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In Reply to: RE: Seems like you agree with me posted by Dave_K on July 15, 2015 at 10:58:51
It doesn't matter that "both cables are driven by the same voltage source at the amp" because both cables are not exposed to the same frequency loads.
If we were to take bi-wiring further, and (passively) bi-amp, the amp connected to the high pass would not be loaded by low frequencies, and obviously the amp connected to the low pass would not be loaded by high frequencies.
Without load no current is generated. If an amp (or cable) is only allowed continuity with a narrow range of frequencies, only those frequencies will generate load and be reproduced by a speaker.
If a speaker used a shunt or low value resistor to "block" a chosen range of frequencies, your use of the word "block" would be accurate, the cable and amp would be loaded by, and conduct the chosen frequencies.
But since chosen frequencies are allowed to 'pass' and the frequencies that are not to pass, are shown an near infinite impedance, no conductance of the chosen frequencies range takes place, the amp and cables are not loaded by the frequencies.
You contradict yourself with "both cables are driven by the same voltage source at the amp, but the current is different because the impedance of the crossover is different.
If the impedance is different, do both cables see the same frequencies load?
Follow Ups:
It doesn't matter that "both cables are driven by the same voltage source at the amp" because both cables are not exposed to the same frequency loads.If we were to take bi-wiring further, and (passively) bi-amp, the amp connected to the high pass would not be loaded by low frequencies, and obviously the amp connected to the low pass would not be loaded by high frequencies.
Without load no current is generated. If an amp (or cable) is only allowed continuity with a narrow range of frequencies, only those frequencies will generate load and be reproduced by a speaker.
No kidding. I said as much in my first post. That is the crux of the issue with Vandersteen's hypothesis.
If a speaker used a shunt or low value resistor to "block" a chosen range of frequencies, your use of the word "block" would be accurate, the cable and amp would be loaded by, and conduct the chosen frequencies.
But since chosen frequencies are allowed to 'pass' and the frequencies that are not to pass, are shown an near infinite impedance, no conductance of the chosen frequencies range takes place, the amp and cables are not loaded by the frequencies.You contradict yourself with "both cables are driven by the same voltage source at the amp, but the current is different because the impedance of the crossover is different.
If the impedance is different, do both cables see the same frequencies load?
I don't see anything contradictory in what I've said. The terms drive and block are commonly used in electronics and my usage of them is consistent with their common meaning. A filter 'blocks' some frequencies and 'passes' others. A source 'drives' a load. The amp is (approximately) a voltage source, the load is a linear motor, and in between the two is a cable and crossover network that acts as a filter. The EMF produced by the amp drives the load through the filter network.
I don't want to debate it further because as far as I can tell we both agree regarding the original topic, and this has just become a pedantic argument over common electronics terms.
Edits: 07/16/15
You wrote "his explanation doesn't work" and the oversimplification "the high pass filter portion of the crossover is blocking low frequency currents from flowing through the tweeter"
Where the reality is the high pass filter is also preventing the high pass cable (bi-wire) fron CONDUCTING high frequencies.
You wrote "the only frequency range where the low and high frequency networks interact is in the crossover range where low and high filter responses overlap" which is not true.
So no, I don't agree with you.
I think it's really lousy to say stupid things about a successful industry professional when you could go to his website and discuss (learn) the finer points with him directly.
I ment to write "Where the reality is the high pass filter is also preventing the high pass cable (bi-wire) from CONDUCTING (low) frequencies".
Apologies
Up until your last post, you have been trying to argue with me while reiterating the same points I originally made in different words, which has been bizarre.
You wrote "his explanation doesn't work" and the oversimplification "the high pass filter portion of the crossover is blocking low frequency currents from flowing through the tweeter"
Yes, that is correct, and if you look at your own previous posts they are saying the exact same thing.
Where the reality is the high pass filter is also preventing the high pass cable (bi-wire) fron CONDUCTING high frequencies.
NO! That makes no sense.
