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speaker cables and jumpers vs bi-wiring - Page 3

post #61 of 256
2/10/10 at 6:28am
Quote:
Originally Posted by ccotenj View Post

you'd want your active xover "upstream"...

i.e. source ---> xover ----> amp ----> speaker

Sorry, yes I meant upstream. The speaker is about as downstream as one can get... Just verbal flub on my part. Actually the listener is downstream as one can get.
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post #62 of 256
2/10/10 at 6:30am
^^^

yea, i knew you knew what you meant... just clarifying for anyone else who might be following along...
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post #63 of 256
2/10/10 at 6:44am
So where do I go if my snake needs oiling?

Actually, interesting that Polk was brought up. At least at some point in the past, He designed his speakers knowing full well they would be hooked up to low end receivers with zip cord. That was taken into account in the driver design, box size, and tuning. i.e. real DF was "no". When hooked up to very high end amps, they were over-damped. Good systems design approach. I would guess marketing forces changed their design goals. I don't like Polk's speakers, but I do respect his engineering ability.
I have not read Totem's WEB as I have had more "Madison Ave. engineering" than I can stand. I have heard their $4000/pair monitors. They definitely know what they are doing.
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post #64 of 256
2/10/10 at 7:01am
Here is the train of thought from some people:

"I provided the schematic for the LSi15 which clearly shows the LSi15 consists of two separate speaker circuits. The tweeter, and the upper/lower driver, woofer combination. When the jumper is installed between the + and - posts these two circuits become one. When the jumper is removed they are two distinct circuits with no path for current flow between them. As mentioned earlier, with the jumpers removed, this is the same as if there are two separate, distinct, and unrelated speakers attached to the amp."

The schematic in question

There isn't a x-over that isn't going to show multiple circuits. That is the nature of the beast. It's like pointing out water is wet. I am not sure how it supports the bi-wire argument he is making.

The sad part is: You can't tell them otherwise. They simply won't stand for it. He could have posted an x-over schematic from a $25 2 way Radio Shack speaker for all intents and purposes.
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post #65 of 256
2/10/10 at 7:20am
Quote:
Originally Posted by Jinjuku View Post

I am not sure if you are referring to electrically isolating the different passive cross over circuits. It can not be done with a bi-wire setup.

Just to be clear: Bi-wire is two sets of wires being ran to the same binding post of amplification. No isolation/separation has taken place. You have only moved the jumper from the binding post on the speakers to the binding post on the amp.

at the speaker, hence the need for jumpers

if you only connect the amp wires to the upper posts, you only get the highs, same goes for the lows...they each feed their respective crossover network...
without the jumper (or being biwired) they are isolated from each other
essentially 2 filters, one hi, one low, the filter network is divided into 2, isolated by the jumper
when biwired the wire acts as the 'jumper'

from my limited understanding of a crossover they reject signals of a certain freq, they will not allow the current to flow...
so if each isolated network rejects the appropriate current, one will allow the high freq to flow, the other the low...

so each wire will only have high or low freq current (below a certain attenuation level)
there are no perfect or 'ideal' filters, lol
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post #66 of 256
2/10/10 at 7:29am
Quote:
Originally Posted by ArthurPE View Post

so each wire will only have high or low freq current (below a certain attenuation level)

each driver will only see the frequencies the xover in the speaker passes....

however, each wire will still be carrying the exact same signal...
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post #67 of 256
2/10/10 at 7:33am
a pic is worth 1000 words, especially in this kind of discussion:



if the upper and lower are isolated:
consider the diagnol lines at the left as the jumpers, 'remove' them, you have 2 sets of connection points, like a bi-wired speaker without the jumpers...
now you have 2 filters, correct?
you must wire to each, in this case paralleled at the amp
good so far?

