I am just reading Merilaïnen's book.
Note sure of that, but negative impedance probably gives its best results to reduce THD at resonance as current driving does it well above resonance.
Sorry, but distortion at resonance is insensitive to output impedance, high, low, or negative. Of course negative output impedance will allow you to achieve very low apparent Q.
In every case, when discussing distortion we must compare outputs of the same level. That is where the confusion comes from. If you simply increase output impedance and allow the bass to peak up the distortion will rise as well, but only because bass output (excursion) is up.
In my testing I used a compressor loop to maintain a flat output level, regardless of amplifier input. Doing so showed that the distortion did not change with driving impedance.
David S.
In every case, when discussing distortion we must compare outputs of the same level. That is where the confusion comes from. If you simply increase output impedance and allow the bass to peak up the distortion will rise as well, but only because bass output (excursion) is up.
In my testing I used a compressor loop to maintain a flat output level, regardless of amplifier input. Doing so showed that the distortion did not change with driving impedance.
David S.
Hi Speaker Dave
In every case, when discussing distortion we must compare outputs of the same level.
Of course.
Sorry, but distortion at resonance is insensitive to output impedance, high, low, or negative. Of course negative output impedance will allow you to achieve very low apparent Q.
This seems to be in contradiction with the findings of many authors :
- Werner & Carrell, Macaulay for negative impedance (seen by Hegglun as a kind of MFB)
- Greiner & Sims, Hawksford (for high impedance) even before applying MFB wich was included in their project to restore a good resonance damping.
I have plans to check both techniques one day.
In every case, when discussing distortion we must compare outputs of the same level.
Of course.
Sorry, but distortion at resonance is insensitive to output impedance, high, low, or negative. Of course negative output impedance will allow you to achieve very low apparent Q.
This seems to be in contradiction with the findings of many authors :
- Werner & Carrell, Macaulay for negative impedance (seen by Hegglun as a kind of MFB)
- Greiner & Sims, Hawksford (for high impedance) even before applying MFB wich was included in their project to restore a good resonance damping.
I have plans to check both techniques one day.
The equation for Qts including all series resistances is well known, and since you supply this quantity in to your handy measurement software it can substract out that term in the equation to give you the "measurement". The program is not reporting the raw data...
-Charlie
I don't have the answer to this, but I am very curious to hear the correct explanation. My only idea was to apply the definition of Q equal to the ratio of reactive power to real power to the electrical model of a driver. However, in that model, there are resistances in series (Re + Rx) and in parallel (Res) with the reactive components (Cmes & Lces), so I wouldn't know exactly how to go about it.
Perhaps there is a much simpler explanation?
Regards,
Pete
Go to the literature:
http://diyaudioprojects.com/Technical/Papers/Direct-Radiator-Loudspeaker-System-Analysis.pdf
Re+Rg is all over the place. They appear together, because they are in series and act in the same way.
You can also read about it here, at Rod Elliot's web site, where he talks about how to measure T-S parameters and gives the relevant equations:
Measuring Loudspeaker Driver Parameters
-Charlie
Thats a bit of a ruse. The resistance we put in series is purely for the purpose of calculating impedance. While calculating impedance the actual series resistance is factored out of the equation leaving the impedance curve and that curve is used to calculate the various Q's.
Regarding Q and drive resistance, Small defines it pretty explicitly. He defines Qtco as total Q (electrical and mechanical combined) for the case where Rg (driving resistance) is zero. When Rg is not zero he calls it simply Qtc. Qm does not change when series resistence is present, but Qe does vary since Rg and Re are in series and simply add. Electrical damping is directly impacted by any series resistance and Qt is the "parallel combination" of the 2 Qs: (Qm x Qe/(Qm + Qe)).
David S.
I'm very familiar with the equation
modified Qes = [(Re + Rx)/ Re] * Qes
What I'm looking for is an explanation as to why increasing the resistance in-series with the driver increases Qes. Someone else posting in this thread has stated that adding a series resistor increases Qes as result of weakening counter EMF. I wonder if John K would agree with that?
Regards, Pete
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I guess you could think of the speaker first with no electrical connections if you want. Then the Q of the resonant system is 100% determined mechanically. Just say the Qms were 1. You can never get higher than 1 using a normal amplifier with whatever positive output impedance. As you decrease amplifier impedance and Re, you gain damping over the mechanical resonance, and the Q goes below 1. If Re and amplifier impedance were 0 you could practically kill all fundamental resonant response. As electrical impedance increases to infinity you lose more and more control over the cone until you're back to the mechanical system and no additional damping.
Interesting.
