I don't know if anyone mentioned this already, but the OP can readily test how damping changes by tapping a woofer with his finger. If the speaker is disconnected, he will observe a purely mechanical response. If the terminals of the disconnected speaker are shortcircuited with a piece of wire, he will observe the full combined mechanical and electromagnetic response. By connecting resistors of progressively higher value across the terminals, he can observe the whole spectrum between the two aforementioned response extremes.
+1 to that.if u have a scope u can do the same with a small battery and toggle switch across the speaker terminals, ie across ground and probe also. Maybe a momemtary push button would work better? 1.5V cell, maybe a pot to lower the dc step voltage, if u find the oscillations to low in amplitude relative to the step injection. Timebase set somewhere around 1/Fs. Try with no additional series R, then with a R from say 1/2Re to maybe 2*Re. Havent tried it yet, but im sure it would work, maybe i missed something. The dc step should be injected at the speaker terminals so that the added series resistor does not attenuate the dc step, OR use a momentary changeover switch, one ole for dc step other to series R and scope
One downside, if it is one is that the speaker is driving a hi Z load. Substantially less current can flow and its the back current or back emf that damps cone movement i think, rather than microamps in the scope circuit...im not sure the actual results would mean alot. I could be wrong lol
One downside, if it is one is that the speaker is driving a hi Z load. Substantially less current can flow and its the back current or back emf that damps cone movement i think, rather than microamps in the scope circuit...im not sure the actual results would mean alot. I could be wrong lol
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derivation of equation for modified Qes
Thanks for your analysis, John K!
The above quote implies that the peak impedance of the driver or Rmax is unchanged by placing resistance in series circuit with the driver. And of course voice coil resistance Re is unchanged by the presence of series resistance. In analyzing the system consisting of the driver and series resistance, then,
R'o = (Rmax + Rx)/(Re + Rx)
where R'o is the ratio of maximum (at resonance) to minimum impedance of the system of series resistance and the driver.
From the equations for deriving Q from the impedance characteristic of a driver,
Qes = Qms/(Ro -1)
Then given that the above can also be applied to the system of series resistance and the driver,
Q'es = Qms/(R'o-1)
Solving for the ratio of electrical Q of the system of series resistor and driver to that of the driver alone,
Q'es/Qes = [Qms/(R'o-1)] * [(Ro-1)/Qms]
Q'es/Qes = (Ro - 1)/(R'o -1)
Ro -1 = (Rmax/ Re) - 1
Ro -1 = (Rmax - Re)/ Re
R'o - 1 = [(Rmax + Rx)/(Re + Rx)] -1
R'o - 1 = (Rmax - Re)/(Re + Rx)
Then finally solving for the ratio of the electrical Qs,
Q'es/Qes = [(Rmax - Re)/Re] * [(Re + Rx)/(Rmax - Re)]
Q'es/Qes = (Re + Rx)/Re
Q'es = (Re + Rx)Qes/Re
The last line above is the equation that I first saw in Vance Dickason's Loudspeaker Design Cookbook. What's above I think helps at least some in understanding what placing resistance in-series with a driver does.
Regards, Pete
Thanks for your analysis, John K!
The above quote implies that the peak impedance of the driver or Rmax is unchanged by placing resistance in series circuit with the driver. And of course voice coil resistance Re is unchanged by the presence of series resistance. In analyzing the system consisting of the driver and series resistance, then,
R'o = (Rmax + Rx)/(Re + Rx)
where R'o is the ratio of maximum (at resonance) to minimum impedance of the system of series resistance and the driver.
From the equations for deriving Q from the impedance characteristic of a driver,
Qes = Qms/(Ro -1)
Then given that the above can also be applied to the system of series resistance and the driver,
Q'es = Qms/(R'o-1)
Solving for the ratio of electrical Q of the system of series resistor and driver to that of the driver alone,
Q'es/Qes = [Qms/(R'o-1)] * [(Ro-1)/Qms]
Q'es/Qes = (Ro - 1)/(R'o -1)
Ro -1 = (Rmax/ Re) - 1
Ro -1 = (Rmax - Re)/ Re
R'o - 1 = [(Rmax + Rx)/(Re + Rx)] -1
R'o - 1 = (Rmax - Re)/(Re + Rx)
Then finally solving for the ratio of the electrical Qs,
Q'es/Qes = [(Rmax - Re)/Re] * [(Re + Rx)/(Rmax - Re)]
Q'es/Qes = (Re + Rx)/Re
Q'es = (Re + Rx)Qes/Re
The last line above is the equation that I first saw in Vance Dickason's Loudspeaker Design Cookbook. What's above I think helps at least some in understanding what placing resistance in-series with a driver does.
