I couldn't quite figure out the VituixCad box simulator. The diffraction is pretty clunky. I have heard of BoxSim, I will try it out. Are the graphs it generates exportable? I assume it can do multiple woofers at once correct?
Got it. I did notice the cap values were much higher.
Cost wise, a higher value electrolytic cap to ground seems basically free compared to a low DCR air core or even iron core inductor.
In general with everything else being equal i guess that's a analogy.
But keep in mind that when you increase current draw, the heating goes up proportionally due to the halved impedance. So it is not 100% correct.
And Power compression is not linear, dependent on the driver, heatsinks, voice coil type/material/venting , motor venting etc.
Well, compared to a single woofer, using two in parallel halves the impedance and doubles the current draw, yes. So it increases the total heat energy dissipated. But that's only because there are two drivers. Looking at each driver on its own, if the input is the same, the heating will be exactly the same as if it were being used on its own. And you're forgetting that the SPL also increases by 6dB. To keep the SPL the same as you'd get from a single woofer, you would need to reduce the drive by 6dB, so each woofer would experience significantly less heating.
Overall, it doesn't matter whether you connect a pair of drivers in series or parallel, you're gaining efficiency because of the doubled cone area. So for the same SPL, you will have less total heat dissipation. And that dissipation is split across two drivers, so looking at them singly, the reduction in heating is even more significant.
Last edited:
I don't mean to derail this conversation for my own selfish needs, but I'd like to ask a question.
When we think of stacking drivers what happens? Imagine a single driver directly on a ground plane in a minimalist box and no other surfaces. I expect to have no bounces. Just like why we would do ground plane measurements. Imagine I then take another identical driver in another minimalist box and stick it directly on top. What's the correct way to model this? Do I now have 2 separate radiating sources? One at the ground plane and one above it? Or do I think of this as a single radiating surface with 2x the vertical radiating surface?
When, if ever do we make the transition of modelling this as a line array with a planar wavefront??
In this case imagine we are only interested in low frequencies at which the woofer would be otherwise producing an omni wavefront. Long before it would beam.
Are my questions confusing enough? 🙂
When we think of stacking drivers what happens? Imagine a single driver directly on a ground plane in a minimalist box and no other surfaces. I expect to have no bounces. Just like why we would do ground plane measurements. Imagine I then take another identical driver in another minimalist box and stick it directly on top. What's the correct way to model this? Do I now have 2 separate radiating sources? One at the ground plane and one above it? Or do I think of this as a single radiating surface with 2x the vertical radiating surface?
When, if ever do we make the transition of modelling this as a line array with a planar wavefront??
In this case imagine we are only interested in low frequencies at which the woofer would be otherwise producing an omni wavefront. Long before it would beam.
Are my questions confusing enough? 🙂
It depends on wavelength you're thinking of, and the listening/measuring distance. You can model it in Edge. 😀
Are you talking about isobaric? Rule of thumb there is you halve the VAS. WINISD can model this configuration. Maybe I'm not envisioning what you're putting out though. Have a picture?
I haven’t red the whole thing but a couple comments.
1/ most t”tower” speakers are ML-TLs and not reflex, a reflex modeler will not work.
2/ Add an additional woofer of the same type to a box, the bass extention will not reach as far down. Double the box volume and otherwise keep the alignment the same, bass response is the same but you have 3dB more potential level.
Did this a long time ago: https://www.t-linespeakers.org/design/multidriver.html
dave
1/ most t”tower” speakers are ML-TLs and not reflex, a reflex modeler will not work.
2/ Add an additional woofer of the same type to a box, the bass extention will not reach as far down. Double the box volume and otherwise keep the alignment the same, bass response is the same but you have 3dB more potential level.
Did this a long time ago: https://www.t-linespeakers.org/design/multidriver.html
dave
For purposes of my post, I only meant to discuss dispersion and reflection as a function of radiating area. Assume a 12" woofer in a sealed cabinet at ground level. Then we add a second identical woofer in an identical cabinet on top.
The bottom woofer would have no floor bounce possible. My guess was that the second woofer would not either until the wavelength approached the driver diameter.
The bottom woofer would have no floor bounce possible. My guess was that the second woofer would not either until the wavelength approached the driver diameter.
"What happens" depends on the wavelength considered.
You would model it as two drivers in a box of double the size.
At low frequencies you can consider the two drivers acting as one.
A "ground plane" measurement actually considers the driver's baffle mounted flush with the ground, pointing at the sky.
At low frequencies, drivers being perpendicular to the ground plane makes no difference.
Line arrays do not have a "planar wavefront", they create destructive interference patterns in the near field, the length of the near field varies with frequency, becoming longer as the frequency increases.
