mixed speakers: voltage VS current

[jeff]

 Imagine you have two identical 8 ohm speakers. You could wire them parallel and use the 4 ohm tap or series and use the 16 ohm tap. Since they're identical you'd be splitting the power between them equally.

 Now image you have two different types{EDIT: different brands but both 8 OHM*}of speaker, and at a certian frequency they have different impedances(since impedances change with frequency from speaker to speaker) At those frequencies where the impedance is different the power will not be split exactally evenly, so here's my question.

 If you wired them in series the current would be the same but each speaker would have different voltages because of the different impedacne at that frequency.
 
If you wired them parallel, the voltages would be the same but the current would not.

So is there a "better" way to wire them?(read least likley to blow a speaker) In other words is it better that each speaker never exceeds half the total voltage, or each speaker never exceeds half the total current over the frequencies in which the two speakers have dissimilar impedances?

Either way one will recive more power(voltage X current) but, with respect to a speaker, does it make a difference if it recieves more power because the voltage is greater or the current is greater?
 
*I meant two different types of 8 ohm speaker(ei celestion/weber) not two different ohms(ei 8 ohm 16 ohm)

[HotBluePlates]

No, there is no "better option".

If you have 8Ω and 4Ω speakers in parallel, the 4Ω speaker will take 2/3 of the power delivered by the amp.

If you place them in series, the 8Ω speaker will take 2/3 of the power delivered by the amp. But your amp will also be down on output power some due to the mismatch of 12Ω on either an 8Ω or 16Ω tap.

The right answer is get the correct speaker impedance. You can get away with a lot (and maybe for a long time), but when things blow you'll know why it happened.

[HotBluePlates]

The speaker builders  mix impedance on speakers all the time, Look at the specs on tweeters versus the mid speakers and the bass speakers, then look at the filters installed to keep the bass out of the tweeters.

We're talking guitar amps, not a stereo with a crossover for the speakers. So let's not confuse the issue with extraneous info.

Now image you have two different types of speaker, and at a certian frequency they have different impedances(since impedances change with frequency from speaker to speaker) At those frequencies where the impedance is different the power will not be split exactally evenly, so here's my question.

If my memory serves correctly, the nominal resistance on a speaker is given for at 400 Hz, and amps voltage test points results are stated at 1KHz, 

Don't get hung up on frequency. You will make the problem appear more complex than it really is.

Assume Speaker A is "8Ω, nominal" (however the manufacturer arrives at that spec) and that Speaker B is "16Ω, nominal." Because you have a guitar amp, with a single mono output and no crossover feeding different frequency ranges to each speaker, then each speaker receives substantially the same input signal.

Because each speaker receives substantially the same signal, there is a essentially constant ratio between the two speakers' impedances; this is 2:1 our example speakers.

As a result, when figuring out power division between the speakers, you can simplify the model. Replace the speakers with a resistor equal to the nominal impedance. Replace the a.c. power output with a d.c. source. Calculate voltages and currents using Kirchhoff Laws and Ohm's Law, as well as the equation for power.

Doing all this is perfectly valid for the scenario described, and removes the smoke and "guru magic" while it leads to quicker understanding.

[jeff]

 Thanks for the replies but I think I was a little unclear. I've EDITed my original post to be clearer.

What I meant is if you have two 8 ohm speakers that are different brands they won't have the same impedances over all frequincies.

 If they are wired in series then both speakers will have the same current going through them(series) but the voltages across each will be different. Basically it's a voltage divider with the resistances changing with the individual frequency resopnces of each speaker.

 If wired parallel the voltage must be the same across each speaker but the current will be divides unequally over frequencies in which the impedance of the two are different.

 I'm not sure which is more important to the operation of speakers.

If both speakers always have the same current with slightly varing voltages,
or if both speakers always have the same voltage with slightly different currents.

 I guess what I want to understand is what drives a speaker: voltage or current. When using to different voice coils both rated at 8 ohms but in reality very different from each other, we can wire to split the power so both speakers always have half the current or split the power so both speakers always have half the voltage.
...does that matter or watts is watts?

[HotBluePlates]

What I meant is if you have two 8 ohm speakers that are different brands they won't have the same impedances over all frequincies.

They are probably similar enough to allow you to call them "the same" with reasonable accuracy. They will evenly split power beyond your ability to tell the difference.

In other words, prove to me that they are not the same impedance at any given time.

Impedance isn't the only thing that decides frequency response anyway; If you'e thinking "the speakers sound different so they're dividing power differently," in all likelihood that's not the cause of the sonic difference.

