I was researching how wiring multiple drivers in series or parallel affect TS parameters. I decided to capture everything in a table and decided to share here in case this is useful to others. Explanations can be found following the table. I will also include tables for the 4-driver parallel-series case and for Dual-voice-coil driver wirings.
Explanations:
Fs:
Wiring drivers together has no effect on the resonant frequency.
Fs = 1/sqrt(2π*Cms*Mms)
With two drivers, the moving mass doubles, but the compliance halves, so these factors cancel.
Re:
Just like any resistor: wired in parallel, equivalent resistance halves; wired in series, equivalent resistance doubles.
Qes:
(Ignoring any external damping factor)
Qes = 2πFs*Mms*Re/((Bl)^2)
In series: The mass doubles, Re doubles, Bl doubles and Fs remains the same. Because Bl is squared in the denominator, these factors cancel.
In parallel: The mass doubles, Re halves and Bl and Fs remain the same. These factors cancel.
Qms:
Qms = 2πFs*Mms/Rms
With two drivers, the moving mass doubles, but so do the mechanical losses (e.g. friction). These factors cancel.
Qts:
Since Qes and Qts remain unchanged, so does Qts.
Qts = 1/(1/Qms + 1/Qes)
Vas:
Intuitively, with two drivers, the volume of air having the same compliance as the driver’s suspension doubles. You need twice as much air to push back against two driver suspensions compared to just one.
Vas = 𝜌c^2*Sd^2*Cms
With two drivers, Sd doubles, but Cms halves. Because Sd is squared, this results in a net factor of 2.
Sd:
The total moving surface area is the sum of each driver’s moving surface area.
Xmax:
Xmax does not change, since each driver's Xmax is still the limiting factor.
Mms & Mmd
The total moving mass is the sum of each driver’s moving mass. (Same for air load)
BL
The magnetic flux density B = Wb/m^2 does not change. There is twice as much magnet flux (due to the additional magnets), but also twice as much cross-sectional area.
In series: the length of wire L in the magnetic field effectively doubles. The voice coils can be thought of as being arranged end-to-end in a magnetic field that is twice as long.
In parallel: the length of wire L in the magnetic field is considered unchanged. The voice coils can be thought of as being arranged overlapping in space (with the same starting point). Honestly, this is a bit unintuitive to me, but the formulas agree.
BL = sqrt(2πFs*Mms*Re/Qes)
With two drivers in series, Mms doubles, Re doubles, and Fs and Qes remain the same. So this grows by a factor of sqrt(4) = 2.
With two drivers in parallel, Mms doubles, Re halves, and Fs and Qes remain the same. So these factors cancel.
Le
Just like any inductors, in series the equivalent voice coil inductance doubles, in parallel it halves.
Cms
With two drivers, the effective compliance is halved. If 1 Newton moves a single cone 1 meter, then moving two cones 1 meter each would require 2 Newtons.
Cms = Vas/(𝜌c^2*Sd^2)
With two drivers, both Vas and Sd double. Because Sd is squared in the denominator, the total factor is 0.5.
Rms
With two drivers, the mechanical resistance (e.g. friction) doubles.
n0
Reference efficiency is defined as:
n0 = 𝜌(BL)^2*Sd^2/(2π*Mms^2*Re)
In series: BL doubles, Sd doubles, Mms doubles and Re Doubles. So the total factor is 4*4/(4*2) = 2. In dB, that is 10*log(2) = 3.01
In parallel: Sd doubles, Mms doubles and Re halves. So the total factor is 4/(4*½) = 2. Again, in dB that is 10*log(2) = 3.01
EBP
EBP = Fs/Qes
With two drivers, both Fs and Qes remain unchanged.
SPL, 1w@1m
(Assuming phase coherence)
dB(1w@1m) = 112.02 + 10log(n0)
With two drivers, n0 doubles, so dB(1w@1m) increases by 3dB.
SPL, 2.83v@1m
(Assuming phase coherence)
dB(2.83v@1m) = dB(1w@1m) + 10log(8/Re)
With two drivers in series: dB(1w@1m) is +3dB, and Re doubles, so 10log(8/Re) is -3dB.
