Bi-wire/bi-amp

[Sparkus]

Honestly, I'm not on a bi-amp bandwagon . I have read lots of posts on whether it makes a difference and it seems to be a common argument, which is why I asked here.
I'm using Pure Direct mode for music so it's just the mains, and I'm no audiophile so I'm not hearing something minute, honestly it made a difference. I'll check a larger single cable just to see, I have lots of available wire so a test won't cost $.
Before the "passive bi-amp" I tried putting copper in place of the brass jumpers, that made no difference.
Yes, if it works don't fix it...I would like to understand the why.

 

[Beave]

Voltage isn't current.

Peng is trying hard to explain it. He must have a lot more patience than I do these days.

In a bi-wire scenario, the voltage on the two amp outputs (high freq and low freq) will be the same. The current, however, that flows in the two sets of wires going to the speakers will be divided into high and low frequency due to the load presented by the speaker drivers and crossovers.

 

[Sparkus]

No sir, voltage is not current...have a firm grasp of that .
I can understand what your saying on the current for the 2 amp outputs. When using a single cable the current for high and low would be combined on one cable.

 

[Sparkus]

I honestly thought that when I put the 14 in.
Something changed...geez o' pete...it's gonna bug the hell out of me.

 

[mtrycrafts]

You don't have to worry about it. A single wire can handle all audible frequencies as it has always done in the past and at present. Otherwise, audio systems would never have existed.

And, it can carry all the inaudible frequencies as well.

 

[mtrycrafts]

.... When using a single cable the current for high and low would be combined on one cable.

Yes, but so what? If you are worried about this effect, why not also worry about all the frequency currents in the high pass section interacting, say 2 kHz on up to 20 kHz?
Or the opposite in the low section 16 Hz to 80 Hz? Or whatever the crossover is at. They are all interacting and we have no issues in hearing the results clearly

 

[Beave]

No sir, voltage is not current...have a firm grasp of that .
I can understand what your saying on the current for the 2 amp outputs. When using a single cable the current for high and low would be combined on one cable.

Sorry, that post was meant for everybody in this thread, not just you.

And yes, what you posted is correct.

 

[highfigh]

I honestly thought that when I put the 14 in.
Something changed...geez o' pete...it's gonna bug the hell out of me.

Are you from Detroit?

 

[highfigh]

LOL!
This is exactly why I chose Mechanical Engineering over Electrical!


I think this is more properly stated as "The amplifier is making available a full-range signal to the crossovers"!
It doesn't send it unless it has a place to go and if the XO has a filter, some of it doesn't have a place to go!
So it was there, so where did it go? Now my head is starting to hurt and I am, once again, glad I chose Mechanical Engineering!

PENG or whoever, please correct me if I am mistaken on any of this, and if you can tell me what happens to the unused signal in terms dummied-down enough for me to understand it, that would be great!

PS: I think it terms of upstream and downstream, but with circuitry, I believe you have to fully understand what is "downstream" before you can know what is happening upstream!

Well, it's not the same as the amplifier saying (in proper British accent) "You, there- take this signal and do with it as you please!", but it really doesn't have anywhere else to go, so......by being connected by wires, it all has a place to go, but some is filtered, as if it were under-age kids outside of a club who were blocked from entering. Not sure if they complain about it, though.

Scientific American had an article titled 'What Happens To The Electrons When They Leave Your Toaster?'- I can't find it, but maybe it's the same result WRT the speaker crossover.

 

[Kvn_Walker]

I think your right.
I think what happened is by running 2 cables I doubled the size of my cable. If I use 12 or 10 awg and use one cable, I bet I'll get the same results. I used 14 awg, as an electrician I tend to think current, I'm guessing less resistance might be the answer here.
It was very noticeable, not subtle at all. What else could have improved the sound so much?

I'm going to ask the question I haven't seen yet... since he's now using 4 full range channels instead of 2, is the speaker not seeing double the power at any given volume? And the fact that he can reach a higher volume without clipping is because each channel is seeing a reduced load? Many people do perceive an increase in volume as an increase in quality.

