robb, alex, etc. - useless, super-theoretical impedance question ...

[music-man]

Sorry RIGHT UP FRONT about the length and technicality of this post ...

 

Yeah, OK I know, driving two 8 ohm cabs in parallel with an amp rated 800W @ 4ohms will theoretically result in each of the two cabs getting 400W apiece. But this sounds too easy.

 

Doesn't the impedance of a cabinet vary with the amplitude and frequency of the signal? I mean it would HAVE to, since at least part of what accounts for the electrical resistance in a cabinet is the actual physical motion of the speaker, which creates a magnetic field, etc. - and the physical motion of the speaker of course varies with freq and amplitude, right?

 

So if impedance actually varies with amplitude and frequency ... then it would seem to me that two "unlike" cabinets (like a 4x10 and a 1x15, which have very different responses to a given signal) would have different impedances at a given frequency/amplitude - even if the impedance differences were subtle.

 

So then ... if that's true, and you're using two unlike cabs (like a 4x10 and 1x15) in parallel, it would seem to me that the two cabinets would be drawing different amounts of power from the amp. I guess you'd end up with a sort of "ratio" of impedances that hovers close to a theoretical "1" but that would vary, based on the efficiency and design of the speakers, and also based on the amplitude/freq of the signal.

 

Is any of this true? (Is anyone still awake?) (Or did I just "discover" a ridiculously fundamental principle that every engineer learns in their first day of E-school?)

 

OK, if this is true, how might it affect the sound of a bridge-mono rig?

 

I mean, I wonder whether shifting impedances might result in "underpowering" one cab at certain frequencies, if the impedance ratio between the two cabs shifts too far (like when you play a huge low note). If so, then you might end up with some pretty inefficient usage of your cabs ... And then perhaps there really is an argument for a stereo amp, which would provide two separate channels of power to two different cabinets ...

[_Sweet Willie_]

Totally interesting question.

 

I'm definitely curious about the answer from anyone who's "in the know" on this kind of stuff.

 

Peace.

 

(BTW, music-man, I sent you a PM a few days ago... )

[Prague]

Yes to all of the above, except it's the 2nd day!

 

Seriously, the impedances do vary even from cone to cone, much less cab to cab. It's not a huge issue, though. Mixing cone sizes is why crossovers were developed, so you're right on that point. Rather than running them full-range with separate amps, you might want to use a crossover to to maximize the best of what each cab has to offer. If they both have the same charateristics (frequency response), you just end up with 2 different tones. For bass, I liked this one cab that had a 15 and 2 10's with a crossover at 200Hz. Lots of low end and clear mids and highs. It was 8 ohms, so we could guess that the 2 8ohm 10's were in series, with a 16ohm 15 being in parallel to those. There are others.

[GeorgeR]

Loudspeakers are extremely volatile and non-linear devices as far as having a fixed impedance rating is concerned.

 

But their analysis is really quite simple when you break it all down to it's constituent components - inductance, capacitance, resistance, and magnetic interactive effects.

 

At DC, they offer a pure resistance, the coil resistance.

 

At low frequencies (below 50 Hz), and low amplitudes (say, below 5 watts), inductive reactance is negligible, and back-EMF from the voice coil starts to oppose the excitation from the amp.

 

At low frequencies (below 50 Hz), and high amplitudes (50% or greater rated power input to speaker), inductive reactance is still negligible, and back-EMF from the voice coil has a major effect on opposing the excitation from the amp.

 

Now, regardless of amplitude, as the frequencies increase, the effects of inductance of the voice coil increase by the following formula:

 

X = 2 * (pi) * F * L

 

where X = inductive reactance, in Ohms

pi = 3.141592653.........

F = frequency in Hertz

L = coil static inductance

 

This becomes compounded with the back-EMF produced by the coil oscillating in a magnetic field.

 

As well, inter winding capacitance of the voice coil will begin to affect the voice coil as the frequency increases.

 

As a specific frequency, called the resonant frequency, the effects of the inter winding capacitance and voice coil inductance will cause the coil's effective impedance to rise as a matter of the ratio of reactance to resistance, called "Q" (not to be confused with mechanical "Q" as defined in the manufacturer's ratings).