You wrote "the only frequency range where the low and high frequency networks interact is in the crossover range where low and high filter responses overlap" which is not true.
Yes, it absolutely is true. This is basic electronic theory at work. And your own previous posts are consistent with this.
I think it's really lousy to say stupid things about a successful industry professional when you could go to his website and discuss (learn) the finer points with him directly.
Just because he's an industry professional doesn't mean he's right. Or in this case, just because he's a world class speaker designer doesn't prove his knowledge of electromagnetics is correct.
His FAQ says this:
"Additional experiments with a Hall Effect probe revealed that high-current bass frequencies created a measurable field around the wires that expanded and collapsed with the signal. We believe that this dynamic field modulates the smaller signals, especially the very low level treble frequencies. With the high-current signal (Bass) separated from the low-current signal (Treble) this small signal modulation was eliminated as long as the cables were separated by at least an inch or two. (To keep the treble cable out of the field surrounding the bass cable.)"
As I stated before, the above hypothesis is not a valid explanation. Faraday's Law establishes that a time varying magnetic field produces an electric field that induces EMF in a circuit. As such, bass currents flowing in the LF speaker cables will generate a magnetic field that induces EMF at bass frequencies in the HF speaker cables. However, this is an entirely moot point because the HF speaker cables are already being driven with a full range signal containing the same bass frequencies by the amp. The high pass filter in the crossover rejects the induced EMF at bass frequencies just as it rejects the amp's EMF at bass frequencies.
There is another problem with his explanation which I didn't even bother to pick on the first time because the point above is sufficient to reject his explanation. The other problem is that he claims there is signal modulation, which is a non-linear process. Audio cables are linear devices and can't produce modulation.
Finally, as I said before, there are a few other explanations for why biwiring would make a difference, starting with the most basic one: the cable impedance is different.
Try this..
Rig up a high pass filter that you can temporarily place at the terminals of one of your speakers, or on a spare woofer if you have one.
Use two individual wires as your speaker cable for this test.
Place a clamp-on ammeter on one of the wires (+ or - but not both) to the speaker.
Play a test CD with a steady low bass tone.
Measure the milliamperes (AC) with, and without the high pass filter over the entire length of the wires.
You will see that with the high pass filter in place the milliamperes are lower, throughout the length of the wires than without the filter.
Do it, and you will see that your statement "however this is an entirely moot point because the high frequency speaker cables are already being driven with a full range signal containing the same bass frequencies by the amp" Is incorrect.
With bi-wiring, the high pass cable is not subjected to the "measurable field" generated by bass frequencies, over the entire length, of the high pass cable.
I saw exactly this same argument occurring on the subject in another forum - at least I think it was the same. Basically if the same current runs through both bi wire cables or if the high pass low pass causes the content of the wires to be different.
I was wondering the answer and I don't know it - but assumed that "nothing exists in the wire until the circuit is closed" and once it is closed with a filter in place then only the unfiltered content will be running in it
If this is NOT true then at least Vandersteen's rationale makes no sense
JaroTheWise
Both cables in a passive bi-wire setup have the same electrical POTENTIAL.
IT's true, "the high pass low pass causes the CURRENT (not content) measured in milliamperes, of the wires to be different".
What exists in the wire untill the circuit is closed is an electrical potential, that can be measured with a VOLT METER or a FREQUENCY METER but not an AMMETER because without electrical continuity (a circuit) to a LOAD there is NO CURRENT or milliamperes.
If you were to separate the two wires on your toaster and place a clamp-on ammeter on one of the wires, you would not see any amperage (current) untill depressing the lever to make toast.
Your statement "and once it is closed with a filter in place then only the unfiltered content (current) will be running in it" is correct.
Regards, Craig
nt
The VOLTAGE delivered to both pairs of speaker terminals by both cables is full range. Due to the different impedances of the crossover networks and the different impedances of the drivers, the CURRENT flowing through both cables is different and not full range.
"nothing exists in the wire until the circuit is closed" is not correct. When connected to a voltage source (i.e. the amp), an electric field exists in the wire regardless of whether current is flowing. The electric field provides the electro-motive force that causes charges to flow in the wire creating current.
