| Home |
| Search |
| Today's Posts |
|
#161
|
|||
|
|||
Bi-wiring - Hogwash?
On Tue, 02 Mar 2004 22:13:39 GMT, "afh3" wrote:
"flint" wrote in message ... Here is my attempt at the "good logical argument." It's called the application of Kirchoff's laws. Since the bi-wired speakers ultimately connect to the same set of output terminals on the voltage source (amplifier/receiver/whatever) all of the points on both pairs of conductors will be at the same electrical potential at any one point in time. Excluding the negligible resistance of the conductors Actually, you can't. That small change is exactly what is being counted on to create the isolation improvement between the halves of the crossover. The idea is to work that increased resistance or more properly, impedance, against the very, very low output impedance of the majority of today's solid-state amplifiers as a Tee network. The possibility of the difference being audible is pretty slim. Regards, Bruce Hitting reply is futile, use the following: ). |
|
#162
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Bruce Burke" wrote in message
... On Tue, 02 Mar 2004 22:13:39 GMT, "afh3" wrote: "flint" wrote in message ... Here is my attempt at the "good logical argument." It's called the application of Kirchoff's laws. Since the bi-wired speakers ultimately connect to the same set of output terminals on the voltage source (amplifier/receiver/whatever) all of the points on both pairs of conductors will be at the same electrical potential at any one point in time. Excluding the negligible resistance of the conductors Actually, you can't. That small change is exactly what is being counted on to create the isolation improvement between the halves of the crossover. The idea is to work that increased resistance or more properly, impedance, against the very, very low output impedance of the majority of today's solid-state amplifiers as a Tee network. Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? -afh3 |
|
#163
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Bruce Burke" wrote in message
... On Tue, 02 Mar 2004 22:13:39 GMT, "afh3" wrote: "flint" wrote in message ... Here is my attempt at the "good logical argument." It's called the application of Kirchoff's laws. Since the bi-wired speakers ultimately connect to the same set of output terminals on the voltage source (amplifier/receiver/whatever) all of the points on both pairs of conductors will be at the same electrical potential at any one point in time. Excluding the negligible resistance of the conductors Actually, you can't. That small change is exactly what is being counted on to create the isolation improvement between the halves of the crossover. The idea is to work that increased resistance or more properly, impedance, against the very, very low output impedance of the majority of today's solid-state amplifiers as a Tee network. Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? -afh3 |
|
#164
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Bruce Burke" wrote in message
... On Tue, 02 Mar 2004 22:13:39 GMT, "afh3" wrote: "flint" wrote in message ... Here is my attempt at the "good logical argument." It's called the application of Kirchoff's laws. Since the bi-wired speakers ultimately connect to the same set of output terminals on the voltage source (amplifier/receiver/whatever) all of the points on both pairs of conductors will be at the same electrical potential at any one point in time. Excluding the negligible resistance of the conductors Actually, you can't. That small change is exactly what is being counted on to create the isolation improvement between the halves of the crossover. The idea is to work that increased resistance or more properly, impedance, against the very, very low output impedance of the majority of today's solid-state amplifiers as a Tee network. Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? -afh3 |
|
#165
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Bruce Burke" wrote in message
... On Tue, 02 Mar 2004 22:13:39 GMT, "afh3" wrote: "flint" wrote in message ... Here is my attempt at the "good logical argument." It's called the application of Kirchoff's laws. Since the bi-wired speakers ultimately connect to the same set of output terminals on the voltage source (amplifier/receiver/whatever) all of the points on both pairs of conductors will be at the same electrical potential at any one point in time. Excluding the negligible resistance of the conductors Actually, you can't. That small change is exactly what is being counted on to create the isolation improvement between the halves of the crossover. The idea is to work that increased resistance or more properly, impedance, against the very, very low output impedance of the majority of today's solid-state amplifiers as a Tee network. Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? -afh3 |
|
#166
|
|||
|
|||
|
Bi-wiring - Hogwash?
"afh3" wrote in message
news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. One more heartbreak of biwiring is that re-establishing the connection between the woofer and tweeter terminals on the back panel of the speaker will always provide flatter frequency response at the speaker's terminals. IOW not only do you buy twice the wire, you throw away just about all of the second wire's potential benefits compared to a more simplistic use of the wire - IOW hooking it in parallel at both ends. In most cases these benefits have vanishing importance, but with biwiring, they are sacrificed, and for what? |
|
#167
|
|||
|
|||
|
Bi-wiring - Hogwash?