Essentially he's just used two more discreet amplifiers, no different than if I bought a second pair of 2200's and separated woofer and tweeter duty. This isn't the same as using the SPEAKER A/B switch on old receivers. My tuppence anyway, if it's even worth that much.

 

[Irvrobinson]

I'm going to ask the question I haven't seen yet... since he's now using 4 full range channels instead of 2, is the speaker not seeing double the power at any given volume? And the fact that he can reach a higher volume without clipping is because each channel is seeing a reduced load?

No, the speaker is seeing the same power level, assuming the woofer amp wasn't clipping when running with bi-amping. Amplifier output level is determined by the input voltage, since amplifiers are fixed gain devices, and the input levels are the same. The power delivered is just divided between the two crossover sections.

As for amp clipping levels, the only advantage is that if the amp channel driving the woofer crossover section clips it won't affect the mid-tweeter amp channel. This could be an advantage when using an under-rated amplifier for the speaker-room combination, but it does seem rather unlikely to be a real-world advantage (unless you're listening at deafening levels).

 

[PENG]

Gents, I think some of you are missing @Sparkus original point/question. He obviously understands a single pair of speaker wires can carry the full band signal.

He only seems concerned about whether the "lower" frequency band signal would interact with the "higher" frequency band signal when the full band signal is flowing in the same pair of wires, would produce a net effects that would result in an audible difference.

So for him, he seems to understand the theory, but he is not fully convince about the effects being "audible" or not. On this forum, I have the impression that >99% of the frequent posters don't believe he needs to worry about that, but that's his original question if I understood correctly.

 

[PENG]

LOL!

I think this is more properly stated as "The amplifier is making available a full-range signal to the crossovers"!
It doesn't send it unless it has a place to go and if the XO has a filter, some of it doesn't have a place to go!

Agreed.

So it was there, so where did it go? Now my head is starting to hurt and I am, once again, glad I chose Mechanical Engineering!

Did you read my post#20, when I sort of anticipated a major misconception that non electrical people may think that electron flow in a wire with the filter and speaker driver connected in series would behave like water flows in a pipe with a filter at the consumer end.

Electric current flow does NOT work like that, you can put the filter physically anywhere along the line/wire, if the filter offers very high impedance to, say frequency below 500 Hz, the amplifier will send little signal current below 500 Hz, period, as soon as steady state condition is reached (very quickly for short lines in audio applications).

PENG or whoever, please correct me if I am mistaken on any of this, and if you can tell me what happens to the unused signal in terms dummied-down enough for me to understand it, that would be great!

Since I supposedly have more patience than @Beave so let me try again. The "unused" part of the signal would be there, measurable as voltage/potential difference but not current if really unused. The frequency of the resulting signal current caused by the voltage will depend on the individual impedance of each of the frequency component of the signal. As often cited, analog filters are not brick walls so "unused" seems like an extreme example, "partially used/or unused" may be more appropriate.

As an numerical example for "a partially used/unused" case, consider the calculation below for a HPF+tweeter circuit.

Signal voltage at amp output = 10 V
Signal frequency band at amp input = Full band
Impedance of the filter and the tweeter (combined) at 1000 Hz = 5000 Ohms

Current flow from amp to speaker binding post = V/Z (Ohm's law),or 10 V/5000 Ohms = 2 mA
You can do this for every single frequency of the signal the same way if you have the needed data of the filter.

 

[PENG]

PS: I think it terms of upstream and downstream, but with circuitry, I believe you have to fully understand what is "downstream" before you can know what is happening upstream!

What is the point here?

 

[Verdinut]

As an numerical example for "a partially used/unused" case, consider the calculation below:

Signal voltage at amp output = 10 V
Signal frequency band at amp input = Full band
Impedance of the filter at 1000 Hz = 5000 Ohms

Current flow from amp to speaker binding post = V/Z (Ohm's law),or 10 V/5000 Ohms = 2 mA
You can do this for every single frequency of the signal the same way if you have the needed data of the filter.