Magnetic induction per Faraday's Law would result in VOLTAGE being induced in the HF cable by currents in the LF cable.
Well you certainly know better than I do so I make no claim to contest what you are saying - I just find the topic interesting to prove or disprove the validity of the rationale of bi-wiring - at least the rationale as stated by Vandersteen and the Hall Effect. Some were arguing I think that the whole signal current existed in the wiring despite the crossovers which would invalidate it - but I assumed that Vandersteen could not be so wrong in stating it, even if he might be wrong in saying the Hall Effect is the major element that is responsible for the benefit
Not to mention that many or most "bi-wire" cables are either one cable with a second set of pigtails at the speaker end or alternatively two runs of cable but put in the same sheath which also would not benefit from eliminating Hall effect factors - unless maybe they were insulated in some manner to prevent it
But I find these posts educational - as I was never schooled in this subject
JaroTheWise
He said he measured the magnetic field with a Hall effect probe.
If you draw out the circuit diagrams for the single wire and bi-wire cases and do a bit of algebra to solve for the total impedance, you will find that the total impedance of the bi-wire setup is different than single wire within the frequency region around the crossover point where the filter responses overlap.
You can also draw a circuit diagram with a voltage source at the woofer representing back-EMF and calculate how much of this reaches the tweeter. You will find the answer is different for the single wire and bi-wire cases within the frequency region around the crossover point.
All the discussion about differences in current in the wires is really besides the point. That's not what makes bi-wiring different than single wiring - at least not from a theoretical POV.
Well the reason that I cite it is just parroting Vandersteen mostly - yet I find his explanation improbable enough to gladly accept other explanations - I think some of these other "causes" are maybe more tied into a specific speaker design as being one that would benefit
Heck if the Hall effect was really it then it sounds like you could take a seperate double biwire and put the wires together and then pull them apart and hear the difference - which I can try but seriously think nothing would be audible
JaroTheWise
Try this..
Rig up a high pass filter that you can temporarily place at the terminals of one of your speakers, or on a spare woofer if you have one.
Use two individual wires as your speaker cable for this test.
Place a clamp-on ammeter on one of the wires (+ or - but not both) to the speaker.
Play a test CD with a steady low bass tone.
Measure the milliamperes (AC) with, and without the high pass filter over the entire length of the wires.
You will see that with the high pass filter in place the milliamperes are lower, throughout the length of the wires than without the filter.
No shit.
(I assume you meant for this experiment to be done on the LF terminals of the speaker)
Do it, and you will see that your statement "however this is an entirely moot point because the high frequency speaker cables are already being driven with a full range signal containing the same bass frequencies by the amp" Is incorrect.
With bi-wiring, the high pass cable is not subjected to the "measurable field" generated by bass frequencies, over the entire length, of the high pass cable.
This is where you are wrong. I assume you understand Ohm's Law, V=IR.
Do you agree that at the amplifier end, the voltage signal applied to both cables is the same? And do you also agree that it is full range, so that the voltage applied to the HF cable at the amp end includes the bass frequencies?
Now do you also agree that the current in the HF cable is NOT full range, because the high pass filter in the crossover is blocking current flow in its stop band?
If we're good so far, then apply Ohm's Law to calculate the voltage drop across the length of the HF speaker cable at a bass frequency, e.g. 100 Hz. Since the current is essentially zero at this frequency, the voltage drop is also essentially zero, and therefore the voltage at the speaker terminals is the same as at the amp end. Ergo, a full range signal is present at the HF speaker terminals.
Running separate wires from the amp to the LF and HF terminals of the speaker does not make the low frequencies magically disappear from the signal at the input to the high pass portion of the crossover, or the high frequencies disappear from the input of the low pass portion.
Do I agree that at the amp, the voltage signal applied to both cables is the same? Yep, absolutely.
Do I also agree the current in the HF cable is NOT full range, because the high pass filter in the crossover is blocking current flow in its stop band? That's for damn sure.
