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#241
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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 |
#242
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Bi-wiring - Hogwash?
afh3 wrote:
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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". |
#243
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Bi-wiring - Hogwash?
afh3 wrote:
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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". |
#244
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Bi-wiring - Hogwash?
afh3 wrote:
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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". |
#245
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Bi-wiring - Hogwash?
afh3 wrote:
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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". |
#246
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Bi-wiring - Hogwash?
afh3 wrote:
"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. OMG, now we know you can't subtract. No wonder . 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. You seem to miss two points: (1) I said "purely academic", and theoretical. I never said that the effects of bi-wiring are necessarily audible. (2) It's not the max. output current capacity that is important. It is the equivalent output impedance of the amp that is important. That is, the small signal impedance. An amp can have a current limit that does not approach thousands of amps, but an output resistance that approaches zero. It is that small signal output impedance you should concern yourself with, when you try to model these effects. When the amp's output impedance is small, Z1 does not see Z2, in your example, because of the shunting effect of the output impedance. |
#247
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Bi-wiring - Hogwash?
afh3 wrote:
"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. OMG, now we know you can't subtract. No wonder . 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. You seem to miss two points: (1) I said "purely academic", and theoretical. I never said that the effects of bi-wiring are necessarily audible. (2) It's not the max. output current capacity that is important. It is the equivalent output impedance of the amp that is important. That is, the small signal impedance. An amp can have a current limit that does not approach thousands of amps, but an output resistance that approaches zero. It is that small signal output impedance you should concern yourself with, when you try to model these effects. When the amp's output impedance is small, Z1 does not see Z2, in your example, because of the shunting effect of the output impedance. |
#248
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Bi-wiring - Hogwash?
afh3 wrote:
"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. OMG, now we know you can't subtract. No wonder . 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. You seem to miss two points: (1) I said "purely academic", and theoretical. I never said that the effects of bi-wiring are necessarily audible. (2) It's not the max. output current capacity that is important. It is the equivalent output impedance of the amp that is important. That is, the small signal impedance. An amp can have a current limit that does not approach thousands of amps, but an output resistance that approaches zero. It is that small signal output impedance you should concern yourself with, when you try to model these effects. When the amp's output impedance is small, Z1 does not see Z2, in your example, because of the shunting effect of the output impedance. |
#249
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Bi-wiring - Hogwash?
afh3 wrote:
"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. OMG, now we know you can't subtract. No wonder . 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. You seem to miss two points: (1) I said "purely academic", and theoretical. I never said that the effects of bi-wiring are necessarily audible. (2) It's not the max. output current capacity that is important. It is the equivalent output impedance of the amp that is important. That is, the small signal impedance. An amp can have a current limit that does not approach thousands of amps, but an output resistance that approaches zero. It is that small signal output impedance you should concern yourself with, when you try to model these effects. When the amp's output impedance is small, Z1 does not see Z2, in your example, because of the shunting effect of the output impedance. |
#250
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Bi-wiring - Hogwash?
"chung" wrote in message vers.com... afh3 wrote: 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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. -afh3 |
#251
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Bi-wiring - Hogwash?
"chung" wrote in message vers.com... afh3 wrote: 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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. -afh3 |
#252
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Bi-wiring - Hogwash?
"chung" wrote in message vers.com... afh3 wrote: 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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. -afh3 |
#253
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Bi-wiring - Hogwash?
"chung" wrote in message vers.com... afh3 wrote: 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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. -afh3 |
#254
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Bi-wiring - Hogwash?
Sorry, typo. Meant to say, "22 here."