I've used amps with output impedances as low as 0.005 ohms. That should do a good job of killing the resonances, right?
I've used amps with output impedances as low as 0.005 ohms. That should do a good job of killing the resonances, right?
The correct answer to my tease is that you start by applying Kickoff's voltage law around the loop. By superposition this can be done first considering only the amplifier output voltage, in the absence of the back emf, and then for the back emf in the absence of the amplifier. What comes out is that the voltage across the voice coil is Vvc = (Re/(Re + Rs)) x (Vg - Vb) where Re is the VC DC resistance, Rs is the series resistance, Vg is the amplifier voltage and Vb is the motor generated back emf. The current through the VC, which is what generates the force to move it is I = Vvc/Re, or
I = 1/(Re+Rs) (Vg - Vb)
If you like you can think current as being composed of two parts, the forward part due to the amplifier and the backward part due to the back emf,
I = Vg/(Re +Rs) -Vb/(Re+Rs)
What this says is that regardless of Rs, the relative contributions of current from the amplifier supplied voltage and that from the back emf change in fixed proportion to each other as Rs changes. Thus the driving force and the damping force retain the same relative strengths.
This goes back to the original post. Adding a series resistance does not change the driver's Q. You can satisfy yourself by measuring the response of a driver with and without the series resistance. BUT, make sure you measure correctly. If the series resistor is placed between the amp + output and the driver + terminal, make sure you place the reference probe at the driver's + terminal so the SPL is measured with respect to the voltage across the VC. If you place the ref probe at the amplifier + terminal you will be measuring the response not of the driver but of the system consisting of the driver in series with the resistor, and it will look as if Qts has increased.
So the often stated adding a series resistance increases Q isn't really correct. Stated correctly it should say that Qts of the system of a resistance in series with a driver increases over that of the driver alone. The difference is subtle, but important, particularly when the pros and cons of passive and active crossovers are considered. In the case of a simple resistor this all works out nicely because the series resistance looks the same when looking forward from the amp as it does when looking back for the driver.
[edit] I just wanted to add that the idea of a system is important when looking at the series resistor, just as when you look at a woofer in a sealed box you are looking at a system. The resistor doesn't change Q of the driver any more than the box changes the driver Q. But the box/woofer system has a different Q than the driver just as the resistor/driver has a different Q then the driver alone. Any time we connect a driver to an amplifier through a passive or active crossover, in a box or on an open baffle, we have to look at what the system does as a whole. Not concern ourselves with what the driver does. After all, we refer to our listening equipment as stereo system, don't we?
I = 1/(Re+Rs) (Vg - Vb)
If you like you can think current as being composed of two parts, the forward part due to the amplifier and the backward part due to the back emf,
I = Vg/(Re +Rs) -Vb/(Re+Rs)
What this says is that regardless of Rs, the relative contributions of current from the amplifier supplied voltage and that from the back emf change in fixed proportion to each other as Rs changes. Thus the driving force and the damping force retain the same relative strengths.
This goes back to the original post. Adding a series resistance does not change the driver's Q. You can satisfy yourself by measuring the response of a driver with and without the series resistance. BUT, make sure you measure correctly. If the series resistor is placed between the amp + output and the driver + terminal, make sure you place the reference probe at the driver's + terminal so the SPL is measured with respect to the voltage across the VC. If you place the ref probe at the amplifier + terminal you will be measuring the response not of the driver but of the system consisting of the driver in series with the resistor, and it will look as if Qts has increased.
So the often stated adding a series resistance increases Q isn't really correct. Stated correctly it should say that Qts of the system of a resistance in series with a driver increases over that of the driver alone. The difference is subtle, but important, particularly when the pros and cons of passive and active crossovers are considered. In the case of a simple resistor this all works out nicely because the series resistance looks the same when looking forward from the amp as it does when looking back for the driver.
[edit] I just wanted to add that the idea of a system is important when looking at the series resistor, just as when you look at a woofer in a sealed box you are looking at a system. The resistor doesn't change Q of the driver any more than the box changes the driver Q. But the box/woofer system has a different Q than the driver just as the resistor/driver has a different Q then the driver alone. Any time we connect a driver to an amplifier through a passive or active crossover, in a box or on an open baffle, we have to look at what the system does as a whole. Not concern ourselves with what the driver does. After all, we refer to our listening equipment as stereo system, don't we?
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Isn't this really just a clever attempt at quibbling over terminology though ? There is always a "system" when electrical Q is being discussed.
The Qes figure quoted for a driver is based on an assumption of an external circuit consisting of a short circuit being present. As soon as you have an external short circuit you no longer have just a driver, you have a "system" consisting of a driver and an external resistance (Rs) of zero ohms, so Qe of a driver alone is technically a meaningless concept.
