Multiple speaker wiring, serial + parallel impedance

[TobiasForsberg]

Hi, I have a question.

Say I have 3 speakers, two are identical, while the third is bigger. They are all 8 Ohms.
Say I wire the two identical speakers in series, and then wire the series in parallel with the third speaker, what happens to the impedance? and how does this affect the sound quality?

 

[Osvaldo de Banfield]

The impedance will be 12 R. 8 in // to 8 gives 4, then in series with 8 gives 12R total. But they must be the same characteristics, you can't wire a tweeter in series with a woofer, as example. All them must be the same type.

 

[Bob Richards]

It's 5.3 ohms total, and the third speaker will be 6dB louder than the two that were wired in series with each other. . The wording of the question was a little confusing.

 

[weltersys]

It's 5.3 ohms total, and the third speaker will be 6dB louder than the two that were wired in series with each other. . The wording of the question was a little confusing.

Bob,

The 16 ohm series pair get half the power (-3dB) of the third 8 ohm speaker in parallel with them, but doubling cone area (radiating surface) adds 3 dB, so the pair is the same sound pressure level as the single larger 8 ohm speaker which gets twice (+3dB) the power.
If the larger speaker is more sensitive than the smaller, it would be louder, but if it were less sensitive, the series pair would be louder.

Regarding sound quality, three speakers will have more comb filtering (peaky off axis response) than one, but if arranged in a vertical line, that won't be much of a problem, depending on size and frequency 😉.

The trio will definitely be louder and more sensitive than one speaker, subjectively louder is usually perceived as "better".

Art

 

[BlueWizard]

Equal Impedance speakers in Parallel =

Rt = R1/n

Where R1 is the impedance of one speaker and "n" is the Number of drivers.

Example - Three 8 ohm speakers in parallel.

Rt = R1/n = 8/3 = 2.667 ohms


Any number of equal or unequal speakers in Series -

Rt = R1 + R2 + R3 + R4 + etc....


TWO drivers of different impedance in Parallel -

Rt = (R1 x R2) / (R1 + R2)

Example - one 8 ohm driver in parallel with one 6 ohm driver -

Rt = (R1 x R2) / (R1 + R2) = (8 x 6) / (8 + 6) = 48/14 = 3.43 ohms


Many unequal drivers in Parallel -

Rt = 1 / [(1/R1) + (1/R2) + (1/R3) + ....]

If you have a common calculator this is easy. Most cheap calculators have a reciprocal key [1/x] meaning ONE divided by 'X' where 'X' is the number.

So, using a calculator with the key strokes shown in square brackets [] and numbers in round () brackets, here is the sequence for 6 ohm with 8 ohms with 16 ohms -

(6) [1/x] [+] (8) [1/x] [+] (16) [1/x] [=] [1/x]

Here is what you should see

6 [1/x] = 0.166666666
[+]
8 [1/x] = 0.125
[+]
16 [1/x] = 0.0625
[=]
= 0.3541666666
[1/x]
= 2.824 ohms


In your case, you have TWO 8 ohm in Series, so -

Rt = R1 + R2 = 8 + 8 = 16 ohms

Then you have another 8 ohms in parallel with that 16 ohm combination, so -

Rt = (R1 x R2) / (R1 + R2) = (16 x 8) / (16 + 8) = 128/24 = 5.33 ohms

However, as others have pointed out, you do not have an even distribution of power or signal.

The two speakers in series only get half the signal each. So, if the signal is 10v, then each series driver only gets 5v. While the larger driver in parallel gets the full 10 volts.

Again, as pointed out, you will not get equal sound out of all the drivers because they are not receiving equal signal.

Steve/bluewizard

 

[Bob Richards]

I could be wrong about the dB - maybe it is 3dB of acoustic power. The series drivers each get half the voltage relative to the 3rd driver. The question is too vague for any other conclusions.

 

[Cal Weldon]

Tobias,

Please show us or tell what speakers you have. This greatly affects how you wire them.

 

[TobiasForsberg]

Hmm... this is pretty interresting. the question was more in general, I don't have the speakers, but I was wondering how uneven/unequal wiring would affect the impedance and power, and also the quality. Have a lot of ideas though, and thinking about how different things should be done. Thanks for the formulas, much needed! 🙂

 

[BlueWizard]

There are combinations of speaker, assuming all the drivers are the same, where they can be wired equally, and have a workable impedance.

For example, it is possible to wire FOUR 8 ohms drivers in a Series/Parallel combination so the final impedance is 8 ohms.

The can be done in two ways, Series/Parallel or Parallel/Series.

Two 8 ohm speakers in Series = 16 ohms.

Two 16 ohm GANGS in Parallel = 16/2 = 8 ohms.

Or

Two 8 ohm speakers in Parallel = 4 ohms

Two GANGS of 4 ohm in Series = 2 x 4 = 8 ohms


The same can be done with NINE 8 ohm drivers.

Three 8 ohm drivers in Parallel = 8/3 = 2.667 ohm

Three GANGS in Series = 3 x 2.667 = 8 ohms

Or

Three 8 ohm drivers in Series = 3 x 8 = 24 ohms

Three GANGS in Parallel = 24/3 = 8 ohms


But there are a limited number of workable combinations. There really is no balanced workable combination of THREE drivers, unless the drivers are 16 ohms each, in which case we have 16 divided by 3 = 5.333 ohms which should be workable with just about any amp. However, it hinges on rare 16 ohm drivers.