[PRR]

> one will recive more power

No. In parallel the power is the same as each alone. In series the power can only *reduce* where one goes high-impedance at resonance.

Yes, it does appear that the power drawn by the array from the amp, and the acoustic output, will be different for series or parallel. I am not prepared to guess which will sound better.

If the speakers are very close together, their bass resonances will shift closer. If they are in the same box the effect will be large, very nearly a single speaker with parameters averaged between the two dissimilar drivers.

> put the higher wattage on upstream of the ground?

Same as a bird in the sky: doesn't know where "ground" is, doesn't matter.

[jeff]

OK so here's my situation:

I have 2 Celestion G12Ls and 2 Celestion Rocket 50s. All marked 8 ohm and I'm putting them in a 4X12 cab.

I could wire the 2 G12L in parallel, the 2 Rocket 50s in parallel and wire those two pairs in series.

OR I could wire one of each in parallel, one of each in parallel, then wire those two pairs in series.

[sluckey]

I'd wire them series/parallel and keep the cab at a nominal 8Ω. Or, I'd wire them parallel/series and keep the cab at a nominal 8Ω. Now, what color wire to use? And probably most important of all,,, which hole to screw them in?     

One twelve is soooo easy.   

OK, now that we know what you really want to know, there are some guys on the forum that can probably give you some good ideas about how to load those speakers in that 4x12 cab. But you certainly don't have to worry about slightly different impedance when mixing speakers. It's all nominal and you wont mess up anything by experimenting.

Visit the Avatar Speakers website. They mix speakers for a living. May be some good info/opinions there.

http://www.avatarspeakers.com/
 

[HotBluePlates]

For all the time you've spent considering the situation, the speakers could already be wired up.

It's not gonna make any real difference in what order you wire any of those. Some may claim different, but they have nothing but voodoo to point to as support for the assertion.

[hesamadman]

I build 3x12 speaker cabinets all the time. Some are THREE different models. Would be slight differences in volumes. But can your ears hear that? Ours can't. And neither can our testing equipment.

[HotBluePlates]

I have 2 Celestion G12Ls and 2 Celestion Rocket 50s.

Celestion doesn't seem to put impedance plots on their (available) spec sheets, and the G12L-35 appears to not be in current production (so Celestion's site has no info on them).

Since I argued for hard fact, I wanted to go grab an actual impedance plot for each to show the (very likely minimal) difference.

[HotBluePlates]

... the eminence site shows impedance plots for different speakers.  ...

Sure does. If you open multiple speaker sheets and flip through them, you'll see very little changes except the bass resonant frequency.

...  I would definitely choose speakers with a broad divergence of resonant frequencies.   ...

Knock yourself out.

If you were building a ported cab, you'd probably want all the speaker to have the same resonant frequency so your cab dimensions and porting works out correctly. "Resonance" in this case is not "sonority" but rather what frequency the cone moves most freely. A mix of speakers with different resonant frequencies means one might behave well, while the others try have their cone jump out of the speaker basket.

This is related to why guys with annoying subwoofers in their cars can get really loud bass with a given power (they use speakers with a certain resonant frequency down in that sub-bass), but also why all their music sounds like it has 1-note bass (because the sub-box is lousy at pumping anything out except than 1 resonant note). Some Hip-Hop & Rap/Dance music exacerbates this by intentionally using certain bass hits that sit in the resonant frequency used by those sub boxes, so you get a chicken & egg effect with the 1-note bass.

HBP, you take all the fun out of doing the calculations, if frequencies can't be used.    ...

Do what you like, but try using frequency to explain why the bass resonant frequency is a large impedance, while the impedance at 300Hz is very much smaller.

The method I suggested is how these problems are solved (in under a couple-minutes); you can try lengthy calculations based on frequency and varied impedance at each, calculating power division at 5-10-100 points, and you will still arrive at the conclusions found with the simple method with a percent of error approaching manufacturing tolerances.

Give it a shot the hard way and see what you come up with.

[Ed_Chambley]

OK so here's my situation:

I have 2 Celestion G12Ls and 2 Celestion Rocket 50s. All marked 8 ohm and I'm putting them in a 4X12 cab.

I could wire the 2 G12L in parallel, the 2 Rocket 50s in parallel and wire those two pairs in series.

OR I could wire one of each in parallel, one of each in parallel, then wire those two pairs in series.

I get this jackplate from AES.  If I have a multitap head, 4, 8 and 16 ohms and wire is so I can use 2 heads at 8 ohms or 1 at 16, but I have to go by what is available to me.