the upper has caps, they present an impedance to low freq, so low freq current can't flow, only high
the lower has inductors, they present a high freq impedance, so high freq current can't flow, only low
so each pair of wires will reject either the hi or low current, depending on it's impedance
in effect, lows will go thru the induct, and highs thru the caps...path of least resistance for a given freq

caps Z ~ inverse of freq, as freq goes down, imped goes up, low freqs are attenuated, highs flow
inductance Z (coils) ~ to freq, as freq goes up, imped goes up, high freqs are attenuated, lows flow
at DC nothing will go thru the caps, and the inductor will appear as a short
at very high freq, the cap will be a short, and the inductor open
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post #68 of 256
2/10/10 at 7:53am
Quote:
Originally Posted by ArthurPE View Post

a pic is worth 1000 words, especially in this kind of discussion:



if the upper and lower are isolated:
consider the diagnol lines at the left as the jumpers, 'remove' them, you have 2 sets of connection points, like a bi-wired speaker without the jumpers...
now you have 2 filters, correct?
you must wire to each, in this case paralleled at the amp
good so far?

the upper has caps, they present an impedance to low freq, so low freq current can't flow, only high
the lower has inductors, they present a high freq impedance, so high freq current can't flow, only low
so each pair of wires will reject either the hi or low current, depending on it's impedance
in effect, lows will go thru the induct, and highs thru the caps...path of least resistance for a given freq

caps Z ~ inverse of freq, as freq goes down, imped goes up, low freqs are attenuated, highs flow
inductance Z (coils) ~ to freq, as freq goes up, imped goes up, high freqs are attenuated, lows flow
at DC nothing will go thru the caps, and the inductor will appear as a short
at very high freq, the cap will be a short, and the inductor open

And your schematic makes no point for your argument. The circuit whether jumpered at the speaker or jumpered at the binding post on the amp is electrically the SAME EXACT CIRCUIT.

I have an actual x-over sitting here I will take some pics and demonstrate.
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post #69 of 256
2/10/10 at 8:18am
Quote:
Originally Posted by Jinjuku View Post

And your schematic makes no point for your argument. The circuit whether jumpered at the speaker or jumpered at the binding post on the amp is electrically the SAME EXACT CIRCUIT.

I have an actual x-over sitting here I will take some pics and demonstrate.

it makes the WHOLE point if you understand the basic circuit theory

are you an electrical engineer?
I'm not trying to be a smart aleck, but it may help me understand how I have to break this down...

the point is, the filters present different impedances to different freqs
so low freq current CAN'T go down the high pass wires, and vice versa...

imagine if you had a 'switch' in the upper network that opened when bass notes played, and closed/shorts when treble notes played...the cap does that proportional to freq...so the wire does not carry low freq current signals, becasue the circuit is effectively 'open'...it is not on on/off thing but has a cut-off freq...only highs pass, only hi freq current flows...

the lower (low pass) does the same for the high freqs
the inductor 'opens' for high freq (and closes/shorts for low) and won't allow the high to pass, so the wire does NOT carry high freq current...

now, I don't care what you 'believe', nor does the physics of the network...
in the wires to the high pass filter, low freq current will be attenuated
in the wires to the low pass filter, high freq current will be attenuated
fact, not conjecture
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post #70 of 256
2/10/10 at 8:31am
Quote:
Originally Posted by ArthurPE View Post

it makes the WHOLE point if you understand the basic circuit theory

are you an electrical engineer?
I'm not trying to be a smart aleck, but it may help me understand how I have to break this down...

the point is, the filters present different impedances to different freqs
so low freq current CAN'T go down the high pass wires, and vice versa...

imagine if you had a 'switch' in the upper network that opened when bass notes played, and closed/shorts when treble notes played...the cap does that proportional to freq...so the wire does not carry low freq current signals, becasue the circuit is effectively 'open'...it is not on on/off thing but has a cut-off freq...only highs pass, only hi freq current flows...

the lower (low pass) does the same for the high freqs
the inductor 'opens' for high freq (and closes/shorts for low) and won't allow the high to pass, so the wire does NOT carry high freq current...

now, I don't care what you 'believe', nor does the physics of the network...
in the wires to the high pass filter, low freq current will be attenuated
in the wires to the low pass filter, high freq current will be attenuated
fact, not conjecture

This has nothing to do with bi-wiring to the same binding post on the Amp. For gods sake... You have not electrically isolated the circuit!