Regards, Pete
I understand what you are saying an I have no problem with the observation that Qes of a system consisting of a woofer in series with a resistor is different than Qes of the woofer. But it is not Qes of the woofer that changes. If you accept that Qes can be extracted form the impedance curve of the woofer then this becomes readily apparent. I have made some measurements to demonstrate this.
This first figure shows the measured impedance of a woofer in green. Then it shows in red a simulation of what the impedance is when a series resistor of 9.1 ohms is placed in series. Obviously in the simulated result the driver's contribution to the total impedance is fixed. Then it shows the measured impedance of the system consisting of the woofer with 9.1 ohm series resistance. Hopefully you will accept that the simulated and measured results are the same, within measurement error, and that this can only be the case if the driver's Z doesn't change. Since the driver's Z doesn't change neither do the driver's T/S parameters. Yes, the system behaves as a higher Q system as reflected by the system impedance, but that is the result of the series resistance interacting with (fixed) driver Z as a voltage divider, as the OP stated. The change in Qes as given by the equation in the previous post is not the change in the driver's Qes but is the Qes of the system which includes the series resistance.
Similarly, the same can be demonstrated when the series resistor is replace by a cap. This is demonstrated in the next figure. Same type of plot shows again that the driver's Z doesn't change, thus the driver parameters don't change.
By the logic above Qes' would equal Qes x (Re + jwC)/ Re. What does that even mean? There is now damping in the complex plain? Isn't that energy storage, not damping? Never the less, that is what will come out of the equation of motion describing the cap in series with the driver system.
The same result applied if an inductor is subtituted fro the resistor, or any series component (like a parallel RLC trap).
This is no different than when you put a driver in a sealed box. Both Fs and Q of the boxed driver will be different than that of the driver alone and it is reflected in the in box impedance. But that does not mean the driver Z changed. What has changes is that a series complinace element has been added to the system. By the same argument that Qes changes with Re you could argue that Qes changes as Qesb = Qes x Sqrt(Cms/C') where C' is the compliance of the box in series with the driver compliance. Where do you think the relationship, Qtc = Qts x sqrt (1+Vas/Vbox) comes from? All these relationships are written down as handy formulas. But I don't hear anyone saying that placing a woofer in a box changes the woofer Qts.
All these convinent relationships come out of writting down the system equation which basicly starts with F = Ma. And as long as all that is added is a series impedance, mechanical, electrical or otherwise, simple scaling of the driver parameters can be used to obtain the system parameters.
This first figure shows the measured impedance of a woofer in green. Then it shows in red a simulation of what the impedance is when a series resistor of 9.1 ohms is placed in series. Obviously in the simulated result the driver's contribution to the total impedance is fixed. Then it shows the measured impedance of the system consisting of the woofer with 9.1 ohm series resistance. Hopefully you will accept that the simulated and measured results are the same, within measurement error, and that this can only be the case if the driver's Z doesn't change. Since the driver's Z doesn't change neither do the driver's T/S parameters. Yes, the system behaves as a higher Q system as reflected by the system impedance, but that is the result of the series resistance interacting with (fixed) driver Z as a voltage divider, as the OP stated. The change in Qes as given by the equation in the previous post is not the change in the driver's Qes but is the Qes of the system which includes the series resistance.
An externally hosted image should be here but it was not working when we last tested it.
Similarly, the same can be demonstrated when the series resistor is replace by a cap. This is demonstrated in the next figure. Same type of plot shows again that the driver's Z doesn't change, thus the driver parameters don't change.
An externally hosted image should be here but it was not working when we last tested it.
By the logic above Qes' would equal Qes x (Re + jwC)/ Re. What does that even mean? There is now damping in the complex plain? Isn't that energy storage, not damping? Never the less, that is what will come out of the equation of motion describing the cap in series with the driver system.
