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#201
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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 |
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#202
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Bi-wiring - Hogwash?
"afh3" wrote in message
news:fNp1c.35731$PR3.677741@attbi_s03 "Arny Krueger" wrote in message ... "afh3" wrote in message news:5mo1c.450966$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. However, the voltage drop will depend on whatever passive or reactive component is connected to each end. This is patently not isolation. Never said it would be effective 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. However, that voltage drop will depend on whatever passive or reactive component is connected to that end, 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. Never said it would be effective isolation. 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. Never said it would be effective isolation. 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. Of course. Biwiring makes no sense. 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. It's so small it doesn't matter either way. 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. |
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#203
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Bi-wiring - Hogwash?
"afh3" wrote in message
news:fNp1c.35731$PR3.677741@attbi_s03 "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. However, the voltage drop will depend on whatever passive or reactive component is connected to each end. This is patently not isolation. Never said it would be effective 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. However, that voltage drop will depend on whatever passive or reactive component is connected to that end, 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. Never said it would be effective isolation. 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. Never said it would be effective isolation. 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. Of course. Biwiring makes no sense. 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. It's so small it doesn't matter either way. 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. |
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#204
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Bi-wiring - Hogwash?
"afh3" wrote in message
news:fNp1c.35731$PR3.677741@attbi_s03 "Arny Krueger" wrote in message ... "afh3" wrote in message news:5mo1c.450966$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. However, the voltage drop will depend on whatever passive or reactive component is connected to each end. This is patently not isolation. Never said it would be effective 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. However, that voltage drop will depend on whatever passive or reactive component is connected to that end, 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. Never said it would be effective isolation. 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. Never said it would be effective isolation. 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. Of course. Biwiring makes no sense. 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. It's so small it doesn't matter either way. 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. |
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#205
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Bi-wiring - Hogwash?
"afh3" wrote in message
news:fNp1c.35731$PR3.677741@attbi_s03 "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. However, the voltage drop will depend on whatever passive or reactive component is connected to each end. This is patently not isolation. Never said it would be effective 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. However, that voltage drop will depend on whatever passive or reactive component is connected to that end, 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. Never said it would be effective isolation. 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. Never said it would be effective isolation. 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. Of course. Biwiring makes no sense. 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. It's so small it doesn't matter either way. 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. |
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#206
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Bi-wiring - Hogwash?
afh3 wrote:
"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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. 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. There seems to be some miscommunication here. Of course, the load impedance impacts the conductor's voltage drop, since the load impedance determines the current flow. |
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#207
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Bi-wiring - Hogwash?
afh3 wrote:
"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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. 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. There seems to be some miscommunication here. Of course, the load impedance impacts the conductor's voltage drop, since the load impedance determines the current flow. |
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#208
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Bi-wiring - Hogwash?
afh3 wrote:
"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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. 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. There seems to be some miscommunication here. Of course, the load impedance impacts the conductor's voltage drop, since the load impedance determines the current flow. |
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#209
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Bi-wiring - Hogwash?
afh3 wrote:
"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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. 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. There seems to be some miscommunication here. Of course, the load impedance impacts the conductor's voltage drop, since the load impedance determines the current flow. |
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#210
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Bi-wiring - Hogwash?
"Arny Krueger" wrote in message
... However, the voltage drop will depend on whatever passive or reactive component is connected to each end. This is patently not isolation. Never said it would be effective isolation. It is not only ineffective isolation, it isn't isolation at all. It's just adding another R value to an RLC circuit. 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. However, that voltage drop will depend on whatever passive or reactive component is connected to that end, The specific value of the voltage drop will, of course, depend on the load and therefore the current flowing in the circuit (duh). But that value *will* be the same at both loads -- minus the difference in the negligible voltage drops of the two conductors. This is the key point: Regardless of what the loads Z1 and Z2 are (call one "woofer" and the other "tweeter"), the voltage measurement accross the loads will *always* be the same (identical, equal, highly exceedingly similar, indistinguishable, without variation, really really a lot alike) as the difference between the voltage drops present on the two conductors of the biwring circuit. Therefore it is not possible to deliver anything different to either load than could be accomplished with a single conductor pair. All bi-wiring is doing is moving the voltage source from the end of the circuit to the middle, and making each (left and right) side of the circuit as long as the original circuit was to begin with. It just doesn't make a difference. It can't. Never did, never will. The physics of electricity just doesn't work that way. The two circuits here are one conductor pair with a voltage supply in the middle and a load on each end. There is no isolation. Both loads see each other just fine. The only outcome of this type of circuit would be the possibility of introducing a variation in the voltage delivered to each load because of the difference in the conductor resistance values used to supply the "woofer" and "tweeter" circuits. At best, it can make no difference. Given conductor variability, it could actually introduce measurable voltage level differences at the loads. This would be a bad. Either way, it ain't a good thing because it introduces the possibility of error without the possibility of improvement. In a bi-amped implementation it is possible to deliver different signal content to the different speaker driver circuits. In a bi-wired configuration it is not. It is that simple. -afh3 |
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#211
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Bi-wiring - Hogwash?