From:
https://www.prosoundtraining.com/2010/03/17/line-array-limitations/
The near field distance can be defined by the following relationship:
D=1.57 L squared/λ
where
D is the distance to the far field transition
L is the physical length of the line source
λ = the wavelength of the frequency in question (all lengths in identical units).
Beyond this distance the listener is in the far field and there is 6 dB drop in level per doubling of distance. The transition distance can be quite long at short wavelengths, that is, high frequencies, but it is shorter at low frequencies. For each octave lower in frequency, the transition distance is cut in half.
Adequately confusing, hope the answers are less so.
Art
Dave,
Yes, speakers that are mass loaded transmission lines are modeled differently than a Helmholtz resonators (bass reflex enclosures), but there are probably more sealed or BR "tower" speakers than ML-TL designs.
Other than the reduction (or increase, if reverse mounted) in box volume from adding a woofer, there is no change in the tuning frequency (Fb) of a bass reflex box, the low frequency extension therefore remains the same.
+6dB more potential SPL, +3dB from sensitivity gained by doubling the radiation area, +3dB from doubling power.
Art
Last edited:
Ah, I thought this was usually done with the driver facing perpendicular to the floor, often turning a speaker on it's side. I did not realize the need to flush mount and point upwards.
Even if the box has the same tuning, the voilume is different so it will be a different alignment.
+3db for double the Sd, end sensitivity depends on how you wire them. If you wire them in series you get -3dB from half the power.
dave
Thanks Art!! So in that formula there's no term for the drivers themselves. That is, there's no place to put in "'I'm using 12" drivers..." but instead only the length of the total array is brought into account. Interesting!
Thanks again.
So based on Arts formula, assuming I have 2 x 12" drivers, one on top of another, for 300 Hz :
(1.57 * 2' * 2') / 1125'/300 Hz = 1.57 * 4 / 3.75 = 1.7'
So, far enough away I'd no longer suffer the constructive and destructive patterns directly, but I'd still potentially suffer any destructive interference from the top woofer reflecting off the floor (or ceiling really)
(1.57 * 2' * 2') / 1125'/300 Hz = 1.57 * 4 / 3.75 = 1.7'
So, far enough away I'd no longer suffer the constructive and destructive patterns directly, but I'd still potentially suffer any destructive interference from the top woofer reflecting off the floor (or ceiling really)
I don't really think I need a picture. Imagine a 12" woofer in a sealed box on the floor, woofer perpendicular to the floor. A typical subwoofer for instance. Then stack a second on top of it. The cabinet volume is irrelevant to the question I'm asking because I'm not asking about volume or bass extention. I'm asking about reflection and radiating surfaces. I guess my question could also be done with woofers mounted on an infinite baffle. Start with an infinite baffle. As close to the floor as possible we put a 12" woofer. bass performance is immaterial. Then we add 1, 2 or 3 woofers vertically.
The bottom woofer we treat as immune to floor bounce since it's exactly or nearly at floor level. The question I'm asking is, if you add woofers, in a box or in an infinite baffle, whether 1 or 5 more on top, if we treat them as separate point sources or as a single driver the height of which is the total height of the drivers. It sounds as if the answer is that they are each point sources (below their beam frequency) and therefore the possibility for reflections off the floor exist.
I'm also not asking about SPL vs. radiating into a half or quarter sphere. Just want to talk about reflections.
Yes, this makes sense. When you add another woofer it really depends on the frequencies that they're playing. Keep the woofers crossed around 200Hz and spaced within 18" of the mid range and it works great. When there are two woofers spaced vertically, you have different responses from each woofer so make one have a floor bounce where the other couples with the floor and you can even it out. You can also integrate the floor bounce frequency into the crossover. I don't know all of the math on it but you can simulate it.
The thing here is that it's not just floor bounce. They "bounce" off of the rear walls, side walls and ceiling. So the goal is to keep the woofers spaced at different intervals from all of these walls so the peaks and dips in the magnitude response are not amplified. If the woofer is equal distant to more than one wall then the dip is even deeper. So, spacing the woofers from each other is less of an issue than how they end up being spaced from a wall. And for the designer, the only boundary that's mostly guaranteed is the floor. Most ceilings are around 8' so we can factor that. Then the side and rear walls are a crap shoot. But if you put at least one woofer near the floor then you move the floor bounce frequency of that woofer above the crossover point and therefore get a more even in room response.
Alison speakers were interesting because they had two woofers but they're both on the floor and at 90 degrees apart. The designer's theories on all of this are interesting and make sense.
First off the 6db of gain is only theoretical, and 3db is coming from doubling the amplifiers current draw. You halve the impedance/double the current draw. That means the amp looses efficiency, there is also extra heat loss in the amp, and in wiring and drivers. Result is that you normally end up in the 5,6-5,9 range, because the amp lost some of it's 'headroom'. Feel free to disagree, but that is a usual effect of it. Oversize the amp enough and you'd get there eventually, when it no longer gets strained. Also mutual coupling between the drivers that give you this, is dependent on wavelength. If you can squeeze them in under 1/2WL distance, at all frequencies in band, then i theory yes.
