Hi folks
I have been lurking here, mostly 'cos I don't think I can make a meaningful contribution. However, as a self taught, practical kind of guy, who's first response to a problem is to fire up the table saw, I just wanted to make a quick comment. Rather than delving deep into theory or simulation, which as I said, I'm not qualified to comment on, wouldn't be worth testing a real system to see if this kind of behaviour exists? I'd do it myself only I don't have the test gear.
al/the guy who dusts the mirrors at the Mount Wilson of diyAudio.
I have been lurking here, mostly 'cos I don't think I can make a meaningful contribution. However, as a self taught, practical kind of guy, who's first response to a problem is to fire up the table saw, I just wanted to make a quick comment. Rather than delving deep into theory or simulation, which as I said, I'm not qualified to comment on, wouldn't be worth testing a real system to see if this kind of behaviour exists? I'd do it myself only I don't have the test gear.
al/the guy who dusts the mirrors at the Mount Wilson of diyAudio.
AndrewT said:
Graph 37 is the output of an excel spreadsheet. All of the waveforms shown are the squares of the currents time the resistance of the cables.
AndrewT said:
My assumption is that the rms powers add up. If they didn't, we'd be breaking a law of thermodynamics..
Yes. It is the voltage intermodulation that must be there in order for the instantaneous power dissipated to ad up to the power delivered by the source.AndrewT said:
The biggest problem I can see is the lack of reporting of this effect. So, either my analysis has some basic flaw, or it has been there all along but nobody has cared look for it, or the tools used to look for it have a hole in them big enough for a 2AB truck to drive through..
I note this right now: under no circumstances can a report of audibility of biwiring be used to justify this analysis. This analysis has been spurred BY anecdotal reports of audibility...it is not valid to use those anecdotes as support of my hypothesis..only measurement is valid..
AndrewT said:
By using DC and an arbitrary high frequency, I believe I have eliminated the need to consider the L and C reactances from the equation. It is certainly possible that this is an incorrect assumption, and that energy storage needs consideration, however, that may not explain the 2AB component.
Cheers, John
We concur in all but the statement ""I therefore doubt the validity of any conlcusions taken from it. ""phase_accurate said:
Only because, the value of instantaneous power is defined by the instantaneous current and the instantaneous voltage. So if the instantaneous power is not what is predicted, then both I and V are also not as predicted. Hence, the dillemma.
Cheers, John
pinkmouse said:
""who's first response to a problem is to fire up the table saw,""
Remind me never to hire you for heart surgery, or to fix a flat..
Yes, that is a very good thing. As I stated within the initial post..
jneutron said:
What we are doing is modelling what the system should do. I have asserted that all in not as it seems in Kansas, and we are discussing the validity of the model.
Measurement of the prediction of the model I am proposing is an absolute must for closure. The level of the effect should be on the order of the sqr of (2ABRc)1/2. Failure to find this error voltage means there is something wrong with the model..sending me back to the blackboard.
Finding this error voltage sends us all back to the blackboard, as, this has never been reported previously..
It would also tend to affirm the validity of bi-wiring based solely on resistive effects of branch circuitry.
Either conclusion is a data point, so I win either way...
HA..Cheers, John
Hi John,
As I understand it your argument hinges on the term 2AB. I should perhaps point out that power is not a linear function of current (or voltage) and hence you can't just add to currents and expect the powers to add as well. Put more plainly: if you increase the current trough a resistance just a little bit then the power will increase a lot more. Superposition does NOT APPLY TO POWER!
Let us simplify the case to 2 DC current sources driving a simple resistor. If I switch on only one of the sources (say A) then the power dissipated in the resistance is obviously sqr(A)*R. If I switch on only source B then this power is sqr(B)*R. Now, if we swith on both current sources is the power then equal to sqr(A)*R + sqr(B)*R? No, it is in fact equal to sqr(A+B)*R = sqr(A)*R + sqr(B)*R + 2*A*B!
So where did this mysterious 2AB come from? Simply from both current sources. They are now together supplying more power than the sum of either one alone acting on the resistance. This is because power does not increase linearly with current (or voltage).
