If I was to make a 2x15" subwoofer, (for concert application), is it plausible to have it end up at 8 ohms? Eminence don't seem to make anything bigger/better than the Delta 12 in 16 ohms, and wiring two 4 ohm drivers in series would seem to be inviting trouble if one of them blows. Or would this be alright?
If you mean that if one goes open circuit they'll both go silent and the amp will see no load, then yes.
Yeah, that's the point - wiring them in series would could cause them both to fail if one did.
PD and some other manufactuers seem to make 16ohm drivers, so maybe it's just that eminence don't stock them.
PD and some other manufactuers seem to make 16ohm drivers, so maybe it's just that eminence don't stock them.
Heres how I would look at it: if you might blow a woofer it is not big enough to handle the job, so go bigger and not worry about it.
Also, if one blows while driving two in parallel, you are going to have seiously reduced bass(-3dB from Sd, -3dB from doubling impedance) and this would suck so going bigger would prevent that. If one blows while wired in series(4ohm stable amp) you can bypass it and and not lose any spl, (-3dB for Sd reduction, +3dB for voltage sensitivity 4 instead of 8ohms). Of course you'll be driving the remaining woofer twice as hard.
Unless Im missing something about concert speakers the logic is sound
To be honest, blowing them's not really a likely occurance. I'm not going to be driving them any harder than they're rated, it was just as a possibility.
So if I ran two 4ohm 1000w speakers in parallel, I'd effectively have a 2000w 8ohm cabinet?
Would there be any difference in efficiency compared to any other combination? Wiring speakers in series in cabinets wouldn't be something that would seem instinctively right.
So if I ran two 4ohm 1000w speakers in parallel, I'd effectively have a 2000w 8ohm cabinet?
Would there be any difference in efficiency compared to any other combination? Wiring speakers in series in cabinets wouldn't be something that would seem instinctively right.
In addition to this, if I had two 8ohm 1000w speakers and wired them in series, what would be the overall power rating of the cabinet?
I'm sorry about the basic questions, I am usually fairly clued up on live sound and the like but this is just something I want to be sure of.
tom daghdha said:
PD, (and most other pro manufacturers), make 16 Ohm drivers specifically so you can quad them up to give a 4 Ohm load.
16 Ohm 2000w.
Thanks for that, it was just the power handling I wasn't sure about. In your opinion, would there be any advantage in either configuration over the other? (2 series wired 4 ohm drivers, or 2 paralleled 16 ohm drivers.)
Hi,
I think this must have been a typo
I agree with Pinky. Wire speakers in parallel not series.
If the amplifier cannot work efficently into a low load value then drive them independantly with a pair of amplifiers.
I think this must have been a typo
you did mean 2ohm when in parallel and 8ohm when in series.
I agree with Pinky. Wire speakers in parallel not series.
If the amplifier cannot work efficently into a low load value then drive them independantly with a pair of amplifiers.
AndrewT said:
Reading that back, I made the mistake of switching parallel and series, yes.
What I was trying to establish was whether you can just add the power ratings when wiring speakers in parallel, assuming drivers are identical? So two 16ohm 1000w drivers in parallel would equal an 8ohm 2000w cabinet?
I'd thought series would be a bad idea, for the reasons you suggested. A failed driver could stop the entire sound.
It just seemed that the 4ohm Eminence drivers were more available than 16ohm versions, so I thought it might have been a common wiring configuration.
I'd rather have 8ohm cabinets, to parallel up to 4ohms.
Hi,
wiring drivers in series is a lot worse than saving one from blowing up when you fry the first.
The problem is each driver sees the other as a series resistance (impedance) that completely screws up the bass response and varies as the resonant peak of one affects the other. The peaks may not even coincide.
wiring drivers in series is a lot worse than saving one from blowing up when you fry the first.
The problem is each driver sees the other as a series resistance (impedance) that completely screws up the bass response and varies as the resonant peak of one affects the other. The peaks may not even coincide.