The amp is seeing the same passive circuit (tweeter, woofers, inductor, resistor, caps) whether you run two wires from the same binding post or jumper at the speaker.

Take the Canare 4S11 cable. Take the negative lines and terminate both with a banana plug. Take both positive lines and terminate with a banana plug.

Now plug one end into your amp and one end into your speaker with a jumper plate connecting the x-overs at the speaker.

Now at the speaker end remove the jumper plate. Take one of negative lines and put it's own banana jack on it. Take one of the positive lines and put a banana jack on it. At this point you don't have two wires going into each banana, we have one wire going into each banana. We are at a total of 6 banana connections (4 at the speaker and 2 at the amp). plug the two extra jacks into the binding post that was previously connected with jumper plate.

Your circuit is electrically 100% the same. Now at the amp add two more banana jacks. Instead of two negative into a single banana jack, instead of two positive into a single banana jack, you have one of each on it's own banana jack. So now you have a total of 8 banana jacks.

Your circuit is still electrically the same. You haven't isolated squat. Why can't you see this?
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post #71 of 256
2/10/10 at 8:35am
Quote:
Originally Posted by ArthurPE View Post


the point is, the filters present different impedances to different freqs
so low freq current CAN'T go down the high pass wires, and vice versa...

OMG... this is all happening on the x-over. It doesn't have anything to do with the bi-wiring. The bi-wire isn't high or low pass. It's simply power from amp to the speaker.

A good x-over design will make sure there is no back EMF from the drivers when they de-engergize (that is back to resting posistion). Again the bi-wire doesn't do anything to prevent that.
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post #72 of 256
2/10/10 at 8:52am
Quote:
Originally Posted by Jinjuku View Post

OMG... this is all happening on the x-over. It doesn't have anything to do with the bi-wiring. The bi-wire isn't high or low pass. It's simply power from the speaker.

A good x-over design will make sure there is no back EMF from the drivers when they de-engergize (that is back to resting posistion). Again the bi-wire doesn't do anything to prevent that.

God has little to with this:
you can thank Léon Charles Thévenin and Gustav Robert Kirchhoff

yes it is, sort of....of a given frequency (actually it's voltage V and current i, the product, V x i, is power)
in the high pass wires, low freq current can't flow
in the low pass wires, high freq can't flow
they are excluded/reject/attenuated...

OK let's look at it like this:
say the amp is making 10W/ch (a given current and voltage, same imped, ie same speaker) the 10W power is comprised of hi and low freq signals, ie, music...
with 1 pair it carries 10W, 1 pair carries it all
with 2 pair it STILL carries 10W, it is somehow magically divided/proportioned between the 2
OK?

what divides it? perhaps the 'impedance' of the filters?
let's say that one pair the low pass carries 7, since bass needs more power and the high carries 10 - 7 ~ 3W...still with me?

now, how would the power be 'proportioned' in such a fashion?
how would the hi pass 'reject' 7W of power?
and the low pass reject 3W of the total 10W signal?
hmmmm
maybe it attenuates the lows (and the low pass accepts them), and vice versa...

now, according to your 'theory', both filters see and absorb the same power...
so you would need 10 + 10 ~ 20W if both filters see ALL of the freqs (current ~ power) in the signal...it doesn't, it only sees what the filters allow to flow through them, not UP to them...then the electrons just congregate there, lol

I'm done, I'll let the readers decide on the merits, and logic presented in the posts...an electrical engineer will understand, and I think anyone with a basic electrical understanding will...

time to listen to my new Miles Davis box set...have a nice day
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post #73 of 256
2/10/10 at 9:07am
Quote:
Originally Posted by ArthurPE View Post

God has little to with this:
you can thank Léon Charles Thévenin and Gustav Robert Kirchhoff

yes it is, sort of....of a given frequency (actually it's voltage V and current i, the product, V x i, is power)
in the high pass wires, low freq current can't flow
in the low pass wires, high freq can't flow
they are excluded/reject/attenuated...