"chung" wrote in message vers.com... afh3 wrote: "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. Sorry, typo. Meant to say, "22 here." OMG, now we know you can't subtract. No wonder . 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. You seem to miss two points: (1) I said "purely academic", and theoretical. I never said that the effects of bi-wiring are necessarily audible. (2) It's not the max. output current capacity that is important. It is the equivalent output impedance of the amp that is important. That is, the small signal impedance. An amp can have a current limit that does not approach thousands of amps, but an output resistance that approaches zero. It is that small signal output impedance you should concern yourself with, when you try to model these effects. When the amp's output impedance is small, Z1 does not see Z2, in your example, because of the shunting effect of the output impedance. No I didn't miss them, I just contend -- like in the case of popular Class A single-ended triode no-feedback designs -- that the output impedance is too high to effectively shunt the two sections of the network. Even a two or one ohm output impedance is too high IMO. Regarding the output current, I was simply trying to define the ideal amplifier that would be necessary to create a realizable difference via bi-wiring. If the extremely low ouput-impedance amp couldn't source a lot of current, then one side of the bi-wiring could affect the other in some cases -- like in the case of a short circuit or very low driver impedance on one side. |
#255
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Bi-wiring - Hogwash?
Sorry, typo. Meant to say, "22 here."
"chung" wrote in message vers.com... afh3 wrote: "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. Sorry, typo. Meant to say, "22 here." OMG, now we know you can't subtract. No wonder . 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. You seem to miss two points: (1) I said "purely academic", and theoretical. I never said that the effects of bi-wiring are necessarily audible. (2) It's not the max. output current capacity that is important. It is the equivalent output impedance of the amp that is important. That is, the small signal impedance. An amp can have a current limit that does not approach thousands of amps, but an output resistance that approaches zero. It is that small signal output impedance you should concern yourself with, when you try to model these effects. When the amp's output impedance is small, Z1 does not see Z2, in your example, because of the shunting effect of the output impedance. No I didn't miss them, I just contend -- like in the case of popular Class A single-ended triode no-feedback designs -- that the output impedance is too high to effectively shunt the two sections of the network. Even a two or one ohm output impedance is too high IMO. Regarding the output current, I was simply trying to define the ideal amplifier that would be necessary to create a realizable difference via bi-wiring. If the extremely low ouput-impedance amp couldn't source a lot of current, then one side of the bi-wiring could affect the other in some cases -- like in the case of a short circuit or very low driver impedance on one side. |
#256
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Bi-wiring - Hogwash?
Sorry, typo. Meant to say, "22 here."
"chung" wrote in message vers.com... afh3 wrote: "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. Sorry, typo. Meant to say, "22 here." OMG, now we know you can't subtract. No wonder . 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. You seem to miss two points: (1) I said "purely academic", and theoretical. I never said that the effects of bi-wiring are necessarily audible. (2) It's not the max. output current capacity that is important. It is the equivalent output impedance of the amp that is important. That is, the small signal impedance. An amp can have a current limit that does not approach thousands of amps, but an output resistance that approaches zero. It is that small signal output impedance you should concern yourself with, when you try to model these effects. When the amp's output impedance is small, Z1 does not see Z2, in your example, because of the shunting effect of the output impedance. No I didn't miss them, I just contend -- like in the case of popular Class A single-ended triode no-feedback designs -- that the output impedance is too high to effectively shunt the two sections of the network. Even a two or one ohm output impedance is too high IMO. Regarding the output current, I was simply trying to define the ideal amplifier that would be necessary to create a realizable difference via bi-wiring. If the extremely low ouput-impedance amp couldn't source a lot of current, then one side of the bi-wiring could affect the other in some cases -- like in the case of a short circuit or very low driver impedance on one side. |
#257
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Bi-wiring - Hogwash?
Sorry, typo. Meant to say, "22 here."