Don't need to go any further because we've already proved that CURRENT is blocked by the high pass filter, and it's CURRENT that creates "measurable fields" that we want to avoid throughout the length of our cables.
In reply to your last paragraph, the frequencies don't disappear, they are attenuated, and not just at the terminals, but as you wrote "Now do you also agree that the current in the HF cable is NOT full range, because the high pass filter in the crossover is blocking current flow in its stop band?" So... How would low frequencies make it to the HF terminals, or the high frequencies to the LF terminals without current?
So... How would low frequencies make it to the HF terminals, or the high frequencies to the LF terminals without current?OK, since we're going to be pedantic and precice, what exactly do you mean by "low frequencies make it to the HF terminals"? Specifically "make it"?
Thanks.
Edits: 07/16/15
"Make it" was my poor substitute for conducted, in reply to your less than ideal term "disappear".
As in, when bi-wiring low frequencies are attenuated over the entire length of high frequency cable, and high frequencies are attenuated over the entire length of low frequency cable.
Good evening
Right. It has to be over the entire length. It's the current that is being attenuated. The voltage is not attenuated until post filter. But the current is attenuated all the way back to the amplifier terminals where the bi-wire split is made.I think Dave is adding discussion about interaction between cables where the individual currents can modulate one another when separated into adjacent cables. This is complex, because we need to know about more than just the distance separating the cables, but the cable geometry including insulation thickness and turns per unit length.
If we're going that far, there are lots of details that would need to be added to the theoretical model if that specific detail were to be added, such as the individual voltage drops *across* each bi-wired cable, which would be different, since the currents going through them are also different, despite each cable having the same impedance. In that case the voltages presented to each filter input are NOT identical, albeit close.
The, introduce two different cables into the model and ALL kinds of mathematical hell breaks loose! ;)
Cheers,
Presto
Edits: 07/17/15
It does not make the full range voltage go away.
It does make the portion of the filtered current go away.
You can't speak of "signal" without confusing the issue.
When you say "signal" are you referring to the full range voltage at the filter terminals? Or the filtered current?
You can't have "unfiltered current" going into the filter unless it is ONLY a shunt filter. If there are series components, then the filter is actually blocking the current below the Fc of the crossover point, with attenuation (impedance) increasing as frequency goes away from Fc.
You most definitely have unfiltered voltage present. Voltage does not get filtered. Current does.
Replace the word "signal" in all of your posts with either the word "current" or "voltage" and the argument would end immediately with either you two agreeing or one of you being wrong. ;)
Cheers,
Presto
You most definitely have unfiltered voltage present *at the input to each filter*.
That clarifies what I said...
It does not make the full range voltage go away.
It does make the portion of the filtered current go away.You can't speak of "signal" without confusing the issue.
When you say "signal" are you referring to the full range voltage at the filter terminals? Or the filtered current?
If we're going to discuss things with pendantry, I guess you have a fair point. But really, which did you think I meant?
You can't have "unfiltered current" going into the filter unless it is ONLY a shunt filter. If there are series components, then the filter is actually blocking the current below the Fc of the crossover point, with attenuation (impedance) increasing as frequency goes away from Fc.Very true, and this is exactly what I stated earlier. We're talking about actual bi-wire capable speakers in this thread, which means parallel LPF and HPF crossover networks. And I don't believe there are any which implement the HPF via shunt inductors only without a series cap. So when I said earlier that the HPF blocks low frequencies, I meant that in the literal sense.
You most definitely have unfiltered voltage present. Voltage does not get filtered. Current does.Not true! Since we're being pedantic now, please reconsider your statement. Voltage most definitely DOES get filtered. If not, the filter wouldn't work. If the full range voltage is present at the output of the high pass filter, bass frequencies and all, what would stop it from producing full range currents through the tweeter? Think about it.
Edits: 07/16/15
Don't move the goalposts. We're talking CABLES. Cables are pre-filter.
Voltage does not get attenuated pre-filter. Only post filter.