So, in that example, the power delivered to the connected speaker driver would be only 0.02 watt.

 

[PENG]

So, in that example, the power delivered to the connected speaker driver would be only 0.02 watt.

Not really, hard to calculate without the missing data but it would be less than that for sure.

 

[KEW]

Agreed.

Did you read my post#20, when I sort of anticipated a major misconception that non electrical people may think that electron flow in a wire with the filter and speaker driver connected in series would behave like water flows in a pipe with a filter at the consumer end.

Electric current flow does NOT work like that, you can put the filter physically anywhere along the line/wire, if the filter offers very high impedance to, say frequency below 500 Hz, the amplifier will send little signal current below 500 Hz, period, as soon as steady state condition is reached (very quickly for short lines in audio applications).

As an numerical example for "a partially used/unused" case, consider the calculation below:

Signal voltage at amp output = 10 V
Signal frequency band at amp input = Full band
Impedance of the filter at 1000 Hz = 5000 Ohms

Current flow from amp to speaker binding post = V/Z (Ohm's law),or 10 V/5000 Ohms = 2 mA
You can do this for every single frequency of the signal the same way if you have the needed data of the filter.

What is the point here?

Thanks PENG!

I suspect your post #20 would make everything clear to me if I properly understood the relevance of current vs voltage!
I think the relevant point (which I am inferring from your posts, but would like to get confirmed) is that if the current is very low, the amp has not seen a significant load.

Let me outline a scenario to make sure I'm on the same page:

Say you have a 100Watt AVR. You also have speakers that have an especially demanding load at 140Hz (inefficient speakers with an impedance drop to 2 Ohms and the Phase is difficult). If you crank it, this load exceeds the ability of the AVR and you get audible distortion, which I presume would show up across many frequencies.
As I understand it, if I were to bi-amp this speaker the amp driving the woofer would see this same load and we would have distortion from that amp. However, the amp driving the mid-range and tweeter would not see this distortion and we would have a "clean" signal for the critical mid-range frequencies.

My question is "is the above a fair assessment of how it would work?" IOW, although both amps are seeing the full range signal, only one is being subjected to the difficult load at 140Hz, correct?

Note that in this scenario, Bi-amping would realize an audible improvement, but you still have an overloaded amp with distortion. Thus, bi-amping "sucks less", but is definitely not a solution!

The appropriate solution would be to use an amp that is capable of dealing with the 140Hz, 2 Ohm load the speaker presents.

Perhaps the most succinct way to put my question is:
Does the filter in an XO
1) reduce the load on the amp (in a bi-amped configuration),or
2) does the amp still do the same amount of work and the filter absorbs the power (like a resistor would - converting the power into heat).

To explain my comment about upstream and downstream - as a mechanical engineer, I want to look at a circuit as having a sequential path - the way you would interpret a Rube Goldberg device! Each sequential action is defined independently of what is next, only dependent on what came before it!
So, my tendency is to look at an electrical circuit as a sequential system, when in actuality, there needs to be awareness of what is downstream in order to properly know what happens upstream.

By using the terms and concepts I think in (not being from an electrical perspective),I am hoping assist others with the same difficulties I have to understand/appreciate these differences between electrical and mechanical systems.

 

[PENG]

Here's the diagram I wanted to draw earlier:

The first diagram shows a stupid but simple high pass filter for the tweeter and an equally stupid low pass filter for the woofer. From the impedance formula you can see that the capacitor offers lower impedance as frequency increases, and the opposite is true for the inductor.

In the 1st circuit diagram, it is easy to see that the current IT1 = IT2 = IT3 = IT1r = IT2r because it is a simple series circuit. So you can clearly understand why there is no difference in terms of "before" or "after", or "Up/Down" stream of the filter, that in this case is just that same stupid capacitor. Practical HPF and LPF do have parallel branches but principles are the same, you can still show the same idea using Ohm's law, Kirchoff's law and may be the superposition theorem as well. For demo, it is much easier and simpler to use a simple series circuit.

