Why Bi-wiring Makes No Sense.

[jneutron]

Come on John! The tweeter filter power in the bi-wire case is given by the equation:

Pfilter1 = Ia * Ib * (Rwoofer + Rwoofer-wire)

It is NOT decoupled AT ALL from the woofer current.

The equations for the mono-wire and the bi-wire can be transformed into one another. I've shown you how to do that.

Clearly you don't understand high school math.

Sigh.

In the biwire case, the tweeter circuit is unaffected by the woofer current.

Review the circuit diagram.

They are both connected to a set of VOLTAGE terminals.

Cheers, John

 

[jneutron]

jneutron is arguing that Omicron's math is wrong, but.... he's the only one presenting any math at all.

You also need to pay attention.

Biwiring removes the woofer current from the tweeter circuit filter equation.

Think it through.. If you have a biwire setup, the two independent circuits are connected at the amp terminals. If you disconnect the woofer circuit from the amp terminal, how does that affect the tweeter filter now???

Answer: It doesn't.

Kurts equations show the connection between the woofer VOLTAGE plus the (woofer current * wire resistance) as affecting the filter circuit of the tweeter.

If you move the woofer connection to the amp, you eliminate the IR portion of the woofer current, leaving only the V portion.

Do not be confused...look at the circuit!

Cheers, John

 

[Omicron]

Sigh.

In the biwire case, the tweeter circuit is unaffected by the woofer current.

Review the circuit diagram.

They are both connected to a set of VOLTAGE terminals.

Cheers, John

You really don't get it do you?

The tweeter filter has to drop de LF voltage over itself. The LF voltage is related to the LF current by the LF impendance, which works out to Rwoofer + Rwoofer-wire.

The current trough the tweeter filter if the HF current. So, summarized for the tweeter filter:

Pfilter = Ufilter * Ifilter = Ilf * (Rwoofer + Rwoofer-wire) * Ihf

The LF current IS present in the power expression for the tweeter filter!

Don't blame me if you draw the wrong conclusions by "just looking" at the schematic.

 

[jneutron]

Actually jneutron is arguing that everyone's math is wrong. Except his off course. Which he doesn't want to show because us mere mortals just need to discover it for ourselves..

No, actually I am saying that you are so intent on looking at the equations, that you are forgetting to look at the circuit.

Biwiring removes the IR woofer drop from affecting the tweeter filter blocking potential. (remember that p=vi part?)

Without the woofer current (the reactive part) having a path to affect the tweeter filter, there is no interaction as a result of the current.

Cheers, John

 

[jneutron]

You really don't get it do you?

The tweeter filter has to drop de LF voltage over itself. The LF voltage is related to the LF current by the LF impendance, which works out to Rwoofer + Rwoofer-wire.

The current trough the tweeter filter if the HF current. So, summarized for the tweeter filter:

Pfilter = Ufilter * Ifilter = Ilf * (Rwoofer + Rwoofer-wire) * Ihf

The LF current IS present in the power expression for the tweeter filter!

Don't blame me if you draw the wrong conclusions by "just looking" at the schematic.

Gents


Show me how that equation relates to the biwire schematic:

http://forums.audioholics.com/forums/attachment.php?attachmentid=4219&d=1172090733

In the biwire case, the tweeter and cap are unaffected by the woofer current.

Cheers, John

ps. I had to paste the url...does anybody know how to put it in as an attachment instead?

 

[trnqk7]

I thought only pure inductance lagged by exactly 90 degrees...hmm. I would suggest the current will lag, but by a smaller amount as the "inductor" has some resistance to it.

I would also suggest that although the woofer current may not directly influence the tweeter current, they are not entirely uncoupled either. Since you only have x current from the single amp, if the woofer draws too much current it will limit the available current to the tweeter (or, although unlikely, vice versa could happen as well).

 

[jneutron]

I thought only pure inductance lagged by exactly 90 degrees...hmm. I would suggest the current will lag, but by a smaller amount as the "inductor" has some resistance to it.

