I guess it's obvious I'm new here. I've seen several subwoofer designs on the 'net, but not any mention of a zobel for them - is it just too obvious to mention, or not a good idea? Sounds like a good idea to me - if they're good for a woofer, why not a sub?
Because over the frequency range of a sub the zobel will only have a very slight effect,
🙂 /sreten.
🙂 /sreten.
Not to mention with a direct connection to the amplifier it is not needed, assuming that you are using an active crossover and a dedicated sub amp.
Zoebels attempt to correct for a driver's rising impedance with frequency so the crossover behaves more like it was driving a resistor, which is what the standard formulas assume.
Trying to explain how it works:
Think of a crossover as a voltage divider. Using a first order low pass as an example, at crossover frequency the impedance of the coil should be the same as the woofer's impedance, dividing the applied voltage in half for a 3db down response.
The impedance of the coil rises with freqency. If the woofer were a purely resistive load, the voltage across it would fall at a nice straight first order slope. Look at a woofer impedance curve and notice that it also rises with frequency. Pick a few spots and calculate the actual drive voltage and compare to the predicted voltage if the woofer impedance was constant. If the crossover frequency is up near the driver's rolloff, the curve difference gets pretty big. The woofer is not attenuated nearly as much as you would expect, and all that nasty cone breakup is at a high level.
If you were using a passive sub crossover, at that top of a subs range, the driver impedance curve is relatively flat and the response closely matches filter response. So as Sreten points out, compensation is not needed. Also wiht an active filter, the woofer drive is the amp output. The line level crossover sets that and the woofer impedance does not affect the crossover slope
The woofer impedance rise is not a purely inductive. The zoebel provides a falling impedance in parallel with the woofer approximately equal the rise. The producet of R and C determines the corner frequency of the. The ratio determines the shape of the slope, or Q.
The published Zoebel formulas will give a decent starting point for impedance compensation. If you want to really nail it, measure the impedance of the driver and Zoebel, and tweak accordingly. Of course you are doing the same sort of thing when you tweak crossover values. I've found it easier to get the crossover sounding right by starting with tweaking the zoebel, YMMV.
Zoebels attempt to correct for a driver's rising impedance with frequency so the crossover behaves more like it was driving a resistor, which is what the standard formulas assume.
Trying to explain how it works:
Think of a crossover as a voltage divider. Using a first order low pass as an example, at crossover frequency the impedance of the coil should be the same as the woofer's impedance, dividing the applied voltage in half for a 3db down response.
The impedance of the coil rises with freqency. If the woofer were a purely resistive load, the voltage across it would fall at a nice straight first order slope. Look at a woofer impedance curve and notice that it also rises with frequency. Pick a few spots and calculate the actual drive voltage and compare to the predicted voltage if the woofer impedance was constant. If the crossover frequency is up near the driver's rolloff, the curve difference gets pretty big. The woofer is not attenuated nearly as much as you would expect, and all that nasty cone breakup is at a high level.
If you were using a passive sub crossover, at that top of a subs range, the driver impedance curve is relatively flat and the response closely matches filter response. So as Sreten points out, compensation is not needed. Also wiht an active filter, the woofer drive is the amp output. The line level crossover sets that and the woofer impedance does not affect the crossover slope
The woofer impedance rise is not a purely inductive. The zoebel provides a falling impedance in parallel with the woofer approximately equal the rise. The producet of R and C determines the corner frequency of the. The ratio determines the shape of the slope, or Q.
The published Zoebel formulas will give a decent starting point for impedance compensation. If you want to really nail it, measure the impedance of the driver and Zoebel, and tweak accordingly. Of course you are doing the same sort of thing when you tweak crossover values. I've found it easier to get the crossover sounding right by starting with tweaking the zoebel, YMMV.
Hi BE,
Allthough impedance compensation helps with crossovers it
also helps with amplifiers by reducing the reactivity of the load,
thus reducing the stress on the amplifier output stage.
Still not relevant in a sub though.
🙂 /sreten.
Allthough impedance compensation helps with crossovers it
also helps with amplifiers by reducing the reactivity of the load,
thus reducing the stress on the amplifier output stage.
Still not relevant in a sub though.
🙂 /sreten.
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