Andrew, that is a commonly thought mistake. The responses are identical in every way whether a single, two parallel or two series drivers are used. This was covered in a thread sometime in the last 6 months.
edit: been trying to find the thread but no luck after a brief try
edit: been trying to find the thread but no luck after a brief try
Something not mentioned is that if you found 2x4ohm drivers and 2x16ohm drivers and all happened to have the same sensitivity (not that they necessarily would), the parallel combination would give the greater overall sensitivity.
Hi Richie,
I would like to read that thread.
I do not put out incorrect information lightly and if I need re-educating then so be it.
A series resistor and driver voice coil will raise the Q of the speaker.
A series resistor and driver voice coil along with a superimposed back EMF will have a complex effect on the speaker Q and other parts of the response.
I cannot understand how a thread could come to a different conclusion.
I would like to read that thread.
I do not put out incorrect information lightly and if I need re-educating then so be it.
A series resistor and driver voice coil will raise the Q of the speaker.
A series resistor and driver voice coil along with a superimposed back EMF will have a complex effect on the speaker Q and other parts of the response.
I cannot understand how a thread could come to a different conclusion.
Andrew like you I prefer to be corrected when appropriate. I assure you 100% what I said is true. Yes a series resistor raises Qes hence Qts also, but a voice coil does not. I have investigated myself when I have built twin driver subwoofers of various designs, as well as seen the theory.
I'm annoyed that I can't find the thread. Basically, DCR doubles but so does Zmax (this is the key point), i.e. the impedance curve is exactly the same shape as with a single driver, just the scale is doubled. As Q is derived from the -3dB points either side of the Zmax peak, ergo Q must be the same as a single driver.
I'm annoyed that I can't find the thread. Basically, DCR doubles but so does Zmax (this is the key point), i.e. the impedance curve is exactly the same shape as with a single driver, just the scale is doubled. As Q is derived from the -3dB points either side of the Zmax peak, ergo Q must be the same as a single driver.
nunayafb said:
OK 🙂
Whether the drivers are in series or parallel electrically, their acoustic interactions would add to increase output by between 3 and 6dB over the single driver.
Assume a total of 8ohms for each example and/or an ideal amplifier.
For the case of a matched volume setting and hence a matched voltage across the driver combos, the series drivers would each see half the voltage and hence, each driver in the series would be down by 6dB over each of the parallelled(sic?) drivers.
Hi Richie,
thanks for being patient with me.
But does it apply to separate drivers that are loosely coupled and/or have production spreads in tolerances?
thanks for being patient with me.
this is exactly the case with twin voice coils that are mechanically coupled together.
But does it apply to separate drivers that are loosely coupled and/or have production spreads in tolerances?
Andrew,
I haven't mastered this myself, but I have some thoughts. Also, I think richie00boy may have more or less said this.
I believe a driver that is fed from a current source can have its Q calculated by way of figuring the extra voltage across it due to the effect of its impedance under the current source, and working backward to get the Q.
The point is, the impedance (for one thing) doesn't fall apart without damping. If it did, I wouldn't measure my impedances with a current source for use with voltage sources.
So, I think it is the fact that a resistor (in theory, or Zo in practice) is of a fixed value and causes a biased voltage perturbation across the spectrum related to the impedance, that causes the Q to change.
In the case of the series speakers, the voltage is perfectly balanced (notwithstanding driver matching as I believe you mentioned).
I haven't mastered this myself, but I have some thoughts. Also, I think richie00boy may have more or less said this.
I believe a driver that is fed from a current source can have its Q calculated by way of figuring the extra voltage across it due to the effect of its impedance under the current source, and working backward to get the Q.
The point is, the impedance (for one thing) doesn't fall apart without damping. If it did, I wouldn't measure my impedances with a current source for use with voltage sources.
So, I think it is the fact that a resistor (in theory, or Zo in practice) is of a fixed value and causes a biased voltage perturbation across the spectrum related to the impedance, that causes the Q to change.
In the case of the series speakers, the voltage is perfectly balanced (notwithstanding driver matching as I believe you mentioned).
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