"chung" wrote in message vers.com... afh3 wrote: "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. Sorry, typo. Meant to say, "22 here." OMG, now we know you can't subtract. No wonder . 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. You seem to miss two points: (1) I said "purely academic", and theoretical. I never said that the effects of bi-wiring are necessarily audible. (2) It's not the max. output current capacity that is important. It is the equivalent output impedance of the amp that is important. That is, the small signal impedance. An amp can have a current limit that does not approach thousands of amps, but an output resistance that approaches zero. It is that small signal output impedance you should concern yourself with, when you try to model these effects. When the amp's output impedance is small, Z1 does not see Z2, in your example, because of the shunting effect of the output impedance. No I didn't miss them, I just contend -- like in the case of popular Class A single-ended triode no-feedback designs -- that the output impedance is too high to effectively shunt the two sections of the network. Even a two or one ohm output impedance is too high IMO. Regarding the output current, I was simply trying to define the ideal amplifier that would be necessary to create a realizable difference via bi-wiring. If the extremely low ouput-impedance amp couldn't source a lot of current, then one side of the bi-wiring could affect the other in some cases -- like in the case of a short circuit or very low driver impedance on one side. |
#258
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message vers.com... afh3 wrote: 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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? |
#259
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message vers.com... afh3 wrote: 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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? |
#260
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message vers.com... afh3 wrote: 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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? |
#261
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message vers.com... afh3 wrote: 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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? |
#262
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Bi-wiring - Hogwash?
afh3 wrote:
Sorry, typo. Meant to say, "22 here." "chung" wrote in message vers.com... afh3 wrote: "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. Sorry, typo. Meant to say, "22 here." Pretty big difference, wouldn't you say? OMG, now we know you can't subtract. No wonder . 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. You seem to miss two points: (1) I said "purely academic", and theoretical. I never said that the effects of bi-wiring are necessarily audible. (2) It's not the max. output current capacity that is important. It is the equivalent output impedance of the amp that is important. That is, the small signal impedance. An amp can have a current limit that does not approach thousands of amps, but an output resistance that approaches zero. It is that small signal output impedance you should concern yourself with, when you try to model these effects. When the amp's output impedance is small, Z1 does not see Z2, in your example, because of the shunting effect of the output impedance. No I didn't miss them, I just contend -- like in the case of popular Class A single-ended triode no-feedback designs -- that the output impedance is too high to effectively shunt the two sections of the network. Uhh, the SET is not a popular amp by any stretch of imagination. Try a solid state amp. Even a two or one ohm output impedance is too high IMO. Of course. A one or two ohm output impedance is too high for most speakers. Good designs have output impedance in the order of 0.1 ohm, or less. Regarding the output current, I was simply trying to define the ideal amplifier that would be necessary to create a realizable difference via bi-wiring. If the extremely low ouput-impedance amp couldn't source a lot of current, You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. then one side of the bi-wiring could affect the other in some cases -- like in the case of a short circuit or very low driver impedance on one side. |
#263
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Bi-wiring - Hogwash?
afh3 wrote:
Sorry, typo. Meant to say, "22 here." "chung" wrote in message vers.com... afh3 wrote: "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. Sorry, typo. Meant to say, "22 here." Pretty big difference, wouldn't you say? OMG, now we know you can't subtract. No wonder . 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. You seem to miss two points: (1) I said "purely academic", and theoretical. I never said that the effects of bi-wiring are necessarily audible. (2) It's not the max. output current capacity that is important. It is the equivalent output impedance of the amp that is important. That is, the small signal impedance. An amp can have a current limit that does not approach thousands of amps, but an output resistance that approaches zero. It is that small signal output impedance you should concern yourself with, when you try to model these effects. When the amp's output impedance is small, Z1 does not see Z2, in your example, because of the shunting effect of the output impedance. No I didn't miss them, I just contend -- like in the case of popular Class A single-ended triode no-feedback designs -- that the output impedance is too high to effectively shunt the two sections of the network. Uhh, the SET is not a popular amp by any stretch of imagination. Try a solid state amp. Even a two or one ohm output impedance is too high IMO. Of course. A one or two ohm output impedance is too high for most speakers. Good designs have output impedance in the order of 0.1 ohm, or less. Regarding the output current, I was simply trying to define the ideal amplifier that would be necessary to create a realizable difference via bi-wiring. If the extremely low ouput-impedance amp couldn't source a lot of current, You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. then one side of the bi-wiring could affect the other in some cases -- like in the case of a short circuit or very low driver impedance on one side. |
#264
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Bi-wiring - Hogwash?