 

Q is defined electrically as:

 

Q = X/R

 

where:

 

"Q" is an artificial quantitive value placed on the "purity" or "quality" of a reactive component, such as a coil or capacitor.

 

"X" may be either Capacitive or Inductive reactance (at resonance they will be equal so it does not matter which you choose)

 

"R" = DC resistance (conductive losses) of device under test.

 

Resonance is defined as when inductive and capacitive reactances match. The formula is:

 

F = 1/ (2 * (pi) * (sq.root of (L*C))

 

An 8 Ohm speaker may, at some point, measure well over 60 Ohms or more, at a specific frequency, for example.

 

At the same time, circulating current between capacitive and inductive components peaks at this resonant frequency, which may cause further excitation of the voice coil and thus further oscillation and thus greater back-EMF, which will in turn increase the effective impedance the speaker presents to the real world.

 

Beyond this resonant frequency, inter winding capacitive effects will serve to shunt coil current around the voice coil, reducing the excitation and lowering the impedance presented to the outside world, by the formula:

 

Xc = 1/(2 * (pi) * F * C)

 

where X = inductive reactance, in Ohms

pi = 3.141592653.........

F = frequency in Hertz

C = coil static inter winding capacitance

 

There are other effects, as well, such as resistance changes due to heating, eddy currents in the magnetic pole and the damping effect those currents have on the voice coil's magnetic field, the slight effects of skin effect on the conductors composing the voice coil itself and, of course, the damping factor of the amplifer connected to said speaker.

 

As well, it's impossible to neglect the effects of cabinet acoustics on the mechanical load the voice coil meets, and the effects of acoustic resonances, what they will do to the voice coil motion, and therefore how the voice coil reacts electrically to being placed in such an environment.

 

But I won't go into them because then it will become more complicated than it needs to be.

 

Hope any of this helps?

[alexclaber]

As I understand it the impedance of a cab does vary massively across its frequency range, due to both the electrical characteristics of the speakers and crossover and the acoustic properties of the design. For instance, if you play a 41Hz tone through a cab that is tuned to 41Hz, the speaker will barely move and almost all the acoustic output of the cab will come out of the port. This natural resonance of the cabinet that is holding the speaker steady drives the impedance up (due to the back EMF I think) and thus reduces the power going into the cabinet (even though the voltage the amp is delivering is unchanged). Consequently the increased acoustic output at cabinet resonance is cancelled out by the increased reluctance to draw power from the amp.

 

The key to understanding the interaction between speakers and amplifiers is to remember that the power amplifier is a VOLTAGE amplifier. i.e. if your preamp puts three spikes of power at 0.5v, 0.7v and 0.2v, and your amplifier has 30dB of gain ( = 1000 times the power), it will put out voltages of 15.8v, 22.1v, 6.3v respectively ( = sq.rt.1000 times as much voltage = 31.6 times).

 

If you connect two different cabs to this amp in parallel and play those 3 spikes at through them, each cabinet will see 15.8v, 22.1v, 6.3v respectively. Each cabinet will draw however much current it needs from the amplifier depending on the frequency of the signal and the corresponding impedance at that frequency. i.e. if you connect cabinets in parallel their varying impedance will have no bearing on the voltage the amplifier delivers to them and thus the power they put out. NB: Daisy chaining cabs is still connecting them in parallel.

 

A problem would arise if you connect two different cabs in series. If one cab is tuned to 40Hz and another to 50Hz, and you play a 40Hz signal through them, the 40Hz cab would be operating at maximum impedance whilst the 50Hz cab would be operating a very low impedance. According to Kirchoff's law (i.e. sum of current at any part of a circuit is constant) the same current would have to flow through both cabs. If we assume that at 40Hz the 40Hz cab's impedance is at 60 ohm and the 50Hz cab's impedance is at 6 ohm (both 8 ohm nominal cabs), then if the amp is putting out 22.1v into them, the current flowing through the cabs is equal to 0.335A. Consequently the 40Hz cab will dissipate 20.1W and the 50Hz cab 2.01W. If these cabs were connected in parallel the 40Hz cab would dissipate 8.14W and the 50Hz cab 81.4W. (Note that connecting them in parallel rather than series halves rather than doubling the nominal impedance hence the difference in power dissipated).