OK let's look at it like this:
say the amp is making 10W/ch (a given current and voltage, same imped, ie same speaker) the 10W power is comprised of hi and low freq signals, ie, music...
with 1 pair it carries 10W, 1 pair carries it all
with 2 pair it STILL carries 10W, it is somehow magically divided/proportioned between the 2
OK?

what divides it? perhaps the 'impedance' of the filters?
let's say that one pair the low pass carries 7, since bass needs more power and the high carries 10 - 7 ~ 3W...still with me?

now, how would the power be 'proportioned' in such a fashion?
how would the hi pass 'reject' 7W of power?
and the low pass reject 3W of the total 10W signal?
hmmmm
maybe it attenuates the lows (and the low pass accepts them), and vice versa...

now, according to your 'theory', both filters see and absorb the same power...
so you would need 10 + 10 ~ 20W if both filters see ALL of the freqs (current ~ power) in the signal...it doesn't, it only sees what the filters allow to flow through them, not UP to them...then the electrons just congregate there, lol

I'm done, I'll let the readers decide on the merits, and logic presented in the posts...an electrical engineer will understand, and I think anyone with a basic electrical understanding will...

time to listen to my new Miles Davis box set...have a nice day


LOL. You have no clue what you are talking about... The only thing that comes into play is the coefficient of the wire since the circuit is still electrically the same.

Ok:

Amp wattage 10
Length of run 8'
Speaker wire gauge 12 AWG
Speaker Load 4 ohm
Wire loss constant 1.77 (ohms per 1000')

Actual power loss 0.0703 watts.

Ohms law at its best. The only thing you you can improve on is the power loss. Think you can hear the 0.0703 watts? You think you can hear the improvement that doubling the wire will make?
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post #74 of 256
2/10/10 at 9:22am
Quote:
Originally Posted by Jinjuku View Post

LOL. You have no clue what you are talking about... The only thing that comes into play is the coefficient of the wire since the circuit is still electrically the same.

Ok:

Amp wattage 10
Length of run 8'
Speaker wire gauge 12 AWG
Speaker Load 4 ohm
Wire loss constant 1.77 (ohms per 1000')

Actual power loss 0.0703 watts.

Ohms law at its best. The only thing you you can improve on is the power loss. Think you can hear the 0.0703 watts? You think you can hear the improvement that doubling the wire will make?

we are not talking power loss, but power allocation per freq band
it's 0.0708 W btw
but you are missing the whole point, but that's cool, becasue you are obviously not capable of grasping it...

how do you explain each wire delivering less than 10W to each network?
what 'decides' which current~power goes where?
or do both deliver 10W for a total of 20W
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post #75 of 256
2/10/10 at 9:25am
Quote:
Originally Posted by ArthurPE View Post


in the high pass wires, low freq current can't flow
in the low pass wires, high freq can't flow
they are excluded/reject/attenuated...

There is no such thing as 'high pass wires'.
There is no such thing as 'low pass wires'.
You have bi-wired, not bi-amped. You have effectively lowered the coefficient by increasing your speaker gauge. You are now delivering the same power, the same signal with less power loss.

You have done ZILCH to isolate the x-over circuits they are still all connected.


Quote:
Originally Posted by ArthurPE View Post

OK let's look at it like this:
say the amp is making 10W/ch (a given current and voltage, same imped, ie same speaker) the 10W power is comprised of hi and low freq signals, ie, music...
with 1 pair it carries 10W, 1 pair carries it all
with 2 pair it STILL carries 10W, it is somehow magically divided/proportioned between the 2
OK?

No it is not magically proportioned between the two. You have simply changed the coefficient of the speaker wire (increased the gauge) and are experiencing less power loss. That is all that you have done.

Quote:
Originally Posted by ArthurPE View Post

what divides it? perhaps the 'impedance' of the filters?
let's say that one pair the low pass carries 7, since bass needs more power and the high carries 10 - 7 ~ 3W...still with me?

The cross over FILTERS the signal power at the terminal plate on the speaker has only changed by the coefficient as effected by a doubling of the the speaker cable.