With two drivers in parallel: dB(1w@1m) is +3dB, and Re halves, so 10log(8/Re) is +3dB.
| SINGLE DRIVER | TWO DRIVERS IN PARALLEL | TWO DRIVERS IN SERIES |
| Fs | Fs | Fs |
| Re | Re/2 | 2*Re |
| Qes | Qes | Qes |
| Qms | Qms | Qms |
| Qts | Qts | Qts |
| Vas | 2*Vas | 2*Vas |
| Sd | 2*Sd | 2*Sd |
| Xmax | Xmax | Xmax |
| Mms | 2*Mms | 2*Mms |
| Mmd | 2*Mmd | 2*Mmd |
| BL | BL | 2*BL |
| Le | Le/2 | 2*Le |
| Cms | Cms/2 | Cms/2 |
| Rms | 2*Rms | 2*Rms |
| n0 (efficiency) | 2*n0 (which is +3db) | 2*n0 (which is +3db) |
| EBP | EBP | EBP |
| SPL, 1w@1m | +3dB | +3dB |
| SPL, 2.83v@1m | +6dB | 0dB |
Explanations:
Fs:
Wiring drivers together has no effect on the resonant frequency.
Fs = 1/sqrt(2π*Cms*Mms)
With two drivers, the moving mass doubles, but the compliance halves, so these factors cancel.
Re:
Just like any resistor: wired in parallel, equivalent resistance halves; wired in series, equivalent resistance doubles.
Qes:
(Ignoring any external damping factor)
Qes = 2πFs*Mms*Re/((Bl)^2)
In series: The mass doubles, Re doubles, Bl doubles and Fs remains the same. Because Bl is squared in the denominator, these factors cancel.
In parallel: The mass doubles, Re halves and Bl and Fs remain the same. These factors cancel.
Qms:
Qms = 2πFs*Mms/Rms
With two drivers, the moving mass doubles, but so do the mechanical losses (e.g. friction). These factors cancel.
Qts:
Since Qes and Qts remain unchanged, so does Qts.
Qts = 1/(1/Qms + 1/Qes)
Vas:
Intuitively, with two drivers, the volume of air having the same compliance as the driver’s suspension doubles. You need twice as much air to push back against two driver suspensions compared to just one.
Vas = 𝜌c^2*Sd^2*Cms
With two drivers, Sd doubles, but Cms halves. Because Sd is squared, this results in a net factor of 2.
Sd:
The total moving surface area is the sum of each driver’s moving surface area.
Xmax:
Xmax does not change, since each driver's Xmax is still the limiting factor.
Mms & Mmd
The total moving mass is the sum of each driver’s moving mass. (Same for air load)
BL
The magnetic flux density B = Wb/m^2 does not change. There is twice as much magnet flux (due to the additional magnets), but also twice as much cross-sectional area.
In series: the length of wire L in the magnetic field effectively doubles. The voice coils can be thought of as being arranged end-to-end in a magnetic field that is twice as long.
In parallel: the length of wire L in the magnetic field is considered unchanged. The voice coils can be thought of as being arranged overlapping in space (with the same starting point). Honestly, this is a bit unintuitive to me, but the formulas agree.
BL = sqrt(2πFs*Mms*Re/Qes)
With two drivers in series, Mms doubles, Re doubles, and Fs and Qes remain the same. So this grows by a factor of sqrt(4) = 2.
With two drivers in parallel, Mms doubles, Re halves, and Fs and Qes remain the same. So these factors cancel.
Le
Just like any inductors, in series the equivalent voice coil inductance doubles, in parallel it halves.
Cms
With two drivers, the effective compliance is halved. If 1 Newton moves a single cone 1 meter, then moving two cones 1 meter each would require 2 Newtons.
Cms = Vas/(𝜌c^2*Sd^2)
With two drivers, both Vas and Sd double. Because Sd is squared in the denominator, the total factor is 0.5.
Rms
With two drivers, the mechanical resistance (e.g. friction) doubles.
n0
Reference efficiency is defined as:
n0 = 𝜌(BL)^2*Sd^2/(2π*Mms^2*Re)
In series: BL doubles, Sd doubles, Mms doubles and Re Doubles. So the total factor is 4*4/(4*2) = 2. In dB, that is 10*log(2) = 3.01
In parallel: Sd doubles, Mms doubles and Re halves. So the total factor is 4/(4*½) = 2. Again, in dB that is 10*log(2) = 3.01
EBP
EBP = Fs/Qes
With two drivers, both Fs and Qes remain unchanged.
SPL, 1w@1m
(Assuming phase coherence)
dB(1w@1m) = 112.02 + 10log(n0)
With two drivers, n0 doubles, so dB(1w@1m) increases by 3dB.
SPL, 2.83v@1m
(Assuming phase coherence)
dB(2.83v@1m) = dB(1w@1m) + 10log(8/Re)
With two drivers in series: dB(1w@1m) is +3dB, and Re doubles, so 10log(8/Re) is -3dB.
With two drivers in parallel: dB(1w@1m) is +3dB, and Re halves, so 10log(8/Re) is +3dB.