 

Assuming both cabs have flat frequency response, then when connected in parallel they'll both have equal acoustic output at 40Hz. When connected in series the cab that is much more efficient at this frequency (the 40Hz cab) is getting 10 times as much power as the 50Hz cab when it should be getting 10 times less power to have equal acoustic output. Consequently the 40Hz cab will be putting out 100 times as much acoustic power as the 50Hz cab, which is 20dB. That's a lot!

 

What a long post, when all I needed to say was, if you connect different cabs in parallel they'll be fine but if you connect them in series you'll get seriously weird things going on.

 

Alex

[rumpelstiltskin.]

Originally posted by C.Alexander Claber:

...if you connect different cabs in parallel they'll be fine but if you connect them in series you'll get seriously weird things going on.

music-man -- what Alex said above is pretty well all you need to worry about. that is, unless you want to really overanalyze all your speakers and choose them in an attempt to match them to your amp for flat frequency response instead of getting the tone you want.

 

your post illustrates why i like a 2-channel setup for 2 cabinets instead of running mono in its various incarnations. i can get cabs of different DCRs (that's DC resistance, kids) -- a 4Ohm and an 8Ohm -- without having to worry. simply adjust volume knob to taste, sort of like a poor man's crossover balance control.

 

or you can just let each cabinet have its own signal and be happy for it. in mono, cabinets are going to be connected in parallel, unless you're really a tweak, so they will draw the current they need, but then you get the same signal to each cabinet and their individual parasitic effects cause different responses.

 

so giving each cabinet its own clean signal prevents interaction between cabinets. that is what i prefer to do. i also prefer to use higher impedance cabinets, because i believe they yield a tighter sound and stress the amplifier less.

 

robb.

[_Sweet Willie_]

So how does this play out in cabs with multiple driver types (e.g., an SWR Triad w/ a 15, 10, and horn or a Bergie 322 w/ a 12, two 10s, and a horn)? And relatedly, how is this affected by "cab within a cab" designs to separate the different types of drivers? I guess in this case, w/ the crossovers (passive, I assume) involved, the frequencies are separated into different bands for the different drivers so that the effect under discussion is lessened. However, I'm not so wise technically, so can anyone put this ball into play for me?

 

Peace.

[GeorgeR]

So how does this play out in cabs with multiple driver types (e.g., an SWR Triad w/ a 15, 10, and horn or a Bergie 322 w/ a 12, two 10s, and a horn)? And relatedly, how is this affected by "cab within a cab" designs to separate the different types of drivers?...

As far as getting a stable impedance out of something like you describe......

http://www.bitemelivebait.com/jpg/worms2.jpg

 

[Talisman]

If you really want to get in depth with e-theory, get into how TUBE amps interact with the various voice coils and cone sizes.

 

GeorgeR is right on the money. The actual impedance of a speaker varies with the signal frequency (the real impedance of your cab is not actually 8 ohms, that's the coil resistance). When you said something about underdriving a cab at certain frequencies, you were referring (in a roundabout way) to a speaker's natural tendency to expel frequencies that are inefficient for it.

 

When you connect two different cabs in parallel, the ballgame changes. The 1x15 cab responds well to lower freqs, so it pulls power away from the 4x10 cab when your playing in the lower register. Likewise, the 4x10 cab responds well to higher freqs, so it pulls power away from the 1x15 cab at those freqs. As long as you connect cabs that react well together, you will get the maximum efficiency out of your rig. The best way to do this is to stick with a certain manufacturer for your cabs.

 

I can explain the power exchange more in depth if you'd like. I try to avoid being TOO long-winded in most cases, but I'll really get down if you want. Just a warning, though: It gets really hairy if you get down to the theory level (if you're inexperienced with electronics).

[alexclaber]

Originally posted by Talisman:

When you connect two different cabs in parallel, the ballgame changes. The 1x15 cab responds well to lower freqs, so it pulls power away from the 4x10 cab when your playing in the lower register. Likewise, the 4x10 cab responds well to higher freqs, so it pulls power away from the 1x15 cab at those freqs. As long as you connect cabs that react well together, you will get the maximum efficiency out of your rig. The best way to do this is to stick with a certain manufacturer for your cabs.