Yep. Absolutely correct.

The closest I get to a pure inductor is a superconducting magnet wound on a G-10 support tube. When I make them on metal, they tend to lag in frequency response from the ideal 90. The worst part is, many metals decrease their resistance at 4 Kelvin, and this causes lots of eddy currents. As a result, there are losses in the tube when I change the coil current. What also happens is some additional phase lag due to the eddies.

For room temperature magnets, the eddy currents can play a major role, expecially at 100 to 1000 hz. One magnet I'm dealing with now has aluminum cooling plates within, and at 100 hz the field lags the current 30 degrees beyond the normal 90.

For speakers, a voice coil is within a metal cylindrical wall, so when energized, the wall eddy currents also cause an electrical lag. While this reduces the measured inductance of the voice coil, it does nothing to reduce the magnetic field lag, which is what drives the force of the coil. I've never heard of it, but if somebody were to lock a VC in place in the gap and measure the induced force due to current, they'd be able to measure the lag caused by the wall eddy currents. The only way I could see removing that effect would be to slice the walls and pole piece to interrupt the eddy currents, just like they do in transformer laminations, while retaining the flux path of the DC circuit.

The cabinet resonates, bass reflex cabs need a few cycles to achieve the steady state response..

All these things cause the current to deviate in time. The assumption that a real speaker will show as resistive is an incorrect one.

I would also suggest that although the woofer current may not directly influence the tweeter current, they are not entirely uncoupled either. Since you only have x current from the single amp, if the woofer draws too much current it will limit the available current to the tweeter (or, although unlikely, vice versa could happen as well).

Again, absolutely correct. But for the purposes of this discussion, I believe all the players agree that assuming the amp has zero output impedance and infinite capability, is not a bad one for now.

Cheers, John

 

[trnqk7]

Scratch that-I just showed myself what was wrong with my first statement. It would take a very crappy inductor or very small inductor for what I said to be entirely true...as a power engineer, we usually deal with slightly "larger" numbers

 

[jneutron]

Scratch that-I just showed myself what was wrong with my first statement. It would take a very crappy inductor or very small inductor for what I said to be entirely true...as a power engineer, we usually deal with slightly "larger" numbers

Don't be so hard on yourself...we are dealing with small inductors...and lossy ones at that..

Cheers, John
ps..

HOWEVER, I NEED TO HAVE A WORD WITH YOU BUDDY BOY!!!!!!


STOP MAKING THE SUBSTATION TRANSFORMERS MORE EFFICIENT!!!!!

You make the bolted fault currents higher and higher, all for the sake of being "green".

I'd love to see a "tree hugger" work for an hour in a 40 cal suit..just because you power guys are trying to save energy....

You've made my job just a wee bit more difficult...

 

[jonnythan]

You also need to pay attention.

Biwiring removes the woofer current from the tweeter circuit filter equation.

Think it through.. If you have a biwire setup, the two independent circuits are connected at the amp terminals. If you disconnect the woofer circuit from the amp terminal, how does that affect the tweeter filter now???

Answer: It doesn't.

And how is that different from if you disconnect the woofer circuit at the binding post?

 

[trnqk7]

It is not any different.

 

[jneutron]

And how is that different from if you disconnect the woofer circuit at the binding post?

For the biwiring case, it doesn't. Remember, just because you disconnect the woofer from the amp terminal, it doesn't mean the woofer voltage signal disappears..that is still there to be seen by the tweeter filter.

Cheers, John

 

[jonnythan]

For the biwiring case, it doesn't. Remember, just because you disconnect the woofer from the amp terminal, it doesn't mean the woofer voltage signal disappears..that is still there to be seen by the tweeter filter.

Cheers, John

Huh?

Yes it does. There's no such thing as a "woofer voltage signal" and if you disconnect the woofer from the amp terminal in a bi-wire configuration, you've changed the circuit significantly. You've taken the parallel woofer circuit completely out of the diagram.