Look at this another way, the same current that leaves the amplifier's red output terminal, flow to the red speaker binding post via the speaker feed wire, through the high pass capacitor, through to the tweeter voice coil, out the return black terminal of the tweeter, continue on the connected black tweeter return wire, and finally back to the black output terminal of the amplifier via the black return speaker feed wire.

You can also clearly see that because the capacitor offers higher impedance to the lower frequencies, this filtered circuit will carry mostly the higher frequency part of the otherwise full band signal.

The same logic obviously applies to the low pass filter/woofer section. So that is basically the simplest passive bi-wire demo circuit diagram I can think of, but again this is for demo only, no one should use such a circuit for even low fi audio. Now if you use a set of jumper wires to link the appropriate speaker binding posts together, you can see that only one amplifier is needed, but in that case the single pair speaker feed wire will carry a current equal to IT1+ITW.

This is not meant to prove whether passive bi-amp offers audible benefits or not. Just that I find it hard not to try and clear the air, when statements were made (unintentionally, I know..) that may lead people to believe that in properly done bi-amp schemes, the two pairs of speaker feed wires would carry the same full band signal current to the speaker binding posts, the fact is, they don't.

 

[PENG]

My question is "is the above a fair assessment of how it would work?" IOW, although both amps are seeing the full range signal, only one is being subjected to the difficult load at 140Hz, correct?

Yes, a voltage source connected to a load of very high impedance, think 10,000 Ohm just to make a point, the load would be negligibly low, even for very low power preamps. That's why you want the input impedance of your power amp to be high, like 15,000 to 25,000 ohms, to make it easy for even a weak preamps to drive.

Note that in this scenario, Bi-amping would realize an audible improvement, but you still have an overloaded amp with distortion. Thus, bi-amping "sucks less", but is definitely not a solution!

"Sucks less" sounds better to me than totally sucks.

The appropriate solution would be to use an amp that is capable of dealing with the 140Hz, 2 Ohm load the speaker presents.

Of course.

Perhaps the most succinct way to put my question is:
Does the filter in an XO
1) reduce the load on the amp (in a bi-amped configuration),or
2) does the amp still do the same amount of work and the filter absorbs the power (like a resistor would - converting the power into heat).

1) Of course, because you now have one amp driving the woofer(s),another driving the mid(s)/tweeter.
2) No, as explained in 1).

Basically in bi-amp, you end up with 2 amps doing the work that would otherwise be done by one amp, so either amp will work as hard.

Edit: One potentially important point to keep in mind is the effects of clipping. In terms of voltage, it has been mentioned multiple times before, that in passive bi-amp, both amplifiers do have to amplifier a full range signal because they share the same input. So if such a "difficult load" cause one amp to clip, it will cause the other to clip as well. In that sense, bi-amp or not, the tweeter will still end up getting the short end of the stick. To me, a 200 W/300 W 8/4 ohm amp is definitely better than two 100/150 W 8/4 ohm amp, all else being equal.

To explain my comment about upstream and downstream - as a mechanical engineer, I want to look at a circuit as having a sequential path - the way you would interpret a Rube Goldberg device! Each sequential action is defined independently of what is next, only dependent on what came before it!
So, my tendency is to look at an electrical circuit as a sequential system, when in actuality, there needs to be awareness of what is downstream in order to properly know what happens upstream.

Okay, so that is exactly the part I anticipated and attempted to address in my post#20. I know it usually help a lot in trying to explain electricity using mechanical and fluid equivalents/analogies. It does not always work though as there are not always good equivalencies. Still, I agree it is good to try that route as much as possible because mechanical things, while can also be as complicated or more so than electrical things, are in most cases easier for people to relate to.

 

[lovinthehd]

I just wanted to congratulate @PENG for his patience and expansion on how it all works....good stuff!