I don't think that this is correct. If I'm not mistaken (assuming the amp can supply sufficient current) the cabs do not pull power away from each other, in fact they do not interact at all when connected in parallel. Thus you can mix and match however you want without any ill effects.

 

Alex

[rumpelstiltskin.]

Alex, your impression is incorrect.

 

while the total current drawn by the speakers is a function of the parallel combination of their impedances, that does not mean the current flowing in both of them is the same. current still takes the path of least resistance, no matter what, so if your 15" cab has a very high impedance at 500Hz and your 10" cab does not, more current will flow through your 10" drivers than your 15" driver(s).

 

take, for example, two "identical" 1N4148 diodes. they have a rating of 100mA (i think -- or at least they can for the purpose of discussion), but you have a 115mA signal you need to send through for whatever reason. if you put two in parallel, they have to be exactly identical in every single way or the current will predominate in one. in fact as the predominant diode heats up and current flows more easily, it will hog even more of the total current. usually a very small series resistor can be used to "regulate" the differences between the diodes in a case like this.

 

two cases of counterexample.

 

robb.

[rumpelstiltskin.]

Originally posted by Sweet Willie:

So how does this play out in cabs with multiple driver types (e.g., an SWR Triad w/ a 15, 10, and horn or a Bergie 322 w/ a 12, two 10s, and a horn)? And relatedly, how is this affected by "cab within a cab" designs to separate the different types of drivers? I guess in this case, w/ the crossovers (passive, I assume) involved, the frequencies are separated into different bands for the different drivers so that the effect under discussion is lessened.

it's not just as simple as the crossovers being in different bands. a really good crossover will manipulate impedances, too -- i.e. it will match the impedances of the different drivers to the intended impedance of the total package. this is alongside separating the input signal by frequency. it should do both at the same time.

 

that is really what is going on when someone says "really good crossover design". it's not just some frequency filter, it's a really clever frequency filter. and it's a lot more complex than throwing a few capacitors, resistors, and maybe an inductor into the mix.

 

robb.

[alexclaber]

Originally posted by robb.:

if your 15" cab has a very high impedance at 500Hz and your 10" cab does not, more current will flow through your 10" drivers than your 15" driver(s).

Yes. However the voltage drop across the 10" cab and the 15" cab will remain the same. So if you connected the 10" cab and 15" cab in parallel to a mono amplifer, they would each draw exactly the same amount of current at any given frequency as they would if each was connected to its own amplifier set at the same gain. Thus they do not interact.

 

Originally posted by robb.:

take, for example, two "identical" 1N4148 diodes. they have a rating of 100mA (i think -- or at least they can for the purpose of discussion), but you have a 115mA signal you need to send through for whatever reason.

A power amplifier is a voltage amplifer not a current amplifier. It has to have sufficient current capacity to supply the impedances at which it's rated but it is still a voltage amplifier. Thus, if you had two "identical" diodes you would be looking to put a signal of a specific voltage across them, not specific current.

 

Originally posted by robb.:

if you put two in parallel, they have to be exactly identical in every single way or the current will predominate in one. in fact as the predominant diode heats up and current flows more easily, it will hog even more of the total current.

Again, the discussion is not one of which component 'hogs more current' but if the difference in impedance affects the voltage drop across them and thus the power dissipated in comparison to whether they were connected individually.

 

If cabinets did interact electrically when connected in parallel it would be much more complicated to add extension speakers and mix and match cabs than it is in reality.

 

Alex

[_Sweet Willie_]

Originally posted by robb.:

it's not just as simple as the crossovers being in different bands. a really good crossover will manipulate impedances, too -- i.e. it will match the impedances of the different drivers to the intended impedance of the total package. this is alongside separating the input signal by frequency. it should do both at the same time.

 

that is really what is going on when someone says "really good crossover design". it's not just some frequency filter, it's a really clever frequency filter. and it's a lot more complex than throwing a few capacitors, resistors, and maybe an inductor into the mix.

 

robb.

Thanks for that. Makes sense. I'm glad that there's more going on than just freq filtering.

 

It's nice to know that some crossovers are actually "clever"!!! (Not sure about that Zeta crossover bass, though...but that's a different story! )

 

Peace.