There is no Qe of the driver by itself, terminals open circuit, only the Qes of the total system, which just so happens to be defined, by convention, with an external resistance of zero ohms in place.
By introducing an external resistor we are simply changing the value of Rs - which must always be present for Qes to exist, by increasing it above zero.
If non-infinite external resistance must be present for Qes to be defined (and thus Qts to also be known) then its wrong to argue that the drivers Qes should be measured in isolation to the external resistor - by measuring across the driver terminals you are cancelling out the effects of the external resistance in the measurement, even though the response of the driver is still modified by the resistance.
By measuring that way, you're calculating the response that the driver would have with zero Rs, even though a non-zero Rs is present, (in other words converting the external resistor into an impedance measuring device) which to use Dave's words in an earlier post is a bit of a ruse.
Fine if you're trying to measure the raw T/S parameters of the driver with an assumed zero Rs, not fine if you're trying to measure the actual Qes and system response with the resistor present.
Qes is changing with added Rs, Qe is not changing, but Qe is a theoretical construct as it is just Qes taken when Rs equals zero, and thus doesn't actually exist when Rs is non-zero.
If Qes is changed by Rs then so is Qts. In every way that matters a black box consisting of the driver and a single series resistor has a higher Qes (and therefore Qts) than the same driver without the external resistor.
We could substitute a second suitably chosen driver with the same Qes without added series resistance and not be able to tell them apart without looking inside the black box. We have no way to tell if we are dealing with Re+Rs or Re alone.
Note that the black box consisting of the driver with a series resistor is also still not a "system" (and doesn't have a Qes) until the circuit is completed by something external.
Of course if you don't just have a single series resistor back to a voltage source, for example an L-Pad which includes a shunt resistor, you have to calculate the "lookback" impedance seen by the driver to calculate the Qes, as you allude to below.
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No, the driver Re is still in the way. The higher the damping factor, the further away you are from changing anything with the amp output impedance.
When we talk about Q we are only thinking about the height and width of the fundamental resonance, although if you could make total drive system impedance 5 mOhm with the same number of turns on the coil you might see some additional ripples in the response, cone ringing at some frequency, due to the neck of the cone being terminated to a moving brick.
Also, a couple volts of drive would barf the cone out of the basket.
Uh, yeah. Must not have had enough coffee this morning. 🙁 I missed the first part of this:
Amazing Jean Michel.
I used a class-D amplifier now my own no feedback PSE. It is not a current drive and not voltage controlled. It transforms the power influenced by driver impedance that needs to be reasonable flat for flat response.
I had a amazing gain in spaciousness of soundstage compared to class-D after optimizing the filter of the loudspeaker. I think the secret lies in the fact it is not voltage or current controlled.
And of course the midrange of the tube amp sounds more natural.
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Frequency response measurements will show why you feel there is a difference in the sound "before" and "after", because there is!
-Charlie
I optimized the filter of the horn impedance is almost flat and more a R than Z. So is the response on both amplifiers is quite the same.
Still there is a big difference in spaciousness of the soundstage Charlie. I know what the transformer behaves like with a curved load so I changed that.
My 2cts here (I've written several posts on the issue elsewhere), whenever you really want to inspect the net effect of different drive impedance alone, for a transducer in an application and changing no other variables, then this is a way to go :
1) Use normal voltage drive (Zout=0) or whatever you want to start with as a baseline. Dial in the target response by choice of cabinet/alignment and EQ. Then record the impulse response of transducer current, and listen/measure the way you are used to.
2) Build an amp with different "drive characteristic" (eg non-zero Zout, like Zout=10R, or Zout=-0.8Re+jwL, ...). Measure transducer current impulse response again (maybe pre-EQ is needed to not overstress etc).
3) Divide the first by the latter IR, then feed the result in a convolution engine that feeds you 2nd amplifier. Test your transducer again (listen and measure).
If all went well (linear operation, enough SNR, robustness of division), you should get the exact same SPL mag&phase response, even the fine grain of it, and in general you'll get the exact same acoustic impulse response, in every way it can be processed and displayed.
This is more or less obvious, as the minutely same current (and voltage) conditions must cause the same acoustic output in a causal world, unless things are so non-linear and unstable that true chaotic behaviour is invoked. For this consideration, let's also assume that VC heating is negleglible (a different phenomen, although relevant for the topic).