Placement of speakers I find if I have a lower wattage, these go on top as they usually have a better upper mid.   Either way, I put the best treble on top and bass on bottom.  Some people cross them, but I find no benefit in this.  Dave at avatar suggested to me to use Alnicos on top and Ceramics on bottom with the wattage rating lower on the Alnicos so they compress easier.  Really nice.

You already have speakers and while 4, 12's are too big for me these days I think you will have a great tone.  4, 12's are hard to beat to me.  Rich and thick.

[jeff]

OK thanks for all the input. Again I probally was overthinking it.

 I stubmled across this: http://edn.com/design/consumer/4423155/Loudspeaker-operation--The-superiority-of-current-drive-over-voltage-drive

 Way over my head but it said this:"The driving force (F), that sets the diaphragm in motion, is proportional to the current (I) flowing through the voice coil according to the well known formula F = BlI where the product Bl is called force factor (B = magnetic flux density; l = wire length in the magnetic field). B is the flux density that exists when the current is zero. (The current always induces its own magnetic field, which may react with adjacent iron, but the effect is not related to this equation.)

This force, then, determines the acceleration (A) of the diaphragm, which in the main operation area (the mass-controlled region) is got from the Newtonian law F = mA. The radiated pressure, in turn, follows the instantaneous acceleration and not the instantaneous displacement, as many mistakenly imagine.

The most remarkable thing here regarding loudspeakers is that the voltage between the ends of the wire does not appear anywhere in these equations. That is, the speaker driver in the end obeys only current, not caring what the voltage across the terminals happens to be."[/b]

So I wired the two G12L in parallel
The two rockets in parallel
Then wired those two sets in series.

This way any variation between the two will manifest as a difference in voltage, but each speaker will carry the same current.

The cab sounds good to me, but I'm the kind of guy who asks "What is the exact height of that bean hill?"

[jeff]

Yeah I drive myself crazy sometimes.

What started all this is I'm downsizing and I had a bandmaster 2X12 cab. The cab had two 12" but it's almost big enough 6X12"(5X12" eaisly). I cut a new baffel board and put all my eggs in one basket. I figure no sense taking up all that space just for 2X12", + another seperate 2X12" cab, when I can store all 4 speakers in that one cab.

[HotBluePlates]

The most remarkable thing here regarding loudspeakers is that the voltage between the ends of the wire does not appear anywhere in these equations. That is, the speaker driver in the end obeys only current, not caring what the voltage across the terminals happens to be."[/i]

So I wired ...

This way any variation between the two will manifest as a difference in voltage, but each speaker will carry the same current.

This is in the realm of how audiophiles take a nugget of truth and expand it to its illogical limits.

The equations only contain a term for current because it is current that physically creates an electromagnetic field around a wire, which when reacting to a static magnetic field causes motion of a voice coil.

• But voltage is also known as electromotive force, which is the force which moves electrons in a directed drift, aka "current".
• In all of the equations, I (current) could be replaced by V/Z (voltage/impedance), which is its equivalent according to Ohm's Law. We don't do this usually because the equations become more clumsy.

Pretending one exists without the other is only useful in a theoretical discussion. The equations don't contain a term for voltage, but they all assume a voltage is present which is sufficiently strong to cause the current specified.

Still doesn't change how you should analyze your speaker situation. It also doesn't help you that the article's central thesis depends on the assumption/assertion that "... the power amplifier acts as a voltage source exhibiting low output impedance and thus strives to force the voltage across the load terminals to follow the applied signal without any regard to what the current through the load will be." Whether an amplifier is a voltage or current source is a function of the output stage design, and what's true for transistors is not necessarily true of tube amps.

Regardless, within limits and with regard to the amplifier topology, any amp could act as a voltage source or a current source, which means the article's assumption may not apply to your amp.

[jeff]

OK thanks for the head check.

 2ND look, you're right, that doesn't make any sense. Like saying it doesn't matter what the voltage across a 10 ohm resistor it only matters what the current is. But.. If you change the voltage the current will change.

 If V=I/R then Vx=Ix/R. You can't say it doesn't matter what V is it only matters what I is.

Thanks.

[jeff]

OK I did some math and I hope I did this right, here's what I got.