"afh3" wrote in message
news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. One more heartbreak of biwiring is that re-establishing the connection between the woofer and tweeter terminals on the back panel of the speaker will always provide flatter frequency response at the speaker's terminals. IOW not only do you buy twice the wire, you throw away just about all of the second wire's potential benefits compared to a more simplistic use of the wire - IOW hooking it in parallel at both ends. In most cases these benefits have vanishing importance, but with biwiring, they are sacrificed, and for what? |
|
#168
|
|||
|
|||
|
Bi-wiring - Hogwash?
"afh3" wrote in message
news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. One more heartbreak of biwiring is that re-establishing the connection between the woofer and tweeter terminals on the back panel of the speaker will always provide flatter frequency response at the speaker's terminals. IOW not only do you buy twice the wire, you throw away just about all of the second wire's potential benefits compared to a more simplistic use of the wire - IOW hooking it in parallel at both ends. In most cases these benefits have vanishing importance, but with biwiring, they are sacrificed, and for what? |
|
#169
|
|||
|
|||
|
Bi-wiring - Hogwash?
"afh3" wrote in message
news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. One more heartbreak of biwiring is that re-establishing the connection between the woofer and tweeter terminals on the back panel of the speaker will always provide flatter frequency response at the speaker's terminals. IOW not only do you buy twice the wire, you throw away just about all of the second wire's potential benefits compared to a more simplistic use of the wire - IOW hooking it in parallel at both ends. In most cases these benefits have vanishing importance, but with biwiring, they are sacrificed, and for what? |
|
#170
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? Hey kids, try this experiment at home. Disconnect one pair of your biwiring connections from one of your speaker inputs and hook it up to an RLC bridge. Notice how you "see" the reactance of the other biwired pair's drivers and crossover? When you reconnect the pair, so does the other driver and crossover. Isolation my ass... I know I missed a couple of days of circuits class at the "U", but this is rediculous. -afh3 |
|
#171
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? Hey kids, try this experiment at home. Disconnect one pair of your biwiring connections from one of your speaker inputs and hook it up to an RLC bridge. Notice how you "see" the reactance of the other biwired pair's drivers and crossover? When you reconnect the pair, so does the other driver and crossover. Isolation my ass... I know I missed a couple of days of circuits class at the "U", but this is rediculous. -afh3 |
|
#172
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? Hey kids, try this experiment at home. Disconnect one pair of your biwiring connections from one of your speaker inputs and hook it up to an RLC bridge. Notice how you "see" the reactance of the other biwired pair's drivers and crossover? When you reconnect the pair, so does the other driver and crossover. Isolation my ass... I know I missed a couple of days of circuits class at the "U", but this is rediculous. -afh3 |
|
#173
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? Hey kids, try this experiment at home. Disconnect one pair of your biwiring connections from one of your speaker inputs and hook it up to an RLC bridge. Notice how you "see" the reactance of the other biwired pair's drivers and crossover? When you reconnect the pair, so does the other driver and crossover. Isolation my ass... I know I missed a couple of days of circuits class at the "U", but this is rediculous. -afh3 |
|
#174
|
|||
|
|||
|
Bi-wiring - Hogwash?
"afh3" wrote in message
news:5mo1c.450966$I06.5065611@attbi_s01 "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. However, the voltage drop will depend on whatever passive or reactive component is connected to each end. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) I wrote "larger wire" when I meant "larger resistance" I think we agree with this correction. Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? Hey kids, try this experiment at home. Disconnect one pair of your biwiring connections from one of your speaker inputs and hook it up to an RLC bridge. Notice how you "see" the reactance of the other biwired pair's drivers and crossover? When you reconnect the pair, so does the other driver and crossover. Isolation my ass... Like I said, "to a microscopic degree". I know I missed a couple of days of circuits class at the "U", but this is ridiculous. No, its the result of a typo. |
|
#175
|
|||
|
|||
|
Bi-wiring - Hogwash?