Which brings us to the part that is different. It is the electrical damping which is applied in different amounts whenever you change drive impedance profile. Damping comes from reacting to VC induced voltage (proportional to VC velocity), extremes range from no reaction at all (pure current drive) to full local degeneration (very low effective Re). Pure current drive means transducer operating purely force controlled, while maximum damping means it is operated in a local feedback, controlling it's velocity, until either mass or spring corner of the cone motion equation is reached, no feedback available beyond these points. To have enough gain -- and frequency range of gain along the 6dB/oct slope of velocity-controllable behaviour -- to degenrate from in a useful way, a driver with a linear and strong motor and light mass is nice to start with. Whenever the VC voltage does not have the expected (as applied) momentary value for whatever reason, huge currents flow trying to rectify the situation. Practical drive impedance will lie somewhere between the extemes, also depending on transducer construction details.
The different damping will cause different reaction to the following events :
a) behaviour near and at excursion overdrive, notably recovery from it.
b) reaction to any superimposed motion of the VC from both self-induced components of motion as well as externally induced motion (as seen from the system). Besides direct driver ill-effetc this includes things like air spring nonlinearity, etc
c) reaction to changes in acoustic loading, or changes of operation conditions in general.
You're not changing the driver itselnf, but the system behaviour beyond small signal linearized response will change, sometimes in relevant (clearly audible) amounts, although basically two setups can measure exactly the same (with the exception of distortion, which most often is lower -- close to 100% of drivers I tested, cones, domes, CDs, AMTs -- the higher the drive-Z, for frequency regions above system resonance).
Bottom line, try find the exact drive impedance profile that makes your transducer shine (sound best) in the application. For woofers, I found standard low-Z amps is OK when using a big series coil, making the Z rise at higher frequencies and the response go down (which we will want in every case).
- Klaus
1) Use normal voltage drive (Zout=0) or whatever you want to start with as a baseline. Dial in the target response by choice of cabinet/alignment and EQ. Then record the impulse response of transducer current, and listen/measure the way you are used to.
2) Build an amp with different "drive characteristic" (eg non-zero Zout, like Zout=10R, or Zout=-0.8Re+jwL, ...). Measure transducer current impulse response again (maybe pre-EQ is needed to not overstress etc).
3) Divide the first by the latter IR, then feed the result in a convolution engine that feeds you 2nd amplifier. Test your transducer again (listen and measure).
If all went well (linear operation, enough SNR, robustness of division), you should get the exact same SPL mag&phase response, even the fine grain of it, and in general you'll get the exact same acoustic impulse response, in every way it can be processed and displayed.
This is more or less obvious, as the minutely same current (and voltage) conditions must cause the same acoustic output in a causal world, unless things are so non-linear and unstable that true chaotic behaviour is invoked. For this consideration, let's also assume that VC heating is negleglible (a different phenomen, although relevant for the topic).
Which brings us to the part that is different. It is the electrical damping which is applied in different amounts whenever you change drive impedance profile. Damping comes from reacting to VC induced voltage (proportional to VC velocity), extremes range from no reaction at all (pure current drive) to full local degeneration (very low effective Re). Pure current drive means transducer operating purely force controlled, while maximum damping means it is operated in a local feedback, controlling it's velocity, until either mass or spring corner of the cone motion equation is reached, no feedback available beyond these points. To have enough gain -- and frequency range of gain along the 6dB/oct slope of velocity-controllable behaviour -- to degenrate from in a useful way, a driver with a linear and strong motor and light mass is nice to start with. Whenever the VC voltage does not have the expected (as applied) momentary value for whatever reason, huge currents flow trying to rectify the situation. Practical drive impedance will lie somewhere between the extemes, also depending on transducer construction details.
The different damping will cause different reaction to the following events :
a) behaviour near and at excursion overdrive, notably recovery from it.
b) reaction to any superimposed motion of the VC from both self-induced components of motion as well as externally induced motion (as seen from the system). Besides direct driver ill-effetc this includes things like air spring nonlinearity, etc
c) reaction to changes in acoustic loading, or changes of operation conditions in general.
You're not changing the driver itselnf, but the system behaviour beyond small signal linearized response will change, sometimes in relevant (clearly audible) amounts, although basically two setups can measure exactly the same (with the exception of distortion, which most often is lower -- close to 100% of drivers I tested, cones, domes, CDs, AMTs -- the higher the drive-Z, for frequency regions above system resonance).
Bottom line, try find the exact drive impedance profile that makes your transducer shine (sound best) in the application. For woofers, I found standard low-Z amps is OK when using a big series coil, making the Z rise at higher frequencies and the response go down (which we will want in every case).
- Klaus
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I'll have to scan my distortion curves since everybody keeps saying that LF distortion varies with drive impedance.
It doesn't.
David S.
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