Ω, and to zero. If you look at the highest frequencies they are in order of the size of the resistors, but in the mid band, the higher the resistor, the lower the impedance.

[shmb]

So it seems to compensate in the upper area but not in the lower area where it probably matters more? So maybe its better just to leave the lower value in there, ....mmm.
I'll keep playing.

Thanks for the detailed responses, very much appreciated.

[Triumph]

Great article!!! Very informing on a accessible way. AllenB thanks.

Can anyone help with article(s) for tutorials concerning:
Introduction to designing crossovers with measurement.

[qusp]

Allen, mate I cant thank you enough for this, i'm going the digital crossover route, but all the same the information held within is invaluable!