"afh3" wrote in message
news:5mo1c.450966$I06.5065611@attbi_s01 "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. However, the voltage drop will depend on whatever passive or reactive component is connected to each end. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) I wrote "larger wire" when I meant "larger resistance" I think we agree with this correction. Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? Hey kids, try this experiment at home. Disconnect one pair of your biwiring connections from one of your speaker inputs and hook it up to an RLC bridge. Notice how you "see" the reactance of the other biwired pair's drivers and crossover? When you reconnect the pair, so does the other driver and crossover. Isolation my ass... Like I said, "to a microscopic degree". I know I missed a couple of days of circuits class at the "U", but this is ridiculous. No, its the result of a typo. |
|
#176
|
|||
|
|||
|
Bi-wiring - Hogwash?
"afh3" wrote in message
news:5mo1c.450966$I06.5065611@attbi_s01 "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. However, the voltage drop will depend on whatever passive or reactive component is connected to each end. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) I wrote "larger wire" when I meant "larger resistance" I think we agree with this correction. Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? Hey kids, try this experiment at home. Disconnect one pair of your biwiring connections from one of your speaker inputs and hook it up to an RLC bridge. Notice how you "see" the reactance of the other biwired pair's drivers and crossover? When you reconnect the pair, so does the other driver and crossover. Isolation my ass... Like I said, "to a microscopic degree". I know I missed a couple of days of circuits class at the "U", but this is ridiculous. No, its the result of a typo. |
|
#177
|
|||
|
|||
|
Bi-wiring - Hogwash?
"afh3" wrote in message
news:5mo1c.450966$I06.5065611@attbi_s01 "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. However, the voltage drop will depend on whatever passive or reactive component is connected to each end. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) I wrote "larger wire" when I meant "larger resistance" I think we agree with this correction. Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? Hey kids, try this experiment at home. Disconnect one pair of your biwiring connections from one of your speaker inputs and hook it up to an RLC bridge. Notice how you "see" the reactance of the other biwired pair's drivers and crossover? When you reconnect the pair, so does the other driver and crossover. Isolation my ass... Like I said, "to a microscopic degree". I know I missed a couple of days of circuits class at the "U", but this is ridiculous. No, its the result of a typo. |
|
#178
|
|||
|
|||
|
Bi-wiring - Hogwash?
afh3 wrote:
"Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. Actually if you read Arny Krueger's response carefully, you'll find that he was correct. The voltage drop is dependent on *both* the wire impedance *and* the load impedance. Just consider the case where the other end of the wire is open. No matter how big the wire impedance is, there is no drop in voltage across the wire. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? This is a case of not ageeing what isolation means. In a single wire case, the smaller the impedance of the wire, the more isolation you have between the drivers, if you define isolation as independence between the two drivers. What this means simply is that if you have a large wire impedance, the voltage at the speaker terminal is less dependent on the current drawn by the drivers. |
|
#179
|
|||
|
|||
|
Bi-wiring - Hogwash?
afh3 wrote:
"Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. Actually if you read Arny Krueger's response carefully, you'll find that he was correct. The voltage drop is dependent on *both* the wire impedance *and* the load impedance. Just consider the case where the other end of the wire is open. No matter how big the wire impedance is, there is no drop in voltage across the wire. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? This is a case of not ageeing what isolation means. In a single wire case, the smaller the impedance of the wire, the more isolation you have between the drivers, if you define isolation as independence between the two drivers. What this means simply is that if you have a large wire impedance, the voltage at the speaker terminal is less dependent on the current drawn by the drivers. |
|
#180
|
|||
|
|||
|
Bi-wiring - Hogwash?
afh3 wrote:
"Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. Actually if you read Arny Krueger's response carefully, you'll find that he was correct. The voltage drop is dependent on *both* the wire impedance *and* the load impedance. Just consider the case where the other end of the wire is open. No matter how big the wire impedance is, there is no drop in voltage across the wire. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? This is a case of not ageeing what isolation means. In a single wire case, the smaller the impedance of the wire, the more isolation you have between the drivers, if you define isolation as independence between the two drivers. What this means simply is that if you have a large wire impedance, the voltage at the speaker terminal is less dependent on the current drawn by the drivers. |
|
#181
|
|||
|
|||
|
Bi-wiring - Hogwash?