I know you know better than this so clearly we are not understanding each other.

 

[jneutron]

Huh?

Yes it does. There's no such thing as a "woofer voltage signal" and if you disconnect the woofer from the amp terminal in a bi-wire configuration, you've changed the circuit significantly. You've taken the parallel woofer circuit completely out of the diagram.

I know you know better than this so clearly we are not understanding each other.

You are not understanding. I'll try to explain.

For the equations posted, two signals were output...an 11 volt low frequency one, and an 11 volt hf one. The peak voltage from the amplifier was 22 volts, the summation of the two.

In a biwire configuration, the tweeter plus filter sees the entire 22 volt signal. This is independent of having a woofer connected to that amplifier output.

Re-examine the biwire and monowire schematic please. I'm sorry I had to put a link in, I don't know how to simply put the pic up, and I can't seem to upload any more..


Cheers, John

 

[trnqk7]

There wouldn't be a "summation" though, the frequency information is all part of the output voltage and is all coming out of the same output terminal. There are not "two little signals" or even "1000's of little signals" (if you were going to go for every frequency being played at the same time) coming out...there is one signal with all that information. Show me if I'm wrong-I'm not preaching gospel...but I'm pretty sure that is how it works.

 

[Omicron]

Gents


Show me how that equation relates to the biwire schematic:

http://forums.audioholics.com/forums/attachment.php?attachmentid=4219&d=1172090733

I'll do better than tat. I'll give you a generic solution for either schematic. With the equations I'm going to derive for you, you can calculate all powers in either circuit. So we will hopefully end this sillyness once and for all. And if you object then show us YOUR equations and stop blabbering!

Let us suppose a circuit "halfway in between" the mono and bi-wire cases. One set of wires leaving the amp. Halfway they split into 2 sets of wires one going to the tweeter, one to the woofer. Say:

Rw1 = wires from the amp to the "in between" node
Rw2 = wires from "in between" node to the tweeter
Rw3 = wires from "in between" node to the woofer

The bi-wired case can then be considered as Rw1 = 0 and conversely the mono wire case can be considered Rw2 and Rw3 = 0. So the resulting set of equations will apply to both. We will also consider voltage drive and all expressions will be done in terms of voltage only (just so you can remove the woofer branch without breaking the equations on an infinity).

So:

Pamp = Pw1 + Pw2 + Pw3 + Pfil-tw + Pfil-wf + Ptw + Pwf

With:

Pamp: power put out by the amp
Pw1, Pw2, Pw3: the wire losses
Pfil-tw: power term of the tweeter filter
Pfil-wf: power term of the woofer filter
Ptw: power term of the tweeter
Pwf: power term of the woofer

Al being functions of time (not talking about average power here).

From this we can say:

Pspeakers = Ptw + Pwf = Pamp - (Pw1 + Pw2 + Pw3 + Pfil-tw + Pfil-wf)

What we want to prove here is that there is no distortion in the speakers and hence the power in the speakers must be given by an expression that looks like this:

Pspeakers = K1*sqr(Va) + K2*sqr(Vb)

With K1 and K2 constants that must be independent of time and independent of either Va or Vb. Va is the HF signal and Vb is the LF signal. Both functions of time (although their exact shape doesn't matter, only that one is pure LF and the other pure HF).