Quote:
Originally Posted by ArthurPE View Post

now, how would the power be 'proportioned' in such a fashion?
how would the hi pass 'reject' 7W of power?
and the low pass reject 3W of the total 10W signal?
hmmmm
maybe it attenuates the lows (and the low pass accepts them), and vice versa...

What you just said is jibberish. You aren't rejecting anything: You either resist(or) or you don't. Inductors also have resistance by nature of their design.


Quote:
Originally Posted by ArthurPE View Post

now, according to your 'theory', both filters see and absorb the same power...
so you would need 10 + 10 ~ 20W if both filters see ALL of the freqs (current ~ power) in the signal...it doesn't, it only sees what the filters allow to flow through them, not UP to them...then the electrons just congregate there, lol

It's not my 'theory'. It is Ohms law.

The cross over will see everything the same with with bi-wire plus the minuscule increase in power due to extra wire gauge.

What are these 'filters' that you keep talking about? The only 'filter' is the crossover. It's still electrically the same. The load on the amp is the same. The circuit is the same...



I'm done, I'll let the readers decide on the merits, and logic presented in the posts...an electrical engineer will understand, and I think anyone with a basic electrical understanding will...

time to listen to my new Miles Davis box set...have a nice day[/quote]
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post #76 of 256
2/10/10 at 9:28am
Quote:
Originally Posted by ArthurPE View Post

we are not talking power loss, but power allocation per freq band
it's 0.0708 W btw
but you are missing the whole point, but that's cool, becasue you are obviously not capable of grasping it...

how do you explain each wire delivering less than 10W to each network?
what 'decides' which current~power goes where?
or do both deliver 10W for a total of 20W

I am at a loss for what to say. You are utterly and completely mistaken. I will have to let me posts stand on their merit.

You have done nothing but speak in jibberish.
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post #77 of 256
2/10/10 at 9:29am
this may help

Zc = 1/jwC

Zi = jwL

w = 2 Pi f

j = sqrt(-1)

it's shown like this at times:

Zc = 1/wC e^(-j PI/2)
Zi = wL e^(j PI/2)
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post #78 of 256
2/10/10 at 9:31am
Quote:
Originally Posted by Jinjuku View Post

I am at a loss for what to say. You are utterly and completely mistaken. I will have to let me posts stand on their merit.

You have done nothing but speak in jibberish.

actually the language I used is called 'engineering' specifically 'circuit theory'
as opposed 'what you believe', even if it contradicts the lwas of nature

I see you did not answer?
what apportions the power between filters?
or do they each see 10W?
if so, they see the same current, if not, some current is rejected from each, meaning the WIRE carries less..
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post #79 of 256
2/10/10 at 9:47am
Quote:
Originally Posted by Jinjuku View Post

I am at a loss for what to say. You are utterly and completely mistaken. I will have to let me posts stand on their merit.

You have done nothing but speak in jibberish.

I think what Arther is saying is that the tweeter side could be wired with 24ga wire while the woofer side may require 16ga for the length. The tweeter side does consume less power. In my main HT system I use Biamped LCR speakers. The woofers are 70w amps and the tweeters are only 10 watts. The sub amps are 350w x2. The 70w and 10w amps are tube hence their rather low power by today's standards but 10w driving a tweeter is plenty!

P.S. I don't beleive in this bi-wiring crap either. Either do it right with a line level crossover and two power amps or forget it.
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post #80 of 256
2/10/10 at 9:49am
Quote:
Originally Posted by ArthurPE View Post

this may help

Zc = 1/jwC

Zi = jwL

w = 2 Pi f

j = sqrt(-1)

it's shown like this at times:

Zc = 1/wC e^(-j PI/2)
Zi = wL e^(j PI/2)

Thanks for the primer on Circuit Capacitance / Inductance etc... It still doesn't explain how the x-over is seeing power any differently from the amplifier.
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post #81 of 256
2/10/10 at 9:58am
Quote:
Originally Posted by Glimmie View Post

I think what Arther is saying is that the tweeter side could be wired with 24ga wire while the woofer side may require 16ga for the length. The tweeter side does consume less power. In my main HT system I use Biamped LCR speakers. The woofers are 70w amps and the tweeters are only 10 watts. The sub amps are 350w x2. The 70w and 10w amps are tube hence their rather low power by today's standards but 10w driving a tweeter is plenty!