afh3 wrote:
"Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. Actually if you read Arny Krueger's response carefully, you'll find that he was correct. The voltage drop is dependent on *both* the wire impedance *and* the load impedance. Just consider the case where the other end of the wire is open. No matter how big the wire impedance is, there is no drop in voltage across the wire. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? This is a case of not ageeing what isolation means. In a single wire case, the smaller the impedance of the wire, the more isolation you have between the drivers, if you define isolation as independence between the two drivers. What this means simply is that if you have a large wire impedance, the voltage at the speaker terminal is less dependent on the current drawn by the drivers. |
|
#182
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Arny Krueger" wrote in message ... "afh3" wrote in message news:5mo1c.450966$I06.5065611@attbi_s01 "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. However, the voltage drop will depend on whatever passive or reactive component is connected to each end. No it won't. Put whatever you want at either end of the conductor pair, and Kirchoff's Laws GUARANTEE that the voltage at both ends will be identical -- minus the difference of the voltage drop in each conductor pair. This is patently not isolation. My ASCII art probably sucks but I'll try anyway. ---------(+)-------- Z1 Z2 ---------(-)-------- Z1 and Z2 are the loads presented by the two separate speaker driver circuits in a biwired arrangement. No values for Z1 or Z2 will ever result in the voltages across Z1 or Z2 varying by more than the difference of the voltage drop of the conductors to the (in this case) left of the power-supply (in the middle) and those on the right. Furthermore, it is obvious from the diagram that the voltage source is still loaded by the parallel values of Z1 and Z2 (along with the conductor resistance that you have in either case, biwired or not), and, most importantly, Z1 still "sees" Z2 and vice-versa. Envision Z1 as a dead short. The voltage measured at Z2 will be equal to the voltage drop of the conductors on the Z1 side of the supply, minus the voltage drop of the conductors on the Z2 side of the supply. If it does not, then you have just invented a perpetual motion machine, or free power -- either of which will make you a fortune. Now, if you were to use significantly different conductors for the left vs. right side of the circuit, you would get different voltages delivered to the Z1 and Z2 loads (but still summing to zero when you include the conductor voltage drops.) Which is, of course, exactly what you don't want. This is why using one low-resistance conductor to deliver the voltage to the load will always be more desirable if the goal is delivering the same signal to both loads -- like this: (+)------------------- Z1 Z2 (-)------------------- Why anyone would want to introduce the possibility of mismatched output voltages being delivered to the loads, by using different conductors for each circuit, is beyond my understanding -- assuming high fidelity is the desired goal. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) I wrote "larger wire" when I meant "larger resistance" I think we agree with this correction. Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? Hey kids, try this experiment at home. Disconnect one pair of your biwiring connections from one of your speaker inputs and hook it up to an RLC bridge. Notice how you "see" the reactance of the other biwired pair's drivers and crossover? When you reconnect the pair, so does the other driver and crossover. Isolation my ass... Like I said, "to a microscopic degree". The only possible change biwiring could make would be to provide a slightly (or significantly, depending the conductor qualities relative to each other) mismatched voltage to each set of driver circuits, due to slightly (or significantly) different conductor-pair voltage drops. If this is considered an improvement, even microscopically, then so be it. It seems contradictory to me. I thought the idea was to maintain balance, not introduce variation. I know I missed a couple of days of circuits class at the "U", but this is ridiculous. No, its the result of a typo. |
|
#183
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Arny Krueger" wrote in message ... "afh3" wrote in message news:5mo1c.450966$I06.5065611@attbi_s01 "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. However, the voltage drop will depend on whatever passive or reactive component is connected to each end. No it won't. Put whatever you want at either end of the conductor pair, and Kirchoff's Laws GUARANTEE that the voltage at both ends will be identical -- minus the difference of the voltage drop in each conductor pair. This is patently not isolation. My ASCII art probably sucks but I'll try anyway. ---------(+)-------- Z1 Z2 ---------(-)-------- Z1 and Z2 are the loads presented by the two separate speaker driver circuits in a biwired arrangement. No values for Z1 or Z2 will ever result in the voltages across Z1 or Z2 varying by more than the difference of the voltage drop of the conductors to the (in this case) left of the power-supply (in the middle) and those on the right. Furthermore, it is obvious from the diagram that the voltage source is still loaded by the parallel values of Z1 and Z2 (along with the conductor resistance that you have in either case, biwired or not), and, most importantly, Z1 still "sees" Z2 and vice-versa. Envision