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| SINGLE DRIVER | FOUR DRIVERS IN A PARALLEL-SERIES (OR S-P) ARRANGEMENT |
| Fs | Fs |
| Re | Re |
| Qes | Qes |
| Qms | Qms |
| Qts | Qts |
| Vas | 4*Vas |
| Sd | 4*Sd |
| Xmax | Xmax |
| Mms | 4*Mms |
| Mmd | 4*Mmd |
| BL | 2*BL |
| Le | Le |
| Cms | Cms/4 |
| Rms | 4*Rms |
| n0 (efficiency) | +6db |
| EBP | EBP |
| SPL, 1w@1m | +6dB |
| SPL, 1w@2.83v | +6dB |
This table follows by multiplying the factors in the “series” and “parallel” columns above.
For Dual voice coils on a single driver:
With a dual voice coil driver, the mechanical properties (Fs, Qms, Sd, Xmax, Mms, Mmd, Cms, Rms) are unaffected by the wiring. Only the electrical properties are affected.
Re and Le behave like normal resistors and inductors.
BL behaves as described above in the two-driver table. It doubles in series and remains unchanged in parallel.
Qes is halved compared to wiring a single coil, which also affects Qts. Adding a resistor in series with one voice coil (but not the other) can be used to achieve any value between [Qes/2, Qes].
Qes = 2πFs*Mms*Re/((Bl)^2)
With two voice coils in series, this changes by a factor of 2/(2^2) = 0.5
With two voice coils in parallel, this changes by a factor of 0.5/(1^2) = 0.5
Efficiency increases by 3.01dB
n0 = 𝜌(BL)^2*Sd^2/(2π*Mms^2*Re)
With two voice coils in series: this changes by a factor of 2^2/2 = 2 or +3.01dB
With two voice coils in parallel: this changes by a factor of 1/0.5 = 2 or +3.01dB
Sensitivity follows from efficiency, and is the same as in the two-driver table.
I believe this is all correct, but please check my work!
Some references:
https://www.birotechnology.com/articles/mass.html
https://en.wikipedia.org/wiki/Thiele/Small_parameters#Small_signal_parameters
https://www.bestcaraudio.com/a-look-at-series-and-parallel-subwoofer-wiring/
| DVC driver with 1 coil (of 2) connected | DVC driver with coils wired in parallel | DVC driver with coils wired in series |
| Fs | Fs | Fs |
| Re | Re/2 | 2*Re |
| Qes | Qes/2 | Qes/2 |
| Qms | Qms | Qms |
| Qts | 1/(1/Qms + 2/Qes) | 1/(1/Qms + 2/Qes) |
| Vas | Vas | Vas |
| Sd | Sd | Sd |
| Xmax | Xmax | Xmax |
| Mms | Mms | Mms |
| Mmd | Mmd | Mmd |
| BL | BL | 2*BL |
| Le | Le/2 | Le*2 |
| Cms | Cms | Cms |
| Rms | Rms | Rms |
| n0 (efficiency) | +3dB | +3dB |
| EBP | EBP | EBP |
| SPL, 1w@1m | +3dB | +3dB |
| SPL, 1w@2.83v | +6db | 0dB |
With a dual voice coil driver, the mechanical properties (Fs, Qms, Sd, Xmax, Mms, Mmd, Cms, Rms) are unaffected by the wiring. Only the electrical properties are affected.
Re and Le behave like normal resistors and inductors.
BL behaves as described above in the two-driver table. It doubles in series and remains unchanged in parallel.
Qes is halved compared to wiring a single coil, which also affects Qts. Adding a resistor in series with one voice coil (but not the other) can be used to achieve any value between [Qes/2, Qes].
Qes = 2πFs*Mms*Re/((Bl)^2)
With two voice coils in series, this changes by a factor of 2/(2^2) = 0.5
With two voice coils in parallel, this changes by a factor of 0.5/(1^2) = 0.5
Efficiency increases by 3.01dB
n0 = 𝜌(BL)^2*Sd^2/(2π*Mms^2*Re)
With two voice coils in series: this changes by a factor of 2^2/2 = 2 or +3.01dB
With two voice coils in parallel: this changes by a factor of 1/0.5 = 2 or +3.01dB
Sensitivity follows from efficiency, and is the same as in the two-driver table.
I believe this is all correct, but please check my work!
Some references:
https://www.birotechnology.com/articles/mass.html
https://en.wikipedia.org/wiki/Thiele/Small_parameters#Small_signal_parameters
https://www.bestcaraudio.com/a-look-at-series-and-parallel-subwoofer-wiring/
Great info, thanks!