So then let me provoke some thought?
-Having 2 set of woofers connected in parallel and series, ignoring limitations and reality for a bit. With resulting 'ideal' imp. of 4 and 8 ohm.
- When the 2 pairs of drivers are creating identical dbSPL. The series circuit needs a higher setting/more gain from the amp, meaning a increased voltage.
-What is the total dissipated power in each circuit ? 🤔
-And what is the relation between voltage and current over each load?
To take a example from a different place altogether.
You have 2 motors outputting 10kw.
Ran at 230 volts in one case and 400 volts in one case.
Which one uses more current?
The one that uses the most current generates proportionally more heat.
The resulting conductor CSA can be reduced for a reason, hint current draw.
Last edited:
@Arez Indeed, amplifiers often struggle when driving the lower impedance of paralleled drivers. Not always though.
Well, even if the amp heats up more when driving a pair of drivers (and it may not, since drive level will be 6dB lower to compensate for the 6dB sensitivity increase of parallel drivers) it's wrong to say that the drivers will heat up more.
I'll say it again: for the same SPL, a pair of drivers, connected either in series or parallel, will use less power than a single driver. This isn't controversial. It's basic physics.
Let's put some numbers on it. Suppose you drive a single 4 ohm driver with a 2V signal (and we'll pretend it's flat 4 ohm impedance - maybe it's a planar magnetic driver 🙂 ). Current draw will be 0.5 amps, power will be 1 watt (amost all which is dissipated as heat).
Now add a second identical driver in series. Impedance is now 8 ohms. With the 2V signal, SPL will remain the same. (Of course, in the real world, the SPL may not be exactly the same, to multiple decimal places, due to driver matching issues,, directivity, etc. But it will be close. Note that the theory here does not depend at all on mutual acoustic coupling effects, and where these exist, they are likley to further increase the acoustic output, not reduce it.) Total current draw (using Ohm's law) will be 0.25 amps now, and total power consumption will now be 0.5 watts. The signal is split evenly between the two drivers, 1V each. Power dissipation in each driver is 0.25 watts. Yes, a quarter of what the single driver had to deal with.
Now lets connect our pair of 4 ohm drivers in parallel. To keep SPL the same, we need to halve the signal to 1V. Total impedance is now 2 ohms. Total current draw is 0.5 amps (same current draw as the single driver had). Total power dissipated is 0.5 watts. Looking at the drivers individually, each one has 1V signal, 4 ohms impedance, so 0.25 amps current, 0.25 watts of power.
In summary, if you use a pair of drivers, then for the same SPL, each driver will have to dissipate only 1/4 of the power that a single driver would., and that's true whether the connection is in series or parallel.
Well, even if the amp heats up more when driving a pair of drivers (and it may not, since drive level will be 6dB lower to compensate for the 6dB sensitivity increase of parallel drivers) it's wrong to say that the drivers will heat up more.
I'll say it again: for the same SPL, a pair of drivers, connected either in series or parallel, will use less power than a single driver. This isn't controversial. It's basic physics.
Let's put some numbers on it. Suppose you drive a single 4 ohm driver with a 2V signal (and we'll pretend it's flat 4 ohm impedance - maybe it's a planar magnetic driver 🙂 ). Current draw will be 0.5 amps, power will be 1 watt (amost all which is dissipated as heat).
Now add a second identical driver in series. Impedance is now 8 ohms. With the 2V signal, SPL will remain the same. (Of course, in the real world, the SPL may not be exactly the same, to multiple decimal places, due to driver matching issues,, directivity, etc. But it will be close. Note that the theory here does not depend at all on mutual acoustic coupling effects, and where these exist, they are likley to further increase the acoustic output, not reduce it.) Total current draw (using Ohm's law) will be 0.25 amps now, and total power consumption will now be 0.5 watts. The signal is split evenly between the two drivers, 1V each. Power dissipation in each driver is 0.25 watts. Yes, a quarter of what the single driver had to deal with.
Now lets connect our pair of 4 ohm drivers in parallel. To keep SPL the same, we need to halve the signal to 1V. Total impedance is now 2 ohms. Total current draw is 0.5 amps (same current draw as the single driver had). Total power dissipated is 0.5 watts. Looking at the drivers individually, each one has 1V signal, 4 ohms impedance, so 0.25 amps current, 0.25 watts of power.
In summary, if you use a pair of drivers, then for the same SPL, each driver will have to dissipate only 1/4 of the power that a single driver would., and that's true whether the connection is in series or parallel.