"Arny Krueger" wrote in message
... However, the voltage drop will depend on whatever passive or reactive component is connected to each end. This is patently not isolation. Never said it would be effective isolation. It is not only ineffective isolation, it isn't isolation at all. It's just adding another R value to an RLC circuit. 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. However, that voltage drop will depend on whatever passive or reactive component is connected to that end, The specific value of the voltage drop will, of course, depend on the load and therefore the current flowing in the circuit (duh). But that value *will* be the same at both loads -- minus the difference in the negligible voltage drops of the two conductors. This is the key point: Regardless of what the loads Z1 and Z2 are (call one "woofer" and the other "tweeter"), the voltage measurement accross the loads will *always* be the same (identical, equal, highly exceedingly similar, indistinguishable, without variation, really really a lot alike) as the difference between the voltage drops present on the two conductors of the biwring circuit. Therefore it is not possible to deliver anything different to either load than could be accomplished with a single conductor pair. All bi-wiring is doing is moving the voltage source from the end of the circuit to the middle, and making each (left and right) side of the circuit as long as the original circuit was to begin with. It just doesn't make a difference. It can't. Never did, never will. The physics of electricity just doesn't work that way. The two circuits here are one conductor pair with a voltage supply in the middle and a load on each end. There is no isolation. Both loads see each other just fine. The only outcome of this type of circuit would be the possibility of introducing a variation in the voltage delivered to each load because of the difference in the conductor resistance values used to supply the "woofer" and "tweeter" circuits. At best, it can make no difference. Given conductor variability, it could actually introduce measurable voltage level differences at the loads. This would be a bad. Either way, it ain't a good thing because it introduces the possibility of error without the possibility of improvement. In a bi-amped implementation it is possible to deliver different signal content to the different speaker driver circuits. In a bi-wired configuration it is not. It is that simple. -afh3 |
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#212
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Bi-wiring - Hogwash?
"Arny Krueger" wrote in message
... However, the voltage drop will depend on whatever passive or reactive component is connected to each end. This is patently not isolation. Never said it would be effective isolation. It is not only ineffective isolation, it isn't isolation at all. It's just adding another R value to an RLC circuit. 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. However, that voltage drop will depend on whatever passive or reactive component is connected to that end, The specific value of the voltage drop will, of course, depend on the load and therefore the current flowing in the circuit (duh). But that value *will* be the same at both loads -- minus the difference in the negligible voltage drops of the two conductors. This is the key point: Regardless of what the loads Z1 and Z2 are (call one "woofer" and the other "tweeter"), the voltage measurement accross the loads will *always* be the same (identical, equal, highly exceedingly similar, indistinguishable, without variation, really really a lot alike) as the difference between the voltage drops present on the two conductors of the biwring circuit. Therefore it is not possible to deliver anything different to either load than could be accomplished with a single conductor pair. All bi-wiring is doing is moving the voltage source from the end of the circuit to the middle, and making each (left and right) side of the circuit as long as the original circuit was to begin with. It just doesn't make a difference. It can't. Never did, never will. The physics of electricity just doesn't work that way. The two circuits here are one conductor pair with a voltage supply in the middle and a load on each end. There is no isolation. Both loads see each other just fine. The only outcome of this type of circuit would be the possibility of introducing a variation in the voltage delivered to each load because of the difference in the conductor resistance values used to supply the "woofer" and "tweeter" circuits. At best, it can make no difference. Given conductor variability, it could actually introduce measurable voltage level differences at the loads. This would be a bad. Either way, it ain't a good thing because it introduces the possibility of error without the possibility of improvement. In a bi-amped implementation it is possible to deliver different signal content to the different speaker driver circuits. In a bi-wired configuration it is not. It is that simple. -afh3 |
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#213
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Bi-wiring - Hogwash?