Funny I was just wondering about this subject. I was wondering besides the final resistance load the amplifier see's, what other things there were to consider between wiring in series vs parallel. It was the idea behind taking advantage of the statements you made above about drivers in phase that are parallel vs series that are 1w@2.83v and that drivers in parallel will have +6db while in series will have +0db.
I recently picked up a pile of cheap JBL GX1200's in that sale to tinker around with. I did an open baffle with two of them and wired them in series since that's all I could do and still use it with my amplifiers (I don't have spare 2ohm stable amps).
I measured them to see what would change:
Here's a single JBL GX1200 with all its free air parameters too.
Then I wired two in series (4ohm each) and the result here on my multi-meter is 8.7ohms. Since the amp see's this as the load the output goes down and per your table I don't gain the increase in SPL from the two drivers ultimately because while I do, it's lost again due to the output loss from the increase in impedance.
Here's the two drivers wired in series on the DATS to show what changed. I guess now I'm curious how to interpret the Q values relative to the information above because they changed but the information says the parameters should cancel. But these changes do have meaning do they not?
Overall my take away from all this is that for sub woofers it's not a good idea to wire in series and it should basically be used to also wire in parallel. It definitely got me thinking. Instead of trying to build things with two drivers in series that are 4ohm and become 8ohm, it's better probably at least in most situations to have 8ohm drivers and wire in parallel to 4ohm to get the increase in SPL as intended.
Very best,
Funny I was just wondering about this subject. I was wondering besides the final resistance load the amplifier see's, what other things there were to consider between wiring in series vs parallel. It was the idea behind taking advantage of the statements you made above about drivers in phase that are parallel vs series that are 1w@2.83v and that drivers in parallel will have +6db while in series will have +0db.
I recently picked up a pile of cheap JBL GX1200's in that sale to tinker around with. I did an open baffle with two of them and wired them in series since that's all I could do and still use it with my amplifiers (I don't have spare 2ohm stable amps).
I measured them to see what would change:
Here's a single JBL GX1200 with all its free air parameters too.
Then I wired two in series (4ohm each) and the result here on my multi-meter is 8.7ohms. Since the amp see's this as the load the output goes down and per your table I don't gain the increase in SPL from the two drivers ultimately because while I do, it's lost again due to the output loss from the increase in impedance.
Here's the two drivers wired in series on the DATS to show what changed. I guess now I'm curious how to interpret the Q values relative to the information above because they changed but the information says the parameters should cancel. But these changes do have meaning do they not?
Overall my take away from all this is that for sub woofers it's not a good idea to wire in series and it should basically be used to also wire in parallel. It definitely got me thinking. Instead of trying to build things with two drivers in series that are 4ohm and become 8ohm, it's better probably at least in most situations to have 8ohm drivers and wire in parallel to 4ohm to get the increase in SPL as intended.
Very best,
There is no loss of efficiency, there is merely a rearranging of the Voltage and current which the amp supplies at the time.
Per the table there's no gain in SPL with two drivers in series with the 1w@2.83v such as with subwoofers. Correct? So the doubling of impedance cancels the potential increase in SPL from the driver? The parallel wiring shows no loss in this scenario and has the +6db increase in SPL which is significant. So am I reading something incorrectly? Seems to me that regardless of it just being a rearrangement of voltage and current, one has +6db APL and one does not. That matters to me.
Very best,
Ah minor typo in the table, that should read 2.83v@1m. I fixed the original post but can’t seem to edit the comments (second & third parts)
No, I think you're focussing on the wrong thing. Acoustically, the gain in SPL is the same either way. You're supplying less power to the speaker but when you turn up the amp to the same level, it's all the same... excursion, individual driver power consumption, amplifier power etc.
The sensitivity is based on Voltage, yet an amplifier can supply the same power through any combination of Voltage and current, in other words, any impedance.
Funny, I knew what you meant, and read it that way, yet still typed the same thing.
Very best,
Ok, maybe I'm focusing on the wrong thing then.
I'm trying to understand how the table shows +6db when wiring in parallel with 2.83V@1m, compared to +0db when wiring in series with 2.83V@1m. Same two drivers. Same amplifier. Same everything. Just the wiring difference (assuming the amp can handle the load stable at whatever it would be, like 2ohm or 4ohm depending on the drivers of course). As a blanket table, this fools me into thinking I get more from parallel wiring since there's no context.
Very best,
You have -6dB power to the drivers due to 1/4 the current into 16 ohms compared to 4 ohms.
Another way of saying the same thing is that the Voltage is halved for each and so the current is halved for each.