Current is attenuated in the whole branch, i.e., in the individual cables - with one feeding each filter.Let's make it even simpler.
In a pair of bi-wire cables, the current to the low pass filter is only low pass current. The current to the high-pass filter is only high-pass current.
The sum of both currents electrically would be equal to the highpass current and the lowpass current combined, which is precisely what you see at the amplifier terminals IF only one amp is used.
If two amps are used, in a passive bi-wire configuration, each amp (like each cable) sees respective filter current only. (High sees high, low sees low).
Cheers,
Presto
Edits: 07/17/15
This should be entirely obvious if you read back through the thread where I've stated it several times, but I'll clarify once again:
I'm talking about the EMF induced on the HF cable from the magnetic field produced by the bass currents in the LF cable.
If you measure the voltage at the HF speaker terminals, it will come from two sources: the amp and induced EMF. Both sources will contain bass frequency content, and the high pass crossover section will filter out the bass frequency content regardless of source.
If the HF speaker cable is exposed to low frequency EMI from any source, it's not going to bother the tweeter as long as the frequencies are well into the stop band of the HF crossover section.
Oh, I am referring to an ideal model where there is only current driven by a voltage source and not induced.
In any case, a case is made for separating high and low pass filters from each other. Some 'philes have heard differences simply by splitting the board in half and moving the respective filters apart. By virtue of this, the cable in a bi-wire config would also move apart. All this said, the theoretical field around two conductors (send/return) in parallel and close proximity is zero. There may be interaction within a single cable jacket with two pairs cables, but if the cables are inches apart, I don't think it matter nearly as much if at all.
Cheers,
Presto
Oh, I am referring to an ideal model where there is only current driven by a voltage source and not induced.
The whole point of this sub-thread was RV's hypothesis regarding the magnetic field interaction between the two cables causing low frequencies to modulate high frequencies.
Oh, okay. To me it sounded like at certain points that things became fuzzy about current versus voltage at various points in the circuit.One thing that came out of this thread that is of interest to me though, ponder this:
Why would having low and high-pass filter currents in the same cable be less deleterious than having two separate, but in proximity, cables for low and high-pass filter currents? Aka, could not the low frequency content also modulate the high frequency content if in the same cable? In that case, the field associated with the lower frequencies is located immediately to the high frequency field.
In the case for separate cables, they can be moved further apart - reducing or even eliminating any possible effects.
Cheers,
Presto
Edits: 07/20/15
Why would having low and high-pass filter currents in the same cable be less deleterious than having two separate, but in proximity, cables for low and high-pass filter currents? Aka, could not the low frequency content also modulate the high frequency content if in the same cable? In that case, the field associated with the lower frequencies is located immediately to the high frequency field.I don't believe the low frequency content in the cable can modulate the high frequency content in the cable. Modulation is a non-linear process. What non-linear property of the cable is there that could produce this modulation?
EDIT: Let me expand on this a bit. Suppose we could replace regular wire with some kind of non-linear semi-conducting material that did cause one frequency band to be modulated by another. Given that a full range voltage is applied to the cable by the amp, and thus a full range electric field is present in the cable, the modulation would occur regardless of whether or not there is induced EMF from the magnetic field of a nearby cable.
Regarding the basic question of how can single wiring be less deleterious, I would expect it's system dependent. The single and bi-wire circuits are different and not equivalent to each other, so it's plausible that with some systems and cables bi-wiring might make things sound worse.
Edits: 07/22/15
The more I think about it the only thing I can up with is that the low frequency content in a full range signal is *supposed* to be there along with the high frequency content where with bi-wire any low frequency current induced in the high frequency cable is *not* supposed to be there.In a moving speaker cone, multiple frequencies being reproduced at once is another matter...
I wonder if cable self inductance can be a possible source of modulation within the cable for a full range signal? (In this case I thought we were dealing with relatively tiny inductance values that had very little effect...)
Cheers,
Presto
Edits: 07/22/15
Self-inductance is still linear
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