In the case you give indeed more power is lost in the cable with both signals present. But also MORE power is dissipated in the load at this point and MORE power is being delivered by the amplifier. ALL off these powers go up with an equivalent '2*A*B' term. Hence there is no distortion...
Kind regards,
Kurt
As I understand it your argument hinges on the term 2AB. I should perhaps point out that power is not a linear function of current (or voltage) and hence you can't just add to currents and expect the powers to add as well. Put more plainly: if you increase the current trough a resistance just a little bit then the power will increase a lot more. Superposition does NOT APPLY TO POWER!
Let us simplify the case to 2 DC current sources driving a simple resistor. If I switch on only one of the sources (say A) then the power dissipated in the resistance is obviously sqr(A)*R. If I switch on only source B then this power is sqr(B)*R. Now, if we swith on both current sources is the power then equal to sqr(A)*R + sqr(B)*R? No, it is in fact equal to sqr(A+B)*R = sqr(A)*R + sqr(B)*R + 2*A*B!
So where did this mysterious 2AB come from? Simply from both current sources. They are now together supplying more power than the sum of either one alone acting on the resistance. This is because power does not increase linearly with current (or voltage).
In the case you give indeed more power is lost in the cable with both signals present. But also MORE power is dissipated in the load at this point and MORE power is being delivered by the amplifier. ALL off these powers go up with an equivalent '2*A*B' term. Hence there is no distortion...
Kind regards,
Kurt
Excellent points.Omicron said:
In fact, it is exactly the issue of power dissipation going up as the square that I am speaking of. Where your thought model differs from mine, is the use of two loads and a branch. If only one load were dissipating power, then what you have stated is exactly correct. However, consider the two branch model I speak of:
If you push only 1 ampere dc through, the cable will dissipate 50 mW. The load in series with the inductor will dissipate 8 watts.
If you push only 1 ampere peak sine through this circuit, the cable will dissipate 50 mW peak, and the load in series with the capacitor will dissipate 8 watts peak.
If you push both signals at the same time, then the cable will see at some points in time, 2 amperes peak, and the resistor will dissipate 200 mW at that instant in time. During the negative peak of the ac signal, the resistor will have no net current, and therefore, zero dissipation.
So, the change from a single cable to two cables changes the amount of power that is being dissipated by the entity which delivers the power to the load.
If the amplifier is sending the same power to the circuit, and the wires to the loads dissipate power differently, how is it the loads can see the exact same energy dissipation regardless of the wiring scheme?
The term 2ABR appears to be re-distributing how the power is being dissipated within the circuit on a time varying basis, but not on an integral one.
Cheers, John
btw, the editing capability of this forum allows changes for 30 minutes I believe. You could have easily fixed your post instead of a second post.
John,
You are closer... The VOLTAGE amplifier must supply the "missing" 2AB power term... and this term is small. Power amplifier is perhaps a poor term... these are voltage amplifiers with (hopefully) low ouput impedance.
You are starting with currents... no-no? You are presuming a transconductance amp.
Start with a voltage (the instaneous sum of sines). Derive currents based on system or branch impedance and the mysteries go away. Take the current(s) and apply them to their respective loads and see what comes out... pure signs.
The only thing I can see is that biwiring simplifies the impedance when modeling the individual crossover networks. This interaction between being cable resistance and and effective crossover frequency was handled well in the link posted earlier: "BI-WIRING". Mutual impedance, in the single wire case, can shift the HP filter lower and the LP filter higher creating an overlap and an apparent lump in output power. This effect diminishes when we size cable such that its resistance is small in comparison load impedance(s)... square one.
You are closer... The VOLTAGE amplifier must supply the "missing" 2AB power term... and this term is small. Power amplifier is perhaps a poor term... these are voltage amplifiers with (hopefully) low ouput impedance.
You are starting with currents... no-no? You are presuming a transconductance amp.
Start with a voltage (the instaneous sum of sines). Derive currents based on system or branch impedance and the mysteries go away. Take the current(s) and apply them to their respective loads and see what comes out... pure signs.