Pulling numbers out of thin air just as an example I imagine two speakers A and B
At a certian frequency speaker A is 10 ohms and speaker B is 20 ohm.
I first imagined them in series.
So turn the amp up til there's 50W across the 30 ohm load.
Doing the math(hopefully correctly) I came up with:

   Speaker A 16.6W
   Speaker B 33.3W

I then imagined them in parallel.
So turn the amp up til there's 50W across the 6.666 load
Doing the math(hopefully correctly) I came up with:

   Speaker A 33.3W
   Speaker B 16.6W

One way speaker A gets 16.6W speaker B gets 33.3W
The other way vice verca

BUT>>>
 Series speaker A gets 16.6W because 12.9V@1.29A
 Parallel speaker A gets 33.3W because 18.25V@1.82A
 Series speaker B gets 33.3W because 25.8V@1.29A
 Parallel speaker B gets 16.6W because 18.25V@.913A

So one way one speaker is getting twice the watts because it's at twice the voltage, the other way the other speaker gets twice the watts because it's at twice the current

What(if anything) does this mean?(besides I have too much time on my hands )

 ???is that right???

[HotBluePlates]

2ND look, you're right, that doesn't make any sense. Like saying it doesn't matter what the voltage across a ...

There's a lot of unstated baggage and implications in that article. If you're not clued-in to what's unsaid, you'll mis-read what is said.

You apply an input signal to a power amp; whether the amp acts like a current source or a voltage source depends on which quantity tracks the input signal in the face of a changing load impedance.

Transistor power amps tend to act like voltage sources; users assume the amp will output more power if they use a 4Ω load instead of an 8Ω load. Because the amplifier tries to maintain a constant voltage output (ratio) based on the input signal, halving the load impedance doubles the current, and double-current times same voltage equals double power.

You know that doesn't work well with tubes, and that we try to match whenever possible because mismatch in either direction yields less power. Mismatching a transistor amp be attaching a higher load yields less output power (but in a predictable way that's often exploited by headphone users), while mismatching lower yields more power until you burn out the output transistors.

It doesn't help that tubes or transistors can be configured as voltage or current sources, based on the nature of the device and how feedback is (or isn't) applied around the amplifier. And discussions like the article assume theoretical perfect sources when in reality we run into limitations pretty quickly.

At a certian frequency speaker A is 10 ohms and speaker B is 20 ohm. ... in series ... there's 50W across the 30 ohm load.
... I came up with:

   Speaker A 16.6W
   Speaker B 33.3W

If power stays constant, then yes, the power is divided as you've shown.

For a series circuit, you'd remember that there's only 1 path for current so all elements of the circuit see the same current. Therefore the one with the highest impedance has the greatest voltage drop and therefore the greatest power (output or dissipation).

... I then imagined them in parallel. ... 50W across the 6.666 load
...

   Speaker A 33.3W
   Speaker B 16.6W

Yes. For a parallel circuit, each branch sees the same applied voltage between their terminals (because each branch has their terminals connected together where the paths split). Now there will be less current in the high impedance path, so same voltage times less current equals less power. Now the low impedance path has the greatest power output/dissipation.

So one way one speaker is getting twice the watts because it's at twice the voltage, the other way the other speaker gets twice the watts because it's at twice the current

What(if anything) does this mean?(besides I have too much time on my hands )

You have confirmed for yourself what I posted originally.    All that's happening is you are re-discovering fundamental electrical principles known as Kirchhoff's Laws, Ohm's Law and the equation for power.

[jeff]

Thanks,

 I really apreciate you taking the time to explain helping me confirm if I have a good understanding. Always a learning experience(not just "this is what you do", but also "this is why you do it")

 Respect
  Jeff 

[6G6]

I think you may be overthinking this, but learning in the process.
You might gain some real world insight by looking for some of the many
discussions about mixing pairs of different speakers in the top and
bottom halves of Super Reverbs.
Check out the FDP for those.
Those often tend to end up with ceramic on the bottom and Alnico on top.

[HotBluePlates]

... 150W the bass speaker model s2010 ... a guitar speaker model 1028K ... a wattage rating of 35w. ...

What is the definition of "Sophistry"?

The 35w 1028K would have already popped while the S2010 is loafing, and would probably jump out of its frame on the first F you played. Anyone silly enough to put them in the same cabinet deserves what they get...

[HotBluePlates]

You are neglecting that at resonance, the impedance is high, but the speaker cone moves most freely. That free motion generally translate as louder acoustic output at resonance (as in "it resonates" or moves freely), even if power into the speaker might be restricted electrically.

You will also find you lack an electrical cause for the rise of impedance at resonance, as neither inductance nor capacitance account for it. As a result, you'll also lack a simple model for how/why the speakers divide the power the way they do.