The same result applied if an inductor is subtituted fro the resistor, or any series component (like a parallel RLC trap).
This is no different than when you put a driver in a sealed box. Both Fs and Q of the boxed driver will be different than that of the driver alone and it is reflected in the in box impedance. But that does not mean the driver Z changed. What has changes is that a series complinace element has been added to the system. By the same argument that Qes changes with Re you could argue that Qes changes as Qesb = Qes x Sqrt(Cms/C') where C' is the compliance of the box in series with the driver compliance. Where do you think the relationship, Qtc = Qts x sqrt (1+Vas/Vbox) comes from? All these relationships are written down as handy formulas. But I don't hear anyone saying that placing a woofer in a box changes the woofer Qts.
All these convinent relationships come out of writting down the system equation which basicly starts with F = Ma. And as long as all that is added is a series impedance, mechanical, electrical or otherwise, simple scaling of the driver parameters can be used to obtain the system parameters.
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I'm in total agreement with you that Qes or electrical Q of the driver is unchanged by the connection of a resistor in-series with it. I wanted to define modified Qes or Qes' as per your terminology as the electrical Q of the system that is the series-resistor and the driver.
Perhaps "modified" Qes is misleading- it's really the electrical Q of something else.
But anyway, my derivation is useful I think in that it shows that the change of Qes flows from the fact that the series resistor causes a reduction of Ro, not because the damping force has been reduced relative to the driving force.
A more direct way to look at it might be to consider that
Qms = [Fs*sqrt(R'o)]/(F2-F1)
then given that Qms is unchanged by assembling a "system" of a resistor in-series with the driver, and R'o < Ro, then the bandwidth which is (F2-F1) must decrease to keep Qms constant. Further then the system Q must be greater than Q of the driver as
Q is proportional to Fs/(F2-F1)
I believe that in the above I must say proportional rather than "equal to", but I'm not certain exactly why.
In my derivation post, my last equation isn't completely consistent, I believe, using the computer terminology "*" meaning times. It should read-
Q'es = (Re + Rx)*Qes/ Re
Regards, Pete
Perhaps "modified" Qes is misleading- it's really the electrical Q of something else.
But anyway, my derivation is useful I think in that it shows that the change of Qes flows from the fact that the series resistor causes a reduction of Ro, not because the damping force has been reduced relative to the driving force.
A more direct way to look at it might be to consider that
Qms = [Fs*sqrt(R'o)]/(F2-F1)
then given that Qms is unchanged by assembling a "system" of a resistor in-series with the driver, and R'o < Ro, then the bandwidth which is (F2-F1) must decrease to keep Qms constant. Further then the system Q must be greater than Q of the driver as
Q is proportional to Fs/(F2-F1)
I believe that in the above I must say proportional rather than "equal to", but I'm not certain exactly why.
In my derivation post, my last equation isn't completely consistent, I believe, using the computer terminology "*" meaning times. It should read-
Q'es = (Re + Rx)*Qes/ Re
Regards, Pete
Erm, isn't Qes already Q, Electrical, System ? Why invent new terminology ?
I put forward my point again - what is the electrical Q of an unconnected driver ? Undefined ? Infinite ? Certainly not the stated Qes figure from its T/S parameters.
The reality is that the difference between Re and Rs is an artificial distinction. It makes no difference how much of the loop resistance comes from the coil and how much comes from the external return path in the circuit.
The current flow that occurs due to generator action of the motor flows through Re and Rs in series, and the sum of the two control the degree of electrical damping.
The return path of the circuit connected to a driver is just as important in defining its electrical Q as the return path of a straight piece of wire is in defining its inductance.
Obviously we need some way to specify the characteristics of the driver in a known state, so the Qes T/S parameter is by convention quoted with an Rs of zero, but that does not mean that Electrical Q does not change with Rs.
Never mind measuring impedance curves, instead of measuring it electrically, why not measure the total damping of the speaker in response to an external mechanical stimulus, with varying amounts of resistance connected across its terminals from zero to infinity. We know the mechanical Q of the driver stays the same and we can measure total Q by looking at the response of the cone to an externally introduced impulse. From this we can derive electrical Q. If electrical Q is not changing with resistance what is ?