"Arny Krueger" wrote in message
... However, the voltage drop will depend on whatever passive or reactive component is connected to each end. This is patently not isolation. Never said it would be effective isolation. It is not only ineffective isolation, it isn't isolation at all. It's just adding another R value to an RLC circuit. 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. However, that voltage drop will depend on whatever passive or reactive component is connected to that end, The specific value of the voltage drop will, of course, depend on the load and therefore the current flowing in the circuit (duh). But that value *will* be the same at both loads -- minus the difference in the negligible voltage drops of the two conductors. This is the key point: Regardless of what the loads Z1 and Z2 are (call one "woofer" and the other "tweeter"), the voltage measurement accross the loads will *always* be the same (identical, equal, highly exceedingly similar, indistinguishable, without variation, really really a lot alike) as the difference between the voltage drops present on the two conductors of the biwring circuit. Therefore it is not possible to deliver anything different to either load than could be accomplished with a single conductor pair. All bi-wiring is doing is moving the voltage source from the end of the circuit to the middle, and making each (left and right) side of the circuit as long as the original circuit was to begin with. It just doesn't make a difference. It can't. Never did, never will. The physics of electricity just doesn't work that way. The two circuits here are one conductor pair with a voltage supply in the middle and a load on each end. There is no isolation. Both loads see each other just fine. The only outcome of this type of circuit would be the possibility of introducing a variation in the voltage delivered to each load because of the difference in the conductor resistance values used to supply the "woofer" and "tweeter" circuits. At best, it can make no difference. Given conductor variability, it could actually introduce measurable voltage level differences at the loads. This would be a bad. Either way, it ain't a good thing because it introduces the possibility of error without the possibility of improvement. In a bi-amped implementation it is possible to deliver different signal content to the different speaker driver circuits. In a bi-wired configuration it is not. It is that simple. -afh3 |
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#214
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Bi-wiring - Hogwash?
afh3 wrote:
"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. As a purely academic exercise, Z1 does not see Z2 and vice versa because the voltage source in the middle, if it's ideal, has zero impedance. Therefore Z2 only sees the impedance of the wire. I'll redraw it slightly he --------|(+)|------- Z1 | | Z2 --------|(-)|------- The impedance across the (+) and (-) terminals is zero for an ideal voltage source. 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. As a purely academic exercise, the voltage measured at Z2 is *independent* of the value of Z1, since the voltage source in the middle is ideal and has zero impedance. If it does not, then you have just invented a perpetual motion machine, or free power -- either of which will make you a fortune. Not sure what you are getting at here. There is no violation of any conservation principles if the voltage across Z2 is independent of the value of Z1. 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.) No they don't sum to zero. If the voltage source is ideal, Z1 and Z2 are isolated. Voltage across Z1 has no impact on voltage across Z2. |
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#215
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Bi-wiring - Hogwash?
afh3 wrote:
"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. As a purely academic exercise, Z1 does not see Z2 and vice versa because the voltage source in the middle, if it's ideal, has zero impedance. Therefore Z2 only sees the impedance of the wire. I'll redraw it slightly he --------|(+)|------- Z1 | | Z2 --------|(-)|------- The impedance across the (+) and (-) terminals is zero for an ideal voltage source. 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. As a purely academic exercise, the voltage measured at Z2 is *independent* of the value of Z1, since the voltage source in the middle is ideal and has zero impedance. If it does not, then you have just invented a perpetual motion machine, or free power -- either of which will make you a fortune. Not sure what you are getting at here. There is no violation of any conservation principles if the voltage across Z2 is independent of the value of Z1. 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.) No they don't sum to zero. If the voltage source is ideal, Z1 and Z2 are isolated. Voltage across Z1 has no impact on voltage across Z2. |
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#216
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Bi-wiring - Hogwash?
afh3 wrote:
"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. As a purely academic exercise, Z1 does not see Z2 and vice versa because the voltage source in the middle, if it's ideal, has zero impedance. Therefore Z2 only sees the impedance of the wire. I'll redraw it slightly he --------|(+)|------- Z1 | | Z2 --------|(-)|------- The impedance across the (+) and (-) terminals is zero for an ideal voltage source. 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. As a purely academic exercise, the voltage measured at Z2 is *independent* of the value of Z1, since the voltage source in the middle is ideal and has zero impedance. If it does not, then you have just invented a perpetual motion machine, or free power -- either of which will make you a fortune. Not sure what you are getting at here. There is no violation of any conservation principles if the voltage across Z2 is independent of the value of Z1. 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.) No they don't sum to zero. If the voltage source is ideal, Z1 and Z2 are isolated. Voltage across Z1 has no impact on voltage across Z2. |
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#217
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Bi-wiring - Hogwash?