The only thing I can see is that biwiring simplifies the impedance when modeling the individual crossover networks. This interaction between being cable resistance and and effective crossover frequency was handled well in the link posted earlier: "BI-WIRING". Mutual impedance, in the single wire case, can shift the HP filter lower and the LP filter higher creating an overlap and an apparent lump in output power. This effect diminishes when we size cable such that its resistance is small in comparison load impedance(s)... square one.
I my view your last premise is wrong. I.e. for a total current of A+B to flow (no matter if they are AC or DC) the amplifier must deliver MORE power to the circuit than simply the sum of the powers needed to let A or B flow alone.
So, yes the losses caused by the resistance of the wire are greater when you apply both signals together. However, this difference is not taken from the power at the load (the power there can't have changed since the same currents are flowing there). Instead the amplifier is required to deliver this extra power. If it can't then the total current will not be A+B but it wil be less.
In this perspective all you do by adding another wire is lowering the losses in the wire (the resistance of the entire thing halves) and thus let the amplifier do a very tiny amount less work to deliver the same amount of power at the load.
PS: this was the very first time I posted on this forum and I've yet to figure out how to edit a post after it was submitted
So, yes the losses caused by the resistance of the wire are greater when you apply both signals together. However, this difference is not taken from the power at the load (the power there can't have changed since the same currents are flowing there). Instead the amplifier is required to deliver this extra power. If it can't then the total current will not be A+B but it wil be less.
In this perspective all you do by adding another wire is lowering the losses in the wire (the resistance of the entire thing halves) and thus let the amplifier do a very tiny amount less work to deliver the same amount of power at the load.
PS: this was the very first time I posted on this forum and I've yet to figure out how to edit a post after it was submitted
poobah said:
Your answer has not addressed the difference.
I used a current source to simplify the discussion. By using current, the cable dissipation is fixed and invariant. I chose that so that everybody would find understanding a bit easier.
Had I chosen voltage as the locked entitiy, the currents would have to be derived.
By choosing current for the though experiment, it is simple to see the reaction of the circuit to both configurations by viewing the input node...the 2ABRc should make itself apparent there, added to the load's reaction to the stimulus.
Had I chosen voltage, then it would be necessary to view the current out of the input node. A bit more difficult to do without disrupting the circuit loop impedance.
Previous threads did not address this 2ABRc power loss resulting from branch analysis. There is no use in re-hashing that analysis here.
Cheers, John
Omicron said:
Good..
So, then what you are saying is this..when the switch from a bi-wire configuration is made to a monowire, that the resultant increase in power dissipation in the resistor is made up by the amplifier simply providing more power..
Let's examine the consequences of that statement..
1. The resistor losses change from that of A2Rc plus B2Rc, to A2Rc plus B2Rc plus 2ABRc.
2. The load losses have remained the same..A2RloadA plus B2RloadB.
3. The amplifier now provides more power to the system, to add up to A2RloadA plus B2RloadB and the cable loss of A2Rc plus B2Rc plus 2ABRc.
Re-organized: The amp power out becomes:A2(RloadA + Rc) plus B2(RloadB + Rc) and the strange term 2ABRc.
How does the amplifier know to provide the additional power component, a mix of the two signals? It knows only current and voltage.
It should be measurable.
Cheers, John
John, the amplifier only "sees" the complete impedance of the entire circuit. This impedance is not going to be the same for the mono and bi-wire cases. Therefore, you will find that for the same currents to flow in both cases you will have to fiddle a tiny bit with the volume knob of the driving amplifier in our thought experiment. The adjustment needed will be exactly that which will give rise to the 2ABR term.
Obviously if you use current sources then the "fiddling with the volume knob" is done automatically. The current source will adjust its output voltage so that the same currents will flow in both cases. Again, this will exactly compensate for the 2ABR term.
Obviously if you use current sources then the "fiddling with the volume knob" is done automatically. The current source will adjust its output voltage so that the same currents will flow in both cases. Again, this will exactly compensate for the 2ABR term.
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