Z1 as a dead short. The voltage measured at Z2 will be equal to the voltage drop of the conductors on the Z1 side of the supply, minus the voltage drop of the conductors on the Z2 side of the supply. If it does not, then you have just invented a perpetual motion machine, or free power -- either of which will make you a fortune. Now, if you were to use significantly different conductors for the left vs. right side of the circuit, you would get different voltages delivered to the Z1 and Z2 loads (but still summing to zero when you include the conductor voltage drops.) Which is, of course, exactly what you don't want. This is why using one low-resistance conductor to deliver the voltage to the load will always be more desirable if the goal is delivering the same signal to both loads -- like this: (+)------------------- Z1 Z2 (-)------------------- Why anyone would want to introduce the possibility of mismatched output voltages being delivered to the loads, by using different conductors for each circuit, is beyond my understanding -- assuming high fidelity is the desired goal. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) I wrote "larger wire" when I meant "larger resistance" I think we agree with this correction. Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? Hey kids, try this experiment at home. Disconnect one pair of your biwiring connections from one of your speaker inputs and hook it up to an RLC bridge. Notice how you "see" the reactance of the other biwired pair's drivers and crossover? When you reconnect the pair, so does the other driver and crossover. Isolation my ass... Like I said, "to a microscopic degree". The only possible change biwiring could make would be to provide a slightly (or significantly, depending the conductor qualities relative to each other) mismatched voltage to each set of driver circuits, due to slightly (or significantly) different conductor-pair voltage drops. If this is considered an improvement, even microscopically, then so be it. It seems contradictory to me. I thought the idea was to maintain balance, not introduce variation. I know I missed a couple of days of circuits class at the "U", but this is ridiculous. No, its the result of a typo. |
|
#184
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Arny Krueger" wrote in message ... "afh3" wrote in message news:5mo1c.450966$I06.5065611@attbi_s01 "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. However, the voltage drop will depend on whatever passive or reactive component is connected to each end. No it won't. Put whatever you want at either end of the conductor pair, and Kirchoff's Laws GUARANTEE that the voltage at both ends will be identical -- minus the difference of the voltage drop in each conductor pair. This is patently not isolation. My ASCII art probably sucks but I'll try anyway. ---------(+)-------- Z1 Z2 ---------(-)-------- Z1 and Z2 are the loads presented by the two separate speaker driver circuits in a biwired arrangement. No values for Z1 or Z2 will ever result in the voltages across Z1 or Z2 varying by more than the difference of the voltage drop of the conductors to the (in this case) left of the power-supply (in the middle) and those on the right. Furthermore, it is obvious from the diagram that the voltage source is still loaded by the parallel values of Z1 and Z2 (along with the conductor resistance that you have in either case, biwired or not), and, most importantly, Z1 still "sees" Z2 and vice-versa. Envision Z1 as a dead short. The voltage measured at Z2 will be equal to the voltage drop of the conductors on the Z1 side of the supply, minus the voltage drop of the conductors on the Z2 side of the supply. If it does not, then you have just invented a perpetual motion machine, or free power -- either of which will make you a fortune. Now, if you were to use significantly different conductors for the left vs. right side of the circuit, you would get different voltages delivered to the Z1 and Z2 loads (but still summing to zero when you include the conductor voltage drops.) Which is, of course, exactly what you don't want. This is why using one low-resistance conductor to deliver the voltage to the load will always be more desirable if the goal is delivering the same signal to both loads -- like this: (+)------------------- Z1 Z2 (-)------------------- Why anyone would want to introduce the possibility of mismatched output voltages being delivered to the loads, by using different conductors for each circuit, is beyond my understanding -- assuming high fidelity is the desired goal. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) I wrote "larger wire" when I meant "larger resistance" I think we agree with this correction. Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? Hey kids, try this experiment at home. Disconnect one pair of your biwiring connections from one of your speaker inputs and hook it up to an RLC bridge. Notice how you "see" the reactance of the other biwired pair's drivers and crossover? When you reconnect the pair, so does the other driver and crossover. Isolation my ass... Like I said, "to a microscopic degree". The only possible change biwiring could make would be to provide a slightly (or significantly, depending the conductor qualities relative to each other) mismatched voltage to each set of driver circuits, due to slightly (or significantly) different conductor-pair voltage drops. If this is considered an improvement, even microscopically, then so be it. It seems contradictory to me. I thought the idea was to maintain balance, not introduce variation. I know I missed a couple of days of circuits class at the "U", but this is ridiculous. No, its the result of a typo. |