Good, let's start:

Pamp = (Va + Vb)*(Ia + Ib)

Using the fact that the filters have either zero or infinite impendance for each signal, we can calculate the resistance seen by the LF and HF signals:

Let us call:

Ra = Rw1 + Rw2 + Rtw -> HF resistance
Rb = Rw1 + Rw3 + Rwf -> LF resistance

Then

Ia = Va / Ra -> HF current
Ib = Vb / Rb -> LF current

Or:

Pamp = (Va + Vb) * ( Va / Ra + Vb / Rb )
Pamp = sqr(Va) / Ra + sqr(Vb) / Rb + Va*Vb*(1/Ra + 1/Rb)

Pamp = sqr(Va) / Ra + sqr(Vb) / Rb + Va*Vb*(Ra+Rb)/(Ra*Rb) [equation 1]

Next, the power in the wires:

Pw1 = Rw1 * sqr(Iw1) = Rw1 * sqr(Ia + Ib) = Rw1 * sqr(Va / Ra + Vb / Rb)
Pw1 = Rw1 * sqr(Va/Ra) + Rw1 * sqr(Vb/Rb) + 2*Rw1*Va*Vb/(Ra*Rb) [equation 2]

The power in the other wires is trivial and:

Pw2 = Rw2 * sqr(Ia) = Rw2 * sqr(Va/Ra) [equation 3]
Pw3 = Rw3 * sqr(Ib) = Rw3 * sqr(Vb/Rb) [equation 4]

The power in the filters can be found:

Pfil-tw = Vfil-tw * Ifil-tw

Since the HF filter cannot drop an HF voltage (it's impedance is 0) it must drop all of the LF voltage (b) remaining at the end of the wires:

Vfil-tw = Vb - Vbw1 - Vbw2 - Vbtw

Note that since the HF filter cannot pass the LF current, wire 2 cannot be carrying any LF current so cannot drop any LF voltage either. Same for the tweeter. Hence Vbw2 = 0 and Vbtw = 0.

Vfil-tw = Vb - Ib*Rw1 - 0 - 0
Vfil-tw = Vb - Ib*Rw1
Vfil-tw = Vb - (Vb/Rb)*Rw1
Vfil-tw = Vb*(1 - Rw1/Rb)

Since the filter cannot pass an LF current:

Ifil-tw = Ia = Va / Ra

So:

Pfil-tw = (Va/Ra)*Vb*(1 - Rw1)/Rb)
Pfil-tw = Va*Vb*(Ra - Rw1)/(Ra*Rb) [equation 5]

In the same way we can calculate the power term for the woofer filter:

Pfil-wf = Va*Vb*(Rb - Rw1)/(Ra*Rb) [equation 6]


We are now in a position to calculate:

Pspeakers = Ptw + Pwf = Pamp - (Pw1 + Pw2 + Pw3 + Pfil-tw + Pfil-wf)

By combining equations 1, 2, 3, 4, 5 and 6 :

We will do this in 3 steps:

1) gather all the terms of sqr(Va):

k1 = 1/Ra - (Rw1 + Rw2)/sqr(Ra)

2) gather all the terms in sqr(Vb):

k2 = 1/Rb - (Rw1 + Rw3)/sqr(Rb)

3) gather all the terms in Va*Vb:

k3 = (Ra+Rb)/(Ra*Rb) - 2*Rw1/(Ra*Rb) - (Ra - Rw1)/(Ra*Rb) - (Rb - Rw1)/(Ra*Rb)

k3 = (Ra + Rb - 2*Rw1 - Ra + Rw1 - Rb + Rw1)/(Ra*Rb)

k3 = 0 / (Ra*Rb)

k3 = 0

This gives us the expression for the power in the speakers:

Pspeakers = Ptw + Pwf = k1*sqr(Va) + k2*sqr(Vb)

As we can see, since k3 always works out to 0 and neither k1 nor k2 contain any dependencies on Va or Vb, the dissipation in the speakers is just a linear combination of the powers we would expect for an undistorted case.