Come on guys. Qes = Q, Electrical, System, the total loop resistance is part of the system, so Qes must change with external resistance as well as voice coil resistance. (Again, without the external circuit there is no electrical damping...)
The only problem here is that the term Qes is being somewhat abused in normal T/S nomenclature, as it is being used to refer to the electrical Q with the specific scenario of external resistance Rs = 0, rather than referring to the actual electrical Q for a given connected circuit, which is what matters in defining the response.
If we accept the common usage of Qes to mean "Electrical Q with Rs = 0" then we also have to accept the distinction between Qes as a driver defining parameter, and actual Electrical Q of the driver in a given circumstance.
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The newcomers amongst us may not be aware that Qes is most likely measured with short test leads, and strictly you should modify Qes from calculation,to add additional series resistance in the wiring and crossover. Aiming to re calculate Qts, from either the original spec Qms or any you obtain by measuring TS parameters yourself. One can go so far as to add resistance is you want to modify the Qes and change Qts as a result. some people do this with small gauge solid core cables.
you can do it in an amp output if you want, and its the most efficient way i would imagine. It without a doubt changes Qes, and im not sure that a new term should be produced involving something like Res = Rvc + Rxo
or the quoted form...I forget lol
I think that Vendors should state that the Qes is calculated with e.g. 0.3 Ohm series resistance included in circuit, from which Qts quoted has been calculated.
the matter of half an ohm, even an ohm, is small in its effect anyway, if its an 8 ohm driver
for example, ive tried this in a fashion, with a tangband w2-800sl. its a small fullrange/widerange driver. with Fs at 160-200hz its not bass capable. And with a Qts of 0.25 or thereabouts, I experimented with a series R to change the Qes. its a **4 ohm driver i think, and to change the Q from ~0.25 to ~0.45 took around **3.3 ohms i think IIRC. by that time I think were nearly at 3dB power loss/6dB volt loss? cant remember which lol. and losing 3dB was the most i was willing to tolerate. worked out reasonably, making a sealed SMALL box slightly better than before 😀
I cant see the point though, unless youre in a similar situation. I mean generally speaking, when i buy a driver i look for a Qts between maybe 0.35 to 0.5 for a broad ranger, 0.4 to 0.48 for a narrower range. Low Vas is nice so it fits in a reasonable box. low Fs is less of an issue. In that situation I dont think id want to add any more resisitance, or id ensure it was low, because thats what im designing for. unless for an OB or something. this is a good reason why some fostex with very low Qts reportedly sing on valve amps with higher impedance. the bass is brought out by the small increase in total Q.
**Or it was an 8 ohm driver and 6.6R ...memory fails me
After all lets face it Qms CAN change, with age, temperature. Maybe not alot, but it can. air pressure could possibly change something (air density?) and alter Vas (?)
you can do it in an amp output if you want, and its the most efficient way i would imagine. It without a doubt changes Qes, and im not sure that a new term should be produced involving something like Res = Rvc + Rxo
or the quoted form...I forget lol
I think that Vendors should state that the Qes is calculated with e.g. 0.3 Ohm series resistance included in circuit, from which Qts quoted has been calculated.
the matter of half an ohm, even an ohm, is small in its effect anyway, if its an 8 ohm driver
for example, ive tried this in a fashion, with a tangband w2-800sl. its a small fullrange/widerange driver. with Fs at 160-200hz its not bass capable. And with a Qts of 0.25 or thereabouts, I experimented with a series R to change the Qes. its a **4 ohm driver i think, and to change the Q from ~0.25 to ~0.45 took around **3.3 ohms i think IIRC. by that time I think were nearly at 3dB power loss/6dB volt loss? cant remember which lol. and losing 3dB was the most i was willing to tolerate. worked out reasonably, making a sealed SMALL box slightly better than before 😀
I cant see the point though, unless youre in a similar situation. I mean generally speaking, when i buy a driver i look for a Qts between maybe 0.35 to 0.5 for a broad ranger, 0.4 to 0.48 for a narrower range. Low Vas is nice so it fits in a reasonable box. low Fs is less of an issue. In that situation I dont think id want to add any more resisitance, or id ensure it was low, because thats what im designing for. unless for an OB or something. this is a good reason why some fostex with very low Qts reportedly sing on valve amps with higher impedance. the bass is brought out by the small increase in total Q.