|
#185
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Arny Krueger" wrote in message ... "afh3" wrote in message news:5mo1c.450966$I06.5065611@attbi_s01 "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. However, the voltage drop will depend on whatever passive or reactive component is connected to each end. No it won't. Put whatever you want at either end of the conductor pair, and Kirchoff's Laws GUARANTEE that the voltage at both ends will be identical -- minus the difference of the voltage drop in each conductor pair. This is patently not isolation. My ASCII art probably sucks but I'll try anyway. ---------(+)-------- Z1 Z2 ---------(-)-------- Z1 and Z2 are the loads presented by the two separate speaker driver circuits in a biwired arrangement. No values for Z1 or Z2 will ever result in the voltages across Z1 or Z2 varying by more than the difference of the voltage drop of the conductors to the (in this case) left of the power-supply (in the middle) and those on the right. Furthermore, it is obvious from the diagram that the voltage source is still loaded by the parallel values of Z1 and Z2 (along with the conductor resistance that you have in either case, biwired or not), and, most importantly, Z1 still "sees" Z2 and vice-versa. Envision Z1 as a dead short. The voltage measured at Z2 will be equal to the voltage drop of the conductors on the Z1 side of the supply, minus the voltage drop of the conductors on the Z2 side of the supply. If it does not, then you have just invented a perpetual motion machine, or free power -- either of which will make you a fortune. Now, if you were to use significantly different conductors for the left vs. right side of the circuit, you would get different voltages delivered to the Z1 and Z2 loads (but still summing to zero when you include the conductor voltage drops.) Which is, of course, exactly what you don't want. This is why using one low-resistance conductor to deliver the voltage to the load will always be more desirable if the goal is delivering the same signal to both loads -- like this: (+)------------------- Z1 Z2 (-)------------------- Why anyone would want to introduce the possibility of mismatched output voltages being delivered to the loads, by using different conductors for each circuit, is beyond my understanding -- assuming high fidelity is the desired goal. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) I wrote "larger wire" when I meant "larger resistance" I think we agree with this correction. Uh, yeah. Go back your assertion that the advantage of this circuit is it's voltage divider configuration. If the source of the "isolation" is supposedly the quality of the circuit as a voltage divider, then how does reducing the amount of voltage division (by increasing conductor diameter, and therefore reducing conductor resistance) -- improve the "isolation" quality? Hey kids, try this experiment at home. Disconnect one pair of your biwiring connections from one of your speaker inputs and hook it up to an RLC bridge. Notice how you "see" the reactance of the other biwired pair's drivers and crossover? When you reconnect the pair, so does the other driver and crossover. Isolation my ass... Like I said, "to a microscopic degree". The only possible change biwiring could make would be to provide a slightly (or significantly, depending the conductor qualities relative to each other) mismatched voltage to each set of driver circuits, due to slightly (or significantly) different conductor-pair voltage drops. If this is considered an improvement, even microscopically, then so be it. It seems contradictory to me. I thought the idea was to maintain balance, not introduce variation. I know I missed a couple of days of circuits class at the "U", but this is ridiculous. No, its the result of a typo. |
|
#186
|
|||
|
|||
|
Bi-wiring - Hogwash?
"chung" wrote in message
rvers.com... afh3 wrote: "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. Actually if you read Arny Krueger's response carefully, you'll find that he was correct. The voltage drop is dependent on *both* the wire impedance *and* the load impedance. Just consider the case where the other end of the wire is open. No matter how big the wire impedance is, there is no drop in voltage across the wire. And similarly, if you read my statement carefully, you'll find that I was correct. The voltage drop on any conductor is proportional only to the resistance of the conductor, regardless of the type or value of the load presented. The total voltage drop for the circuit will obviously include the load's contribution, but this is not being argued here. The notion that somehow the load will impact the conductor's voltage drop characteristics is completely false. This is a case of not ageeing what isolation means. In a single wire case, the smaller the impedance of the wire, the more isolation you have between the drivers, if you define isolation as independence between the two drivers. What this means simply is that if you have a large wire impedance, the voltage at the speaker terminal is less dependent on the current drawn by the drivers. Agreed, but back to the point in contention he How is dropping some of the voltage across the conductor providing for a higher fidelity experience? If this were true, I would think we'd all be running chains of wet toothpicks in place of [pick your brand and price point] conductors. -afh3 |
|
#187
|
|||
|
|||
|
Bi-wiring - Hogwash?