Next, Speakers have Crossover Networks that filter and direct specific frequencies to specific speakers. So, an 8 ohm woofer, 8 mid-range, and an 8 ohm tweeters is 8 ohms because each driver deals only with a limited range of frequencies that is filtered out of the other drivers.

You are asking a very vague question, and as such, there can be no specific answer. If you give us an application of an idea for a speaker, we can help figure out what needs to be done.

But with your limited question, the answer is 5.333 ohms and unbalanced signal to the drivers, which equals unbalanced output from the drivers.

The better the question, the better the answer.

Steve/bluewizard

 

[BlueWizard]

Here is a graphic indicating 4 equal drivers combined for a final impedance equal to one driver.

If necessary, I could draw a similar diagram for NINE 8 ohms speakers.

Obviously, for practical reasons, you rarely see more than FOUR bass drivers.

Something with 9 or more drivers would be moving into the area of Line Arrays, which is a potentially workable idea, but a slightly different design process.

 

[weltersys]

Obviously, for practical reasons, you rarely see more than FOUR bass drivers.

Obviously you are not familiar with one of the most popular bass cabinets of all time, the Ampeg 8x10" used with the SVT amp, EIGHT 32 ohm drivers in parallel for a four ohm load 😉.

 

[BlueWizard]

I was not under the impression we were talking about Guitar Amp Speaker Cabinets. I assume Stereo Music speakers.

Further it is Eight 32 ohm drivers, find me any 32 ohm drivers anywhere?

Next, these are not full range speakers. Great for creating music, but suck for RE-Producing music.

Show me a stereo speaker that is not a Line Array, that uses more than FOUR bass drivers?

 

[tinitus]

... I was wondering how uneven/unequal wiring would affect the impedance and power, and also the quality.

Have a lot of ideas though, and thinking about how different things should be done.

since the drivers are of different size this would normally involve a crossover

...3 speakers, two are identical, while the third is bigger.

...the two identical speakers in series, and then wire the series in parallel ...

 

[BlueWizard]

Hmm... this is pretty interresting. the question was more in general, I don't have the speakers, but I was wondering how uneven/unequal wiring would affect the impedance and power, and also the quality. Have a lot of ideas though, and thinking about how different things should be done. Thanks for the formulas, much needed! 🙂

You must have some concept or basic idea that lead you to ask the question. If you can expand on what you are trying to achieve, we can give you more specific information toward achieving that goal.

Here is an Ohm's Law chart from my post in another forum. All the formulas you need are contained on this one chart.

An externally hosted image should be here but it was not working when we last tested it.

If you prefer the chart in Black and White, a simply Google Image search will turn of dozens of variations. I keep a B/W version of this tape up on my wall for quick and easy reference.

One thing to keep in mind, is that Power and Voltage are not linear. A bit more Voltage can mean a lot more power.

Here are the RMS voltages for various power levels to an 8 ohm speaker -

40w = 17.89v
50w = 20.00v
75w = 24.49v
100w = 28.28v

Notice TWICE the power (50w vs 100w) in only 8.28 volts more.

If you understand how Series and Parallel drivers or speakers divide up the power and voltage, you can determine what combinations have equal or unequal power or voltage distribution.

Let me go through an example to illustrate. Let's say we have two 8 ohm speakers in Series with a 16 ohm speaker (32 ohms total)

----[8 ohm]----[8 ohm]----[16 ohm]----

Let's say we apply 10v. The 16 ohm speaker is going to get half the voltage because it is half to total impedance. That means the other two 8 ohm speakers are going to divide up the remaining half of the signal voltage.

So, now we have -

---[8 ohms = 2.5v]---[8 ohms = 2.5v]---[16 ohms = 5v]---

From the chart we see that Power is Voltage Squared divided by Impedance -

P = V²/R

So, 2.5v and 8 ohms = 2.5²/8 = 0.78 watts

Then 5v and 16 ohms = 5²/16 = 1.5625 watts

So -

---[8 ohms = 0.78w]---[8 ohms = 0.78w]---[16 ohms = 1.5625w]---

Again, without some specifics, even if those specifics are just a general design concept or an idea, there is little we can say beyond giving you the tools to figure this out on your own.

In your original example of two 8 ohms in series with another 8 ohm driver, it breaks down like this (sorry for the text graphics) -

---[R1=8 ohm]----[R2=8 ohm]-----
|--------[R3= 8 ohms]-----------|

If we apply 10v, then the full 10v is applied to R3.

R1 and R2 divide the 10v equally, so they get 5v each.

---[R1=8 ohm=5v]----[R2=8 ohm=5v]-----
|----------[R3= 8 ohms=10v]------------|

So, P = v²/R

Pr1 = 5²/8 = 3.125w

Pr3 = 10²/8 = 12.5w

Now we have -

---[R1=8 ohm=3.125w]----[R2=8 ohm=3.125w]-----
|-------------[R3= 8 ohms=12.5w]----------------|

As we have said before, the total circuit has a combined impedance of 5.33 ohms.

I'm not sure what this tells you or how it helps you, but this is how the power and voltage break down.

As to the actual acoustical output, that will depend on the size of the drivers and their rated Sensitivity. However, with lesser voltage, the two 8 ohm in series won't put out any more than a single identical driver on its own. However, the two in Series will have half the excursion, twice the power handling capacity, and very likely lower distortion.

It's complex.

Steve/bluewizard