afh3 wrote:
"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. As a purely academic exercise, Z1 does not see Z2 and vice versa because the voltage source in the middle, if it's ideal, has zero impedance. Therefore Z2 only sees the impedance of the wire. I'll redraw it slightly he --------|(+)|------- Z1 | | Z2 --------|(-)|------- The impedance across the (+) and (-) terminals is zero for an ideal voltage source. 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. As a purely academic exercise, the voltage measured at Z2 is *independent* of the value of Z1, since the voltage source in the middle is ideal and has zero impedance. If it does not, then you have just invented a perpetual motion machine, or free power -- either of which will make you a fortune. Not sure what you are getting at here. There is no violation of any conservation principles if the voltage across Z2 is independent of the value of Z1. 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.) No they don't sum to zero. If the voltage source is ideal, Z1 and Z2 are isolated. Voltage across Z1 has no impact on voltage across Z2. |
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#218
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... afh3 wrote: "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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). I now regret inadvertantly putting that word "only" in the paragraph above. 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. There seems to be some miscommunication here. Of course, the load impedance impacts the conductor's voltage drop, since the load impedance determines the current flow. The (somewhat more carefully phrased) statement that I made above is still correct here though. The total voltage drop characteristic of a conductor is only impacted by it's own resistance. While it's true that the absolute value of that drop is a function of the current flow, only the resistance of the specific conductor is a constant -- and therefore the conductor's only "characteristic" with regard to the voltage drop seen for a given current flow. -afh3 |
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#219
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... afh3 wrote: "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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). I now regret inadvertantly putting that word "only" in the paragraph above. 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. There seems to be some miscommunication here. Of course, the load impedance impacts the conductor's voltage drop, since the load impedance determines the current flow. The (somewhat more carefully phrased) statement that I made above is still correct here though. The total voltage drop characteristic of a conductor is only impacted by it's own resistance. While it's true that the absolute value of that drop is a function of the current flow, only the resistance of the specific conductor is a constant -- and therefore the conductor's only "characteristic" with regard to the voltage drop seen for a given current flow. -afh3 |
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#220
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... afh3 wrote: "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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). I now regret inadvertantly putting that word "only" in the paragraph above. 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. There seems to be some miscommunication here. Of course, the load impedance impacts the conductor's voltage drop, since the load impedance determines the current flow. The (somewhat more carefully phrased) statement that I made above is still correct here though. The total voltage drop characteristic of a conductor is only impacted by it's own resistance. While it's true that the absolute value of that drop is a function of the current flow, only the resistance of the specific conductor is a constant -- and therefore the conductor's only "characteristic" with regard to the voltage drop seen for a given current flow. -afh3 |
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#221
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... afh3 wrote: "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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). I now regret inadvertantly putting that word "only" in the paragraph above. 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. There seems to be some miscommunication here. Of course, the load impedance impacts the conductor's voltage drop, since the load impedance determines the current flow. The (somewhat more carefully phrased) statement that I made above is still correct here though. The total voltage drop characteristic of a conductor is only impacted by it's own resistance. While it's true that the absolute value of that drop is a function of the current flow, only the resistance of the specific conductor is a constant -- and therefore the conductor's only "characteristic" with regard to the voltage drop seen for a given current flow. -afh3 |
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#222
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Bi-wiring - Hogwash?
"chung" wrote in message
rvers.com... As a purely academic exercise, Z1 does not see Z2 and vice versa because the voltage source in the middle, if it's ideal, has zero impedance. Therefore Z2 only sees the impedance of the wire. I'll redraw it slightly he --------|(+)|------- Z1 | | Z2 --------|(-)|------- The impedance across the (+) and (-) terminals is zero for an ideal voltage source. OMG. Nice try, but she don't work that way. Look here for an explanation. http://hyperphysics.phy-astr.gsu.edu.../visource.html and subsequently he http://hyperphysics.phy-astr.gsu.edu...evenin.html#c1 and he http://hyperphysics.phy-astr.gsu.edu...norton.html#c1 to see why that just ain't so. Then: Get a battery. Connect it to the middle of a conductor pair as shown above. Put a light on one end of the conductor pair (Z1). Short the other end (Z2 = 0). Did the light go out? Thought so... Since neither of us have amplifiers that approach the zero internal impedance of an ideal voltage source, nor the infinite internal impedance of an ideal current source, I guess we'll just have to settle for reality. -afh3 |
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#223
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Bi-wiring - Hogwash?