"chung" wrote in message
rvers.com... afh3 wrote: "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. Actually if you read Arny Krueger's response carefully, you'll find that he was correct. The voltage drop is dependent on *both* the wire impedance *and* the load impedance. Just consider the case where the other end of the wire is open. No matter how big the wire impedance is, there is no drop in voltage across the wire. And similarly, if you read my statement carefully, you'll find that I was correct. The voltage drop on any conductor is proportional only to the resistance of the conductor, regardless of the type or value of the load presented. The total voltage drop for the circuit will obviously include the load's contribution, but this is not being argued here. The notion that somehow the load will impact the conductor's voltage drop characteristics is completely false. This is a case of not ageeing what isolation means. In a single wire case, the smaller the impedance of the wire, the more isolation you have between the drivers, if you define isolation as independence between the two drivers. What this means simply is that if you have a large wire impedance, the voltage at the speaker terminal is less dependent on the current drawn by the drivers. Agreed, but back to the point in contention he How is dropping some of the voltage across the conductor providing for a higher fidelity experience? If this were true, I would think we'd all be running chains of wet toothpicks in place of [pick your brand and price point] conductors. -afh3 |
|
#188
|
|||
|
|||
|
Bi-wiring - Hogwash?
"chung" wrote in message
rvers.com... afh3 wrote: "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. Actually if you read Arny Krueger's response carefully, you'll find that he was correct. The voltage drop is dependent on *both* the wire impedance *and* the load impedance. Just consider the case where the other end of the wire is open. No matter how big the wire impedance is, there is no drop in voltage across the wire. And similarly, if you read my statement carefully, you'll find that I was correct. The voltage drop on any conductor is proportional only to the resistance of the conductor, regardless of the type or value of the load presented. The total voltage drop for the circuit will obviously include the load's contribution, but this is not being argued here. The notion that somehow the load will impact the conductor's voltage drop characteristics is completely false. This is a case of not ageeing what isolation means. In a single wire case, the smaller the impedance of the wire, the more isolation you have between the drivers, if you define isolation as independence between the two drivers. What this means simply is that if you have a large wire impedance, the voltage at the speaker terminal is less dependent on the current drawn by the drivers. Agreed, but back to the point in contention he How is dropping some of the voltage across the conductor providing for a higher fidelity experience? If this were true, I would think we'd all be running chains of wet toothpicks in place of [pick your brand and price point] conductors. -afh3 |
|
#189
|
|||
|
|||
|
Bi-wiring - Hogwash?
"chung" wrote in message
rvers.com... afh3 wrote: "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. Actually if you read Arny Krueger's response carefully, you'll find that he was correct. The voltage drop is dependent on *both* the wire impedance *and* the load impedance. Just consider the case where the other end of the wire is open. No matter how big the wire impedance is, there is no drop in voltage across the wire. And similarly, if you read my statement carefully, you'll find that I was correct. The voltage drop on any conductor is proportional only to the resistance of the conductor, regardless of the type or value of the load presented. The total voltage drop for the circuit will obviously include the load's contribution, but this is not being argued here. The notion that somehow the load will impact the conductor's voltage drop characteristics is completely false. This is a case of not ageeing what isolation means. In a single wire case, the smaller the impedance of the wire, the more isolation you have between the drivers, if you define isolation as independence between the two drivers. What this means simply is that if you have a large wire impedance, the voltage at the speaker terminal is less dependent on the current drawn by the drivers. Agreed, but back to the point in contention he How is dropping some of the voltage across the conductor providing for a higher fidelity experience? If this were true, I would think we'd all be running chains of wet toothpicks in place of [pick your brand and price point] conductors. -afh3 |
|
#190
|
|||
|
|||
|
Bi-wiring - Hogwash?
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:
"Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) Uh, yeah. I suspect that the author meant a larger AWG number. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
|
#191
|
|||
|
|||
|
Bi-wiring - Hogwash?
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:
"Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) Uh, yeah. I suspect that the author meant a larger AWG number. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
|
#192
|
|||
|
|||
|
Bi-wiring - Hogwash?
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:
"Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) Uh, yeah. I suspect that the author meant a larger AWG number. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
|
#193
|
|||
|
|||
|
Bi-wiring - Hogwash?