Some amplifiers may like 4 ohm loads or some 16 ohm loads.. but in any case the power to the drivers and from the amplifiers can be set the same for a direct comparison of this parameter (to +6dB). The efficiency will remain the same.
Another way of saying the same thing is that the Voltage is halved for each and so the current is halved for each.
Some amplifiers may like 4 ohm loads or some 16 ohm loads.. but in any case the power to the drivers and from the amplifiers can be set the same for a direct comparison of this parameter (to +6dB). The efficiency will remain the same.
Thanks, it makes more sense in context, the idea of having to change voltage for the load explains it better. It was just confusing on the table for me out of context. Solved. 🙂
Very best,
I think you’re both right.
When there are 8ohm and 4ohm variants of the same driver, typically they have the same SPL at 1w@1m. The 4ohm has +3db SPL @ 2.83v@1m vs the 8ohm. See for example https://aespeakers.com/shop/ibht-woofers/ib12ht/
Per the table, wiring the two 4ohm drivers in series will give an additional +3db @ 1W@1m and +0db 2.83v@1m. Final R is 8ohms.
So total 4-4:
R: 8 ohms
SPL @ 1w@1m: +3db (vs a single 8 ohm driver)
SPL @ 2.83v@1m: +3db (vs a single 8 ohm driver)
For 8ohm drivers in parallel, per the table there is +3db @ 1w@1m and +6db @ 2.83v@1m. The final R is 4ohm.
So total 8||8:
R: 4 ohms
SPL @ 1w@1m: +3db (relative to a single 8 ohm driver)
SPL @ 2.83v@1m: +6db (relative to a single 8 ohm driver)
The 3db difference @2.83 can purely be attributed to the lower R value in the parallel case. You are right that this can be advantageous if sensitivity is your concern, since most amps are rated at a higher wattage for 4ohms vs 8ohms.
When there are 8ohm and 4ohm variants of the same driver, typically they have the same SPL at 1w@1m. The 4ohm has +3db SPL @ 2.83v@1m vs the 8ohm. See for example https://aespeakers.com/shop/ibht-woofers/ib12ht/
Per the table, wiring the two 4ohm drivers in series will give an additional +3db @ 1W@1m and +0db 2.83v@1m. Final R is 8ohms.
So total 4-4:
R: 8 ohms
SPL @ 1w@1m: +3db (vs a single 8 ohm driver)
SPL @ 2.83v@1m: +3db (vs a single 8 ohm driver)
For 8ohm drivers in parallel, per the table there is +3db @ 1w@1m and +6db @ 2.83v@1m. The final R is 4ohm.
So total 8||8:
R: 4 ohms
SPL @ 1w@1m: +3db (relative to a single 8 ohm driver)
SPL @ 2.83v@1m: +6db (relative to a single 8 ohm driver)
The 3db difference @2.83 can purely be attributed to the lower R value in the parallel case. You are right that this can be advantageous if sensitivity is your concern, since most amps are rated at a higher wattage for 4ohms vs 8ohms.
Thanks; I guess I need to see if this is true in practice and just test an amplifier with them wired in series vs parallel and set the volume pot to max and do a sweep. If the SPL is different, well, that will answer things. I realize its a system and so the amp matters in this as not all amps are happy at 2ohm compared to 4ohm. Most of my amps are all 4~8ohm happy. I have one amp that is ok with 2ohm I could test with. I figure if I max the volume pot, the only thing that is changing is the results based on the wiring. Right? Then again I bet the amp behaves differently at 2ohm than 4ohm than 8ohm. Bleh.
Very best,
Setting volume to max and doing a sweep sounds like a good way to destroy your speakers ;-)
You should be able to leave it on any volume setting and compare SPL. My understanding is that the volume knob sets the voltage, not the power.
My amp isn't that powerful. The drivers I'm talking about are sub drivers. Ok, if I can leave it parked at a specific place on the knob and then just test 2ohm vs 8ohm, with two drivers testing parallel vs series, I don't know if it will really be showing the wiring difference or the amp difference with those two loads.
Very best,
You can build amps which prefer loads in the hundreds of ohms (where attaching 8 ohms to them is pointless), but if you had a speaker array using multiple drivers and used them in series, you'd be good. The amp may have plenty of Voltage to offer and not too much current. Not all amps are even Voltage sources.. but getting back to the point, you don't lose efficiency by choosing this method alone.
@gregulator if you want a thread split or some editing, PM me.
I have wondered that myself, particularly when it comes to altering a Zobel in an MTM, for instance. You take a driver such as the one you have, or an 8 ohm driver such as the Dayton DC160-8, wired in parallel. It also shows a large impedance increase as the frequency rises.