"chung" wrote in message
rvers.com... As a purely academic exercise, Z1 does not see Z2 and vice versa because the voltage source in the middle, if it's ideal, has zero impedance. Therefore Z2 only sees the impedance of the wire. I'll redraw it slightly he --------|(+)|------- Z1 | | Z2 --------|(-)|------- The impedance across the (+) and (-) terminals is zero for an ideal voltage source. OMG. Nice try, but she don't work that way. Look here for an explanation. http://hyperphysics.phy-astr.gsu.edu.../visource.html and subsequently he http://hyperphysics.phy-astr.gsu.edu...evenin.html#c1 and he http://hyperphysics.phy-astr.gsu.edu...norton.html#c1 to see why that just ain't so. Then: Get a battery. Connect it to the middle of a conductor pair as shown above. Put a light on one end of the conductor pair (Z1). Short the other end (Z2 = 0). Did the light go out? Thought so... Since neither of us have amplifiers that approach the zero internal impedance of an ideal voltage source, nor the infinite internal impedance of an ideal current source, I guess we'll just have to settle for reality. -afh3 |
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#224
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Bi-wiring - Hogwash?
"chung" wrote in message
rvers.com... As a purely academic exercise, Z1 does not see Z2 and vice versa because the voltage source in the middle, if it's ideal, has zero impedance. Therefore Z2 only sees the impedance of the wire. I'll redraw it slightly he --------|(+)|------- Z1 | | Z2 --------|(-)|------- The impedance across the (+) and (-) terminals is zero for an ideal voltage source. OMG. Nice try, but she don't work that way. Look here for an explanation. http://hyperphysics.phy-astr.gsu.edu.../visource.html and subsequently he http://hyperphysics.phy-astr.gsu.edu...evenin.html#c1 and he http://hyperphysics.phy-astr.gsu.edu...norton.html#c1 to see why that just ain't so. Then: Get a battery. Connect it to the middle of a conductor pair as shown above. Put a light on one end of the conductor pair (Z1). Short the other end (Z2 = 0). Did the light go out? Thought so... Since neither of us have amplifiers that approach the zero internal impedance of an ideal voltage source, nor the infinite internal impedance of an ideal current source, I guess we'll just have to settle for reality. -afh3 |
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#225
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Bi-wiring - Hogwash?
"chung" wrote in message
rvers.com... As a purely academic exercise, Z1 does not see Z2 and vice versa because the voltage source in the middle, if it's ideal, has zero impedance. Therefore Z2 only sees the impedance of the wire. I'll redraw it slightly he --------|(+)|------- Z1 | | Z2 --------|(-)|------- The impedance across the (+) and (-) terminals is zero for an ideal voltage source. OMG. Nice try, but she don't work that way. Look here for an explanation. http://hyperphysics.phy-astr.gsu.edu.../visource.html and subsequently he http://hyperphysics.phy-astr.gsu.edu...evenin.html#c1 and he http://hyperphysics.phy-astr.gsu.edu...norton.html#c1 to see why that just ain't so. Then: Get a battery. Connect it to the middle of a conductor pair as shown above. Put a light on one end of the conductor pair (Z1). Short the other end (Z2 = 0). Did the light go out? Thought so... Since neither of us have amplifiers that approach the zero internal impedance of an ideal voltage source, nor the infinite internal impedance of an ideal current source, I guess we'll just have to settle for reality. -afh3 |
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#226
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... afh3 wrote: "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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. |
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#227
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... afh3 wrote: "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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. |
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#228
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... afh3 wrote: "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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. |
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#229
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... afh3 wrote: "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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. |
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#230
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... As a purely academic exercise, Z1 does not see Z2 and vice versa because the voltage source in the middle, if it's ideal, has zero impedance. Therefore Z2 only sees the impedance of the wire. I'll redraw it slightly he --------|(+)|------- Z1 | | Z2 --------|(-)|------- The impedance across the (+) and (-) terminals is zero for an ideal voltage source. OMG. Nice try, but she don't work that way. OMG, did you notice that I said purely academic? Yes, that's how it works. You can have an amp that approaches an ideal voltage source. In fact, the error can be made as small as you want. Look here for an explanation. http://hyperphysics.phy-astr.gsu.edu.../visource.html and subsequently he http://hyperphysics.phy-astr.gsu.edu...evenin.html#c1 and he http://hyperphysics.phy-astr.gsu.edu...norton.html#c1 to see why that just ain't so. Sorry, I am a practicing electrical engineer, for almost 30 years. I understand all these things, and there is absolutly no contradiction between what I said and theory. Then: Get a battery. Connect it to the middle of a conductor pair as shown above. Put a light on one end of the conductor pair (Z1). Short the other end (Z2 = 0). Did the light go out? Thought so... That's because a battery does not have a small enough output impedance. Now try a real power source, and let's use conductor's with, say 0.1 ohm. Even if the load is a short, the output voltage will stay the same: it will simply force the current to flow through that conductor. Since neither of us have amplifiers that approach the zero internal impedance of an ideal voltage source, nor the infinite internal impedance of an ideal current source, I guess we'll just have to settle for reality. Did you miss what I said about academic? But more importantly, real amps approach an ideal source, and that's where biwiring can lead to a "theoretical" advantage, although not one that can be audible necessarily. -afh3 |