afh3 wrote:
Sorry, typo. Meant to say, "22 here." "chung" wrote in message vers.com... afh3 wrote: "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. Sorry, typo. Meant to say, "22 here." Pretty big difference, wouldn't you say? OMG, now we know you can't subtract. No wonder . 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. You seem to miss two points: (1) I said "purely academic", and theoretical. I never said that the effects of bi-wiring are necessarily audible. (2) It's not the max. output current capacity that is important. It is the equivalent output impedance of the amp that is important. That is, the small signal impedance. An amp can have a current limit that does not approach thousands of amps, but an output resistance that approaches zero. It is that small signal output impedance you should concern yourself with, when you try to model these effects. When the amp's output impedance is small, Z1 does not see Z2, in your example, because of the shunting effect of the output impedance. No I didn't miss them, I just contend -- like in the case of popular Class A single-ended triode no-feedback designs -- that the output impedance is too high to effectively shunt the two sections of the network. Uhh, the SET is not a popular amp by any stretch of imagination. Try a solid state amp. Even a two or one ohm output impedance is too high IMO. Of course. A one or two ohm output impedance is too high for most speakers. Good designs have output impedance in the order of 0.1 ohm, or less. Regarding the output current, I was simply trying to define the ideal amplifier that would be necessary to create a realizable difference via bi-wiring. If the extremely low ouput-impedance amp couldn't source a lot of current, You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. then one side of the bi-wiring could affect the other in some cases -- like in the case of a short circuit or very low driver impedance on one side. |
#265
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Bi-wiring - Hogwash?
afh3 wrote:
Sorry, typo. Meant to say, "22 here." "chung" wrote in message vers.com... afh3 wrote: "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. Sorry, typo. Meant to say, "22 here." Pretty big difference, wouldn't you say? OMG, now we know you can't subtract. No wonder . 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. You seem to miss two points: (1) I said "purely academic", and theoretical. I never said that the effects of bi-wiring are necessarily audible. (2) It's not the max. output current capacity that is important. It is the equivalent output impedance of the amp that is important. That is, the small signal impedance. An amp can have a current limit that does not approach thousands of amps, but an output resistance that approaches zero. It is that small signal output impedance you should concern yourself with, when you try to model these effects. When the amp's output impedance is small, Z1 does not see Z2, in your example, because of the shunting effect of the output impedance. No I didn't miss them, I just contend -- like in the case of popular Class A single-ended triode no-feedback designs -- that the output impedance is too high to effectively shunt the two sections of the network. Uhh, the SET is not a popular amp by any stretch of imagination. Try a solid state amp. Even a two or one ohm output impedance is too high IMO. Of course. A one or two ohm output impedance is too high for most speakers. Good designs have output impedance in the order of 0.1 ohm, or less. Regarding the output current, I was simply trying to define the ideal amplifier that would be necessary to create a realizable difference via bi-wiring. If the extremely low ouput-impedance amp couldn't source a lot of current, You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. then one side of the bi-wiring could affect the other in some cases -- like in the case of a short circuit or very low driver impedance on one side. |
#266
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Bi-wiring - Hogwash?
"chung" wrote in message vers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: 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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? Yes, simple enough - and the current will be half, and the voltage will still equal the current times the resistance. I think (hope) we're saying the same thing. |
#267
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Bi-wiring - Hogwash?
"chung" wrote in message vers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: 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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? Yes, simple enough - and the current will be half, and the voltage will still equal the current times the resistance. I think (hope) we're saying the same thing. |
#268
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Bi-wiring - Hogwash?
"chung" wrote in message vers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: 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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? Yes, simple enough - and the current will be half, and the voltage will still equal the current times the resistance. I think (hope) we're saying the same thing. |
#269
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Bi-wiring - Hogwash?
"chung" wrote in message vers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: 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. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? Yes, simple enough - and the current will be half, and the voltage will still equal the current times the resistance. I think (hope) we're saying the same thing. |
#270
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Bi-wiring - Hogwash?
"chung" wrote in message vers.com... You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. Yes. But without the ability to source a lot of current, I'm sure you can see why we don't use 741's for output stages of amplifiers. I neither stated nor implied that one required the other -- simply that a good power amplifier design would have both. Regarding the SET amp not being popular, try rec.audio.tubes with that line... |
#271
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Bi-wiring - Hogwash?