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:
"Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. How does introducing additional resistance to the path of each individual conductor (all of which are still tied to a common output) provide "isolation"? It does isolate the two drivers from each other to a microscopic degree. Larger wire will give more isolation. However this is a poor way to provide isolation between the drivers, which is the crossover's job. Increased speaker cable impedance can easily have audible effects on the basic frequency response of the speakers, biwired or not. Larger wires will introduce less voltage drop across the conductor and somehow this will "give more isolation". (?) Uh, yeah. I suspect that the author meant a larger AWG number. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
|
#194
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Stewart Pinkerton" wrote in message ... On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Sigh. Regardless of the difference in current flowing in the two circuit paths (with a common supply), the total voltage drop summation for any complete circuit is always equal to the supply voltage. You aren't seriously contending that the voltage present on either end of the same conductor pair is different by anything other than the value of the voltage drop across the pair, are you? -afh3 |
|
#195
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Stewart Pinkerton" wrote in message ... On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Sigh. Regardless of the difference in current flowing in the two circuit paths (with a common supply), the total voltage drop summation for any complete circuit is always equal to the supply voltage. You aren't seriously contending that the voltage present on either end of the same conductor pair is different by anything other than the value of the voltage drop across the pair, are you? -afh3 |
|
#196
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Stewart Pinkerton" wrote in message ... On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Sigh. Regardless of the difference in current flowing in the two circuit paths (with a common supply), the total voltage drop summation for any complete circuit is always equal to the supply voltage. You aren't seriously contending that the voltage present on either end of the same conductor pair is different by anything other than the value of the voltage drop across the pair, are you? -afh3 |
|
#197
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Stewart Pinkerton" wrote in message ... On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: "Arny Krueger" wrote in message ... "afh3" wrote in message news:IOn1c.450824$I06.5064681@attbi_s01 Is it not true that the voltage drop across the conductors is only proportional to the resistance of the conductor itself, and not impacted by the load presented at the output end? Nope, its a voltage divider and the impedance of he wire and the load are relevant. The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Sigh. Regardless of the difference in current flowing in the two circuit paths (with a common supply), the total voltage drop summation for any complete circuit is always equal to the supply voltage. You aren't seriously contending that the voltage present on either end of the same conductor pair is different by anything other than the value of the voltage drop across the pair, are you? -afh3 |
|
#198
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Stewart Pinkerton" wrote in message ... On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Stewart, I've read your posting here before and I know you understand this, so I've got to believe that you either WAY misread what I wrote above, or something is fundamentally wrong with how I was communicating my point. It is simply a fact that (to paraphrase directly what I said previously): "The voltage at the end of one of [a] conductor pair will be exactly the same as the voltage at end of the [same] conductor pair, minus the difference in the voltage drop across [the] pair of conductors REGARDLESS of what passive or reactive component is connected to each end." The reason for this is that biwiring is just one pair of conductors with a voltage supply in the middle and a load on each end. The only possible difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair. -afh3 |
|
#199
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Stewart Pinkerton" wrote in message ... On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Stewart, I've read your posting here before and I know you understand this, so I've got to believe that you either WAY misread what I wrote above, or something is fundamentally wrong with how I was communicating my point. It is simply a fact that (to paraphrase directly what I said previously): "The voltage at the end of one of [a] conductor pair will be exactly the same as the voltage at end of the [same] conductor pair, minus the difference in the voltage drop across [the] pair of conductors REGARDLESS of what passive or reactive component is connected to each end." The reason for this is that biwiring is just one pair of conductors with a voltage supply in the middle and a load on each end. The only possible difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair. -afh3 |
|
#200
|
|||
|
|||
|
Bi-wiring - Hogwash?
"Stewart Pinkerton" wrote in message ... On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Stewart, I've read your posting here before and I know you understand this, so I've got to believe that you either WAY misread what I wrote above, or something is fundamentally wrong with how I was communicating my point. It is simply a fact that (to paraphrase directly what I said previously): "The voltage at the end of one of [a] conductor pair will be exactly the same as the voltage at end of the [same] conductor pair, minus the difference in the voltage drop across [the] pair of conductors REGARDLESS of what passive or reactive component is connected to each end." The reason for this is that biwiring is just one pair of conductors with a voltage supply in the middle and a load on each end. The only possible difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair. -afh3 |
| Reply |
| Thread Tools | |
| Display Modes | |
|
|
Similar Threads
|
||||
| Thread | Forum | |||
| Help - wiring identity Pioneer OEM CD changer | Car Audio | |||
| VW Factory CD Changer - Wiring Diagram? | Car Audio | |||
| Wiring for component "drawers"? | General | |||
| Honda Acura external amp wiring (repost) | Car Audio | |||
| Acura/Bose Amp wiring | Car Audio | |||
Linear Mode