P.S. I don't beleive in this bi-wiring crap either. Either do it right with a line level crossover and two power amps or forget it.

When both the 24 GA and 16 GA are wired to the same terminal/binding post on the amp you have bi-wired you have only done one thing: Increased the wire gauge.

You have NOT:

1. Changed the electrical circuit
2. Isolated a cross over
3. Changed the load presented to the amplifier (the load of the entire speaker will still be there)

If you have active cross over going to independent channels of amplification then you have:

1. Changed the electrical cirucuit
2. Isolated a cross over
3. Changed the load presented to the amplifier

And is doesn't take ANY function of Z to prove this.
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post #82 of 256
2/10/10 at 9:59am
Arthur,

If you can't explain it to all in laymens' terms then you can't explain it at all.
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post #83 of 256
2/10/10 at 10:05am
Quote:
Originally Posted by Jinjuku View Post

Arthur,

If you can't explain it to all in laymens' terms then you can't explain it at all.

Let me try. I have a very modest house. I own a table lamp and an electric cloths dryer. My electrical service has a 1/0 gauge wire. My dryer of course has a 10ga cord. Does that mean I need a 10ga cord on my table lamp too?
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post #84 of 256
2/10/10 at 10:15am
Quote:
Originally Posted by Glimmie View Post

Let me try. I have a very modest house. I own a table lamp and an electric cloths dryer. My electrical service has a 1/0 gauge wire. My dryer of course has a 10ga cord. Does that mean I need a 10ga cord on my table lamp too?

What does your example have to do with the price of tea in China? We are not talking about wire gauge here.

We are talking about the perceived benefits of:

1. Bi-Wire (NOT bi-amp)
2. What the circuit looks like when wired with one pair and jumped at the speaker or running two pair of wires and jumped at the same output terminal on the amp.

The effect is still the same: you have a summed load that is the same discounting the negligible gain in wattage due to increase wire gauge.
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post #85 of 256
2/10/10 at 10:23am
If you are arguing the bi-wire some how eliminates back EMF (again from a driver de-energizing) then we are at an impasse.
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post #86 of 256
2/10/10 at 10:55am
Quote:
Originally Posted by Jinjuku View Post

Thanks for the primer on Circuit Capacitance / Inductance etc... It still doesn't explain how the x-over is seeing power any differently from the amplifier.

sure it does, it explains it all...

as the freq into a capacitor increases (gets larger) its impedance decreases
it is inversely proportional, ie, 1/f, and allows the highs through...
as the freq gets smaller/lower, the impedance gets larger, blocking the lows...
if the lows can't get thru, neither can their associated current/power, it is 'stuck' at the amp...it can't flow in the wire...

vice versa for inductance...

a cap is used to make DC, it filters out the AC, the AC can't pass (actually the size of the cap among other things determines how low the AC is blocked)
but if freq = 0, then Z = infinity, and NO AC (doesn't have to be 60hz, can be any freq associated with music) will pass...

a transformer (coil or inductor) can't pass DC, only AC...
hence the prevalent use of AC for power systems, you can step up/down the voltages (and iverse for the current) which you can't easily do with DC...
if f approached infinity, so does Z the impedance, so it blocks those high freqs, ie, the current associated with them...the current can't just flow down

the wire and stop at the cap or inductor, Kirchhoff, if current flows into a closed loop, it must flow out...

