I have wanted to try a first-order series crossover for some time, and to this end I decided to build a FAST/WAW out of drivers and parts already at hand. I have a fullrange with a nominal 8 ohm rating and a woofer with a nominal 4 ohm rating that mate well in terms of SPL at my chosen crossover point, and I have coils and capacitors enough to make a series crossover. The woofer has a rising response in the bass so BSC will not be necessary, and I want to try the crossover without impedance compensation on either driver.
I'm having difficulty analyzing what happens in a first-order series crossover when using 4 ohm and 8 ohm drivers together. All the examples I have seen in my research have shown 8 ohm drivers only, and I'm trying to realize what would happen with the coil and capacitor when they shunt their respective drivers at different impedances. It seems that I should choose the coil value based on the four ohm woofer and the capacitor value based on the 8 ohm fullrange, but it also seems that the electrical crossover slopes would have wildly changed overlapped responses. My understanding is that a first-order series crossover compensates for changes in impedance values by shifting the crossover point slightly, but again that assumes drivers with the same impedance ratings.
I'm hoping that some experienced members can help to ease my confusion. Again, my goal is to build this speaker with parts I have in stock if possible. If it proves impractical to use these drivers with a series crossover I will buy the components to make a standard parallel crossover design.
I'm having difficulty analyzing what happens in a first-order series crossover when using 4 ohm and 8 ohm drivers together. All the examples I have seen in my research have shown 8 ohm drivers only, and I'm trying to realize what would happen with the coil and capacitor when they shunt their respective drivers at different impedances. It seems that I should choose the coil value based on the four ohm woofer and the capacitor value based on the 8 ohm fullrange, but it also seems that the electrical crossover slopes would have wildly changed overlapped responses. My understanding is that a first-order series crossover compensates for changes in impedance values by shifting the crossover point slightly, but again that assumes drivers with the same impedance ratings.
I'm hoping that some experienced members can help to ease my confusion. Again, my goal is to build this speaker with parts I have in stock if possible. If it proves impractical to use these drivers with a series crossover I will buy the components to make a standard parallel crossover design.
I'll start the ball rolling by paraphrasing the words of G.A. Briggs - he of Wharfedale fame.
As each crossover element is in series with one driver and in parallel with the other, their reactance figure should be similar at the crossover frequency.
Their values cannot be changed independently as they can with a parallel network.
As each crossover element is in series with one driver and in parallel with the other, their reactance figure should be similar at the crossover frequency.
Their values cannot be changed independently as they can with a parallel network.
Our first MTM WAW was just such a thing. 8Ω FR midTweeter, a pair of 6.5” midBass in parallel = 4Ω.
Just means the inductor will be half the value of is it was an 8Ω woofer.
dave
Just means the inductor will be half the value of is it was an 8Ω woofer.
dave
To the OP, you should model the xover and not use calculators. This currently is a recipe with failure and disappointment being likely.
The Magnepan 3.7 used a series crossover with 3 ohm tweeter, 8 ohm midrange, and 4.6 ohm woofer, so it has been done in commercial products as well.
If you're trying to do this/understand it in your head or with basic equations, I'd suggest modeling it. Series crossovers are inherently a bit unruly due to the interactions that happen. How you adjust things can also swing the load more inductive or capacitive. Edit: I think wolf_teeth beat me to it by a minute 🙂
With a parallel crossover you have far more latitude to make things do what you want. With series you don't have as much control over independent responses. With a simple series crossover the model and result will likely be pretty close. The more shaping/adjusting you need to do, the more unpredictable things get and the more likely you'll need to adjust things on the fly with the model as more of a guideline - but an important one.
If you're trying to do this/understand it in your head or with basic equations, I'd suggest modeling it. Series crossovers are inherently a bit unruly due to the interactions that happen. How you adjust things can also swing the load more inductive or capacitive. Edit: I think wolf_teeth beat me to it by a minute 🙂
With a parallel crossover you have far more latitude to make things do what you want. With series you don't have as much control over independent responses. With a simple series crossover the model and result will likely be pretty close. The more shaping/adjusting you need to do, the more unpredictable things get and the more likely you'll need to adjust things on the fly with the model as more of a guideline - but an important one.
Last edited:
I haven't a clue what a WAW is, but if it is what used to be known as a D'Appolito configuration, why not connect the two 4 ohm woofers in series so you get 8 ohm?
If I did the mathematics correctly, you can still get a normal first-order crossover response when the woofer and tweeter impedances are different:
Choose L = Rwoofer/(2π fcrossover)
Choose C = L/(Rwoofer Rtweeter)
Of course you have all the usual advantages and disadvantages of a first-order series filter.
Choose L = Rwoofer/(2π fcrossover)
Choose C = L/(Rwoofer Rtweeter)
Of course you have all the usual advantages and disadvantages of a first-order series filter.
WAW = Woofer Assisted Wideband (XO 200-500ish). D’Appolito popularized the MTM = MidBassTweeterMidBass (XO 1.2-3k) and the name stuck — he was far from the first.
dave
I've done a few and the standard parallel formula seems (is heard) to apply, a small tweak by ear to optimize sometimes helpful. My easy-to-remember, do-in-head math: first-order LPF~(160hz*ohm)/mH and HPF~(160khz/ohm)/uF; second-order fix mH*uF per first-order (note frequency~5khz/sqrt(mH*uF) if equal impedance) then shape curves multiplying mH and dividing uF by same amount, sqrt(2 3 4) for Butterworth/Bessel/Linkwitz-Riley.
Driver impedances are not flat and can rise a lot at high frequency where XO occurs. First-order series is supposed to be less affected by impedance changes than parallel XO, though they theoretically sim' indistinguishable.... (I keep being told).
Driver impedances are not flat and can rise a lot at high frequency where XO occurs. First-order series is supposed to be less affected by impedance changes than parallel XO, though they theoretically sim' indistinguishable.... (I keep being told).
Last edited:
Parallel XO parts size can just be wired in series and one gets the basic XO, but if you juggle the values maintaining the same XO point you adjust the zeta = 𝝵 which is an adjustment of the shape of the roll-offs.
dave
dave
Last edited:
In a series crossover, the drivers are in parallel for the purposes of damping of the filter. The impedance variations are shared/apportioned in that respect.
TubeCAD has quite a long tour on series XOs.
https://www.tubecad.com/2017/11/blog0402.htm
https://www.tubecad.com/2017/11/blog0403.htm
If he covers zeta he doesn’t use the term.
dave
https://www.tubecad.com/2017/11/blog0402.htm
https://www.tubecad.com/2017/11/blog0403.htm
If he covers zeta he doesn’t use the term.
dave
This actually applies to second order filters, not first order. (excluding coil DCR & variable impedance correction)
Good golly gents.
The answer to the OP's question is.......the system impedance will transition from 4 ohms to 8 ohms in the crossover range.
Very simple.
Dave.
The answer to the OP's question is.......the system impedance will transition from 4 ohms to 8 ohms in the crossover range.
Very simple.
Dave.
YES >
The question was very clear & simple from the start.
Halve the inductance value that would be used for an 8ohm woofer.
(tweak L for woofer and C for tweeter if necessary 🙂
The question was very clear & simple from the start.
Halve the inductance value that would be used for an 8ohm woofer.
(tweak L for woofer and C for tweeter if necessary 🙂