**Or it was an 8 ohm driver and 6.6R ...memory fails me
After all lets face it Qms CAN change, with age, temperature. Maybe not alot, but it can. air pressure could possibly change something (air density?) and alter Vas (?)
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The amount of Back EMF generated is dependent on the magnitude of the forward current passing through the voice coil. If the output voltage of the amp driving the speaker is held constant and a resistor placed in-series with the driver, then forward current is reduced by the factor Re /(Re + Rs). As the magnitude of Back EMF is reduced by the same factor, the proportion between the driving force acting on the voice coil and damping effected by the speaker functioning as a generator (back EMF) stays the same. That is, Qes of the speaker is unchanged. Yes, electrical damping has been reduced. But the magnitude of electrical damping and Qes are not synonymous. Qes is a measure of the magnitude of back EMF produced for a given forward current.
When using the current source technique for measuring impedance vs frequency of a loudspeaker, usually the series resistor is 1 k Ohm. But 2 k ohm or 500 Ohm would work just as well. In all three cases of using vastly different values of series resistance, the measured value of Qes is identical. Wouldn't that seem to indicate that the speaker has an intrinsic value of Qes that isn't dependent on the value of Rs? This is exactly what John K. demonstrated.
Regards, Pete
I've made an analysis of driver Z from the equation of motion for the driver and also looked a little at distortion reduction at low frequency resulting from current drive. If anyone is interested I'll post it. It is rather long so unless some of you want to see it I'm not going to bore you with it.
Not really. The curves are taken at a fixed level of 80, then 90 dB at 1m. This includes achieving 90 dB at 20 Hz, which ends up being all distortion, so we are well into serious nonlinearity. The interesting byproduct of using a compressor for flat output is that you can track the level of fundamental and see what the maximum pure tone output of a unit is.
It turns out that there is a maximum pure tone output (1st harmonic) and it doesn't matter what the input level is, the 1st harmonic output is fixed. Higher input increases odd harmonic distortion level but not the fundamental. In fact there becomes a 12dB/Octave asymptote that defines maximum fundamental output for all bass frequencies for a given woofer.
This means that any woofer could be defined by its maximum pure output at, say, 50Hz, and thereby compared to any other woofer with one number.
I'll scan the curves this weekend.
Regards,
David S.
Cool David, looking forward to seeing it. At first glance that fits well with my experience, so I'd love to learn more about it.
Hi John,
This sounds interesting, and there are lots of claims about "current drive". I always value your analytical approach to issues relating to loudspeaker design. So, yes, please post it!
-Charlie
Regarding a phenomenological basis for the limit on the fundamental - is this related to the maximum excursion capability of the driver, e.g. when you are driving the unit so much that you are pushing the VC out of the gap you are getting some kind of "mechanical clipping" behavior and further increases in power are only causing the actual cone response to be more "square wave" in nature, with a limited maximum amplitude?
-Charlie
Yes, I'm sure it is related to maximum excursion and comes from the major factors of voice coil geometry and suspenison nonlinearity. Typically VC overhang is the biggest single factor. Gander's paper (JBL) covers the subject well.
What was interesting was that you could put in about any drive level and see constant fundamental output level. It a bit like seeing hard clipping on a sine wave. As you crank it up to a greater percentage of clipping the odd harmonic levels go up but the filtered sine wave output would always be the same level as a sine wave just prior to clipping. i.e. driving it harder doesn't get any more fundamental output. Just more distortion output.
David S.
It is interesting, but also logical. Why wouldnt a mechanical transducers' output of a sine when driven to mechanical overload, follow exactly the same mathematical roots, under similar conditions?
@ Pano Most definately guitar overdrive comes to mind, and the fact that some vintage guitar speakers are 'dirtier' than others, green backs come to mind....extremely minute Xmax but 97dB/W....!? 10 watts of so each and they start overdriving themselves, WITHOUT the amps contribution.
Yep, perfectly logical once the measurements showed what was going on (which is not to say that I expected it in advance!).
David S.
well its one of those things that seems like an obvious conclusion, one of those things one believes he shouldve been able to have thought out in advance, and tested to prove, but in practice IME never seems to actually happen lol. Lastly its logic seems follows occams razor. Simplest logical explanation, is the most likely.
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