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#231
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... As a purely academic exercise, Z1 does not see Z2 and vice versa because the voltage source in the middle, if it's ideal, has zero impedance. Therefore Z2 only sees the impedance of the wire. I'll redraw it slightly he --------|(+)|------- Z1 | | Z2 --------|(-)|------- The impedance across the (+) and (-) terminals is zero for an ideal voltage source. OMG. Nice try, but she don't work that way. OMG, did you notice that I said purely academic? Yes, that's how it works. You can have an amp that approaches an ideal voltage source. In fact, the error can be made as small as you want. Look here for an explanation. http://hyperphysics.phy-astr.gsu.edu.../visource.html and subsequently he http://hyperphysics.phy-astr.gsu.edu...evenin.html#c1 and he http://hyperphysics.phy-astr.gsu.edu...norton.html#c1 to see why that just ain't so. Sorry, I am a practicing electrical engineer, for almost 30 years. I understand all these things, and there is absolutly no contradiction between what I said and theory. Then: Get a battery. Connect it to the middle of a conductor pair as shown above. Put a light on one end of the conductor pair (Z1). Short the other end (Z2 = 0). Did the light go out? Thought so... That's because a battery does not have a small enough output impedance. Now try a real power source, and let's use conductor's with, say 0.1 ohm. Even if the load is a short, the output voltage will stay the same: it will simply force the current to flow through that conductor. Since neither of us have amplifiers that approach the zero internal impedance of an ideal voltage source, nor the infinite internal impedance of an ideal current source, I guess we'll just have to settle for reality. Did you miss what I said about academic? But more importantly, real amps approach an ideal source, and that's where biwiring can lead to a "theoretical" advantage, although not one that can be audible necessarily. -afh3 |
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#232
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... As a purely academic exercise, Z1 does not see Z2 and vice versa because the voltage source in the middle, if it's ideal, has zero impedance. Therefore Z2 only sees the impedance of the wire. I'll redraw it slightly he --------|(+)|------- Z1 | | Z2 --------|(-)|------- The impedance across the (+) and (-) terminals is zero for an ideal voltage source. OMG. Nice try, but she don't work that way. OMG, did you notice that I said purely academic? Yes, that's how it works. You can have an amp that approaches an ideal voltage source. In fact, the error can be made as small as you want. Look here for an explanation. http://hyperphysics.phy-astr.gsu.edu.../visource.html and subsequently he http://hyperphysics.phy-astr.gsu.edu...evenin.html#c1 and he http://hyperphysics.phy-astr.gsu.edu...norton.html#c1 to see why that just ain't so. Sorry, I am a practicing electrical engineer, for almost 30 years. I understand all these things, and there is absolutly no contradiction between what I said and theory. Then: Get a battery. Connect it to the middle of a conductor pair as shown above. Put a light on one end of the conductor pair (Z1). Short the other end (Z2 = 0). Did the light go out? Thought so... That's because a battery does not have a small enough output impedance. Now try a real power source, and let's use conductor's with, say 0.1 ohm. Even if the load is a short, the output voltage will stay the same: it will simply force the current to flow through that conductor. Since neither of us have amplifiers that approach the zero internal impedance of an ideal voltage source, nor the infinite internal impedance of an ideal current source, I guess we'll just have to settle for reality. Did you miss what I said about academic? But more importantly, real amps approach an ideal source, and that's where biwiring can lead to a "theoretical" advantage, although not one that can be audible necessarily. -afh3 |
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#233