Let's see what k1 and k2 work out to for the mono and bi-wire cases:

For mono:

Rw = Rw1
Rw2 = Rw3 = 0 so
Ra = Rw1 + Rw2 + Rtw = Rw + Rtw
Rb = Rw1 + Rw3 + Rwf = Rw + Rwf

k1 = 1/(Rw + Rtw) - Rw/sqr(Rw + Rtw)
k2 = 1/(Rw + Rwf) - Rw/sqr(Rw + Rwf)

For the bi-wire case:

Rw1 = 0
Rw2 = Rw3 = Rw so
Ra = Rw1 + Rw2 + Rtw = Rw + Rtw
Rb = Rw1 + Rw3 + Rwf = Rw + Rwf

k1 = 1/(Rw + Rtw) - Rw/sqr(Rw + Rtw)
k2 = 1/(Rw + Rwf) - Rw/sqr(Rw + Rwf)

Surprise surprise...since k1 and k2 are the same the power in the speakers is also exactly the same in both cases! Not even a difference in magnitude. If you use the same wire resistances, obviously. Otherwise there will be a difference in magnitude, but in magnitude ONLY! Nothing that cannot be fixed by turning the amp up a tad.

Note: As a simple sanity check we will put some numbers in our equations as see if they hold up:

Let's say:

Rw1 = 2 ohms
Rw2 = 4 ohms
Rw3 = 8 ohms
Rtw = 12 ohms
Rwf = 16 ohms
Va = 3 V (instanteneous value, not RMS or average or peak)
Vb = 5 V (instanteneous value, not RMS or average or peak)

Then:

Ra = 2 + 4 + 12 = 18 ohms
Rb = 2 + 8 + 16 = 26 ohms

k1 = 1/18 - (2 + 4)/sqr(18) = 0.037037
k2 = 1/26 - (2 + 8)/sqr(26) = 0.023669

Pamp = sqr(3) / 18 + sqr(5) / 26 + 3*5*(18+26)/(18*26) = 2.871 W
Pw1 = 2 * sqr(3) / sqr(18) + 2 * sqr(5) / sqr(26) + 2*2*3*5/(18*26) = 0.257 W
Pw2 = 4 * sqr(3/18) = 0.111 W
Pw3 = 8 * sqr(5/26) = 0.296 W
Pfil-tw = 3*5*(18 - 2)/(18*26) = 0.513 W
Pfil-wf = 3*5*(26 - 2)/(18*26) = 0.769 W
Pspeakers = 0.037037*sqr(3) + 0.023669*sqr(5) = 0.925 W

And indeed:

2.871 = 0.257 + 0.111 + 0.296 + 0.513 + 0.769 + 0.925

 

[Omicron]

All these things cause the current to deviate in time. The assumption that a real speaker will show as resistive is an incorrect one

A speaker shows up mainly as resistive. Sure, not a perfect resistor by far. But it generally behaves a lot more like a resistor than as an inductor!!

Average power dissipation is given by:

Pavg = Vrms * Irms * cos(phi)

Since cos(90 degrees) is zero a purely inductive speaker wouldn't be able to dissipate any power. Hence it wouldn't be able to produce any sound energy either.

A good speaker is designed to be as close to a resistance as it possibly can be.

 

[Halon451]

It's fun to watch engineers do battle. They spill numbers, not blood.

Hey trnqk7 - PM sent.

 

[Omicron]

It's fun to watch engineers do battle. They spill numbers, not blood.

Hey trnqk7 - PM sent.

Unfortunately I seem to be the only one spilling numbers.

I leave the conclusion of what that implies up to you

 

[jneutron]

Unfortunately I seem to be the only one spilling numbers.

I leave the conclusion of what that implies up to you

I noticed that you did not answer the question..

You claim that the woofer current alters the biwire circuit tweeter (anything).


Prove it.

Stop with the equation BS...

Prove that the woofer current has anything to do with the tweeter circuit in the biwire configuration..

C'mon, try...

Look, we all see it, we know the answer.

So, what is it????


Show that the tweeter filter power has ANYTHING to do with the woofer impedance in a biwire config..Anything..

Didja notice this time??

Can it be any simpler that this?

Go ahead, try to divert...

Or, answer the question..

You crack me up...

I'da canned your behind years ago..if ya worked for me..

Cheers, John