This law is also called Kirchhoff's point rule, Kirchhoff's junction rule (or nodal rule), and Kirchhoff's first rule. The principle of conservation of electric charge implies that:

At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node.
Adopting the convention that every current flowing towards the node is positive and that every current flowing away is negative (or the other way around), this principle can be stated as:

sum (all current in a node/network/junction) = 0
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post #87 of 256
2/10/10 at 11:10am
Quote:
Originally Posted by ArthurPE View Post

sure it does, it explains it all...

as the freq into a capacitor increases (gets larger) its impedance decreases
it is inversely proportional, ie, 1/f, and allows the highs through...
as the freq gets smaller/lower, the impedance gets larger, blocking the lows...
if the lows can't get thru, neither can their associated current/power, it is 'stuck' at the amp...it can't flow in the wire...

vice versa for inductance...

a cap is used to make DC, it filters out the AC, the AC can't pass (actually the size of the cap among other things determines how low the AC is blocked)
but if freq = 0, then Z = infinity, and NO AC (doesn't have to be 60hz, can be any freq associated with music) will pass...

a transformer (coil or inductor) can't pass DC, only AC...
hence the prevalent use of AC for power systems, you can step up/down the voltages (and iverse for the current) which you can't easily do with DC...
if f approached infinity, so does Z the impedance, so it blocks those high freqs, ie, the current associated with them...the current can't just flow down

the wire and stop at the cap or inductor, Kirchhoff, if current flows into a closed loop, it must flow out...

This law is also called Kirchhoff's point rule, Kirchhoff's junction rule (or nodal rule), and Kirchhoff's first rule. The principle of conservation of electric charge implies that:

At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node.
Adopting the convention that every current flowing towards the node is positive and that every current flowing away is negative (or the other way around), this principle can be stated as:

sum (all current in a node/network/junction) = 0

Arthur,

It still doesn't explain:

1 Load that the AMP sees is the same
2 That the tweeter circuit and woofer circuit are electrically isolated. They
are not.

Nothing is 'Stuck at the Amp' why can't you understand that? All you are describing is a circuit (the x-over) doing it's job. The function is the same whether two circuits are summed at the amp (bi-wired) or summed at the speaker (jumper). Your difference is an additional 8 feet of cable and and increase in the wire gauge.
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post #88 of 256
2/10/10 at 11:28am
Quote:
Originally Posted by Jinjuku View Post

Arthur,

It still doesn't explain:

1 Load that the AMP sees is the same
2 That the tweeter circuit and woofer circuit are electrically isolated. They
are not.

Nothing is 'Stuck at the Amp' why can't you understand that? All you are describing is a circuit (the x-over) doing it's job. The function is the same whether two circuits are summed at the amp (bi-wired) or summed at the speaker (jumper). Your difference is an additional 8 feet of cable and and increase in the wire gauge.

Basically you are correct. What ever miniscule electrical changes that take place from bi-wiring are grossly insignificant in the application of a HiFi speaker.

Of course audiophiles like to believe anything can be heard withtheir trained ears and thus the controversy.
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post #89 of 256
2/10/10 at 11:28am
Quote:
Originally Posted by ccotenj View Post

each driver will only see the frequencies the xover in the speaker passes....

however, each wire will still be carrying the exact same signal...

really?

so the full freq current goes down the wire up to the filter, and stops?
blocked by the cap...where does the excess or rejected current go?
it just sits in the wire? the wire has enough capacitance to 'store' it?
that violates the conservation of charge theory...
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post #90 of 256
2/10/10 at 11:34am
Quote:
Originally Posted by Jinjuku View Post

Arthur,

It still doesn't explain:

1 Load that the AMP sees is the same
2 That the tweeter circuit and woofer circuit are electrically isolated. They
are not.

Nothing is 'Stuck at the Amp' why can't you understand that? All you are describing is a circuit (the x-over) doing it's job. The function is the same whether two circuits are summed at the amp (bi-wired) or summed at the speaker (jumper). Your difference is an additional 8 feet of cable and and increase in the wire gauge.

I think what you mean to say is you don't understand it:

the load is not the same
instead of 2 filters in parallel, it sees 2 seperate filters

are you saying if an amp puts out 1A into a single wire set-up it puts out 2A in 2 wires? yes or no...

if yes, how can this be? wouldn't the power double because the impedance is the same (as you said, same load) and P = I^2 Z

if no, say 1/2A to each, for a total of 1A, ie, same power/load
does the 1/2A include all freqs for each wire?
if this is true, and it sees 1/2 the power for all freqs, wouldn't the volume output be less after filtering the lows out for the high pass?
so bi-wired, according to you, needs more power for the same volume...
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