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... As a purely academic exercise, Z1 does not see Z2 and vice versa because the voltage source in the middle, if it's ideal, has zero impedance. Therefore Z2 only sees the impedance of the wire. I'll redraw it slightly he --------|(+)|------- Z1 | | Z2 --------|(-)|------- The impedance across the (+) and (-) terminals is zero for an ideal voltage source. OMG. Nice try, but she don't work that way. OMG, did you notice that I said purely academic? Yes, that's how it works. You can have an amp that approaches an ideal voltage source. In fact, the error can be made as small as you want. Look here for an explanation. http://hyperphysics.phy-astr.gsu.edu.../visource.html and subsequently he http://hyperphysics.phy-astr.gsu.edu...evenin.html#c1 and he http://hyperphysics.phy-astr.gsu.edu...norton.html#c1 to see why that just ain't so. Sorry, I am a practicing electrical engineer, for almost 30 years. I understand all these things, and there is absolutly no contradiction between what I said and theory. Then: Get a battery. Connect it to the middle of a conductor pair as shown above. Put a light on one end of the conductor pair (Z1). Short the other end (Z2 = 0). Did the light go out? Thought so... That's because a battery does not have a small enough output impedance. Now try a real power source, and let's use conductor's with, say 0.1 ohm. Even if the load is a short, the output voltage will stay the same: it will simply force the current to flow through that conductor. Since neither of us have amplifiers that approach the zero internal impedance of an ideal voltage source, nor the infinite internal impedance of an ideal current source, I guess we'll just have to settle for reality. Did you miss what I said about academic? But more importantly, real amps approach an ideal source, and that's where biwiring can lead to a "theoretical" advantage, although not one that can be audible necessarily. -afh3 |
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#234
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Bi-wiring - Hogwash?
"chung" wrote in message ers.com... afh3 wrote: "chung" wrote in message rvers.com... afh3 wrote: "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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. Strictly speaking, as I stated above, it *is* equal to the current times the conductor impedance, *regardless*. Do the math to see. -afh3 |
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#235
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Bi-wiring - Hogwash?
"chung" wrote in message ers.com... afh3 wrote: "chung" wrote in message rvers.com... afh3 wrote: "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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. Strictly speaking, as I stated above, it *is* equal to the current times the conductor impedance, *regardless*. Do the math to see. -afh3 |
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#236
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Bi-wiring - Hogwash?
"chung" wrote in message ers.com... afh3 wrote: "chung" wrote in message rvers.com... afh3 wrote: "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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. Strictly speaking, as I stated above, it *is* equal to the current times the conductor impedance, *regardless*. Do the math to see. -afh3 |
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#237
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Bi-wiring - Hogwash?
"chung" wrote in message ers.com... afh3 wrote: "chung" wrote in message rvers.com... afh3 wrote: "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. No, the voltage drop across the conductor is proportional to both the impedance of the conductor *and* the current flowing through it. As I said earlier, if the load is infinite, then there is no voltage drop across the conductor regardless of what the impedance is. Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. Strictly speaking, as I stated above, it *is* equal to the current times the conductor impedance, *regardless*. Do the math to see. -afh3 |
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#238
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Bi-wiring - Hogwash?
"chung" wrote in message ers.com... Sorry, I am a practicing electrical engineer, for almost 30 years. I understand all these things, and there is absolutly no contradiction between what I said and theory. Ok. 2004 - 1982 = 32 here. That's because a battery does not have a small enough output impedance. Now try a real power source, and let's use conductor's with, say 0.1 ohm. Even if the load is a short, the output voltage will stay the same: it will simply force the current to flow through that conductor. OK, let's try a "real" power source. Let's say it's at an arbitrary 24v DC PS. With your example, the PS would have to be able to source 240A to keep the voltage constant. Oh yeah, that's "real" alright -- only 5760 watts delivered -- about 50+ amps sourced from the mains assuming the PS is running from a (US standard) 117 VAC outlet. You must have some pretty hefty breakers installed at your house. Know of anything like this type of performance found in audio equipment, or for that matter virtually any "real" consumer application? Certainly, some very high-performance designs can achieve output impedance values of .1 ohms and lower, the majority seem to be in the .5-1-2 ohm range -- with a typical single-ended-triode design (without feedback) at about 3 ohms. If we use these real-world values in the Norton or Thevenin model of the amp/speaker circuit, it is easy to see why bi-wiring just won't matter. If your amp can source triple-digit amps worth of current at a constant voltage, and has an output impedance of about 0.02 ohms, then maybe, just maybe, bi-wiring could make a difference measurable via instrumentation -- but very unlikely audibly detectable. Since neither of us have amplifiers that approach the zero internal impedance of an ideal voltage source, nor the infinite internal impedance of an ideal current source, I guess we'll just have to settle for reality. Did you miss what I said about academic? But more importantly, real amps approach an ideal source, and that's where biwiring can lead to a "theoretical" advantage, although not one that can be audible necessarily. No, I didn't miss the academic references. Since most audio equipment is real as opposed to imaginary, I kept my responses within those bounds. -afh3 |
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Bi-wiring - Hogwash?
"chung" wrote in message ers.com... Sorry, I am a practicing electrical engineer, for almost 30 years. I understand all these things, and there is absolutly no contradiction between what I said and theory. Ok. 2004 - 1982 = 32 here. That's because a battery does not have a small enough output impedance. Now try a real power source, and let's use conductor's with, say 0.1 ohm. Even if the load is a short, the output voltage will stay the same: it will simply force the current to flow through that conductor. OK, let's try a "real" power source. Let's say it's at an arbitrary 24v DC PS. With your example, the PS would have to be able to source 240A to keep the voltage constant. Oh yeah, that's "real" alright -- only 5760 watts delivered -- about 50+ amps sourced from the mains assuming the PS is running from a (US standard) 117 VAC outlet. You must have some pretty hefty breakers installed at your house. Know of anything like this type of performance found in audio equipment, or for that matter virtually any "real" consumer application? Certainly, some very high-performance designs can achieve output impedance values of .1 ohms and lower, the majority seem to be in the .5-1-2 ohm range -- with a typical single-ended-triode design (without feedback) at about 3 ohms. If we use these real-world values in the Norton or Thevenin model of the amp/speaker circuit, it is easy to see why bi-wiring just won't matter. If your amp can source triple-digit amps worth of current at a constant voltage, and has an output impedance of about 0.02 ohms, then maybe, just maybe, bi-wiring could make a difference measurable via instrumentation -- but very unlikely audibly detectable. Since neither of us have amplifiers that approach the zero internal impedance of an ideal voltage source, nor the infinite internal impedance of an ideal current source, I guess we'll just have to settle for reality. Did you miss what I said about academic? But more importantly, real amps approach an ideal source, and that's where biwiring can lead to a "theoretical" advantage, although not one that can be audible necessarily. No, I didn't miss the academic references. Since most audio equipment is real as opposed to imaginary, I kept my responses within those bounds. -afh3 |
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#240
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Bi-wiring - Hogwash?
"chung" wrote in message ers.com... Sorry, I am a practicing electrical engineer, for almost 30 years. I understand all these things, and there is absolutly no contradiction between what I said and theory. Ok. 2004 - 1982 = 32 here. That's because a battery does not have a small enough output impedance. Now try a real power source, and let's use conductor's with, say 0.1 ohm. Even if the load is a short, the output voltage will stay the same: it will simply force the current to flow through that conductor. OK, let's try a "real" power source. Let's say it's at an arbitrary 24v DC PS. With your example, the PS would have to be able to source 240A to keep the voltage constant. Oh yeah, that's "real" alright -- only 5760 watts delivered -- about 50+ amps sourced from the mains assuming the PS is running from a (US standard) 117 VAC outlet. You must have some pretty hefty breakers installed at your house. Know of anything like this type of performance found in audio equipment, or for that matter virtually any "real" consumer application? Certainly, some very high-performance designs can achieve output impedance values of .1 ohms and lower, the majority seem to be in the .5-1-2 ohm range -- with a typical single-ended-triode design (without feedback) at about 3 ohms. If we use these real-world values in the Norton or Thevenin model of the amp/speaker circuit, it is easy to see why bi-wiring just won't matter. If your amp can source triple-digit amps worth of current at a constant voltage, and has an output impedance of about 0.02 ohms, then maybe, just maybe, bi-wiring could make a difference measurable via instrumentation -- but very unlikely audibly detectable. Since neither of us have amplifiers that approach the zero internal impedance of an ideal voltage source, nor the infinite internal impedance of an ideal current source, I guess we'll just have to settle for reality. Did you miss what I said about academic? But more importantly, real amps approach an ideal source, and that's where biwiring can lead to a "theoretical" advantage, although not one that can be audible necessarily. No, I didn't miss the academic references. Since most audio equipment is real as opposed to imaginary, I kept my responses within those bounds. -afh3 |
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