"chung" wrote in message vers.com... You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. Yes. But without the ability to source a lot of current, I'm sure you can see why we don't use 741's for output stages of amplifiers. I neither stated nor implied that one required the other -- simply that a good power amplifier design would have both. Regarding the SET amp not being popular, try rec.audio.tubes with that line... |
#272
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Bi-wiring - Hogwash?
"chung" wrote in message vers.com... You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. Yes. But without the ability to source a lot of current, I'm sure you can see why we don't use 741's for output stages of amplifiers. I neither stated nor implied that one required the other -- simply that a good power amplifier design would have both. Regarding the SET amp not being popular, try rec.audio.tubes with that line... |
#273
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Bi-wiring - Hogwash?
"chung" wrote in message vers.com... You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. Yes. But without the ability to source a lot of current, I'm sure you can see why we don't use 741's for output stages of amplifiers. I neither stated nor implied that one required the other -- simply that a good power amplifier design would have both. Regarding the SET amp not being popular, try rec.audio.tubes with that line... |
#274
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message vers.com... You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. Yes. But without the ability to source a lot of current, I'm sure you can see why we don't use 741's for output stages of amplifiers. I neither stated nor implied that one required the other -- simply that a good power amplifier design would have both. Regarding the SET amp not being popular, try rec.audio.tubes with that line... And I'm sure somewhere there is a forum or newsgroup where the flat-earth theory is very popular... |
#275
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message vers.com... You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. Yes. But without the ability to source a lot of current, I'm sure you can see why we don't use 741's for output stages of amplifiers. I neither stated nor implied that one required the other -- simply that a good power amplifier design would have both. Regarding the SET amp not being popular, try rec.audio.tubes with that line... And I'm sure somewhere there is a forum or newsgroup where the flat-earth theory is very popular... |
#276
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message vers.com... You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. Yes. But without the ability to source a lot of current, I'm sure you can see why we don't use 741's for output stages of amplifiers. I neither stated nor implied that one required the other -- simply that a good power amplifier design would have both. Regarding the SET amp not being popular, try rec.audio.tubes with that line... And I'm sure somewhere there is a forum or newsgroup where the flat-earth theory is very popular... |
#277
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message vers.com... You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. Yes. But without the ability to source a lot of current, I'm sure you can see why we don't use 741's for output stages of amplifiers. I neither stated nor implied that one required the other -- simply that a good power amplifier design would have both. Regarding the SET amp not being popular, try rec.audio.tubes with that line... And I'm sure somewhere there is a forum or newsgroup where the flat-earth theory is very popular... |
#278
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Bi-wiring - Hogwash?
"chung" wrote in message
vers.com... afh3 wrote: "chung" wrote in message vers.com... You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. Yes. But without the ability to source a lot of current, I'm sure you can see why we don't use 741's for output stages of amplifiers. I neither stated nor implied that one required the other -- simply that a good power amplifier design would have both. Regarding the SET amp not being popular, try rec.audio.tubes with that line... And I'm sure somewhere there is a forum or newsgroup where the flat-earth theory is very popular... I'm no tube bigot, but there are certainly some very fine Class A SET amps out there that offer pretty great performance. There are those who will have nothing else (negative feedback being completely evil and all) because of the softer clipping and numerous other reasons. I still prefer solid state, but even among those you have to spend quite a bit of money to get output impedance values below or approaching .1 ohms. While Krells, Meridians, CJs and other quality solid-state amps can handily beat even that value, the ability for this to provide audible "isolation" in a biwired configuration is questionable at best. So, I call biwiring "Hogwash". I don't do it. Do you? -afh3 |
#279
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Bi-wiring - Hogwash?
"chung" wrote in message
vers.com... afh3 wrote: "chung" wrote in message vers.com... You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. Yes. But without the ability to source a lot of current, I'm sure you can see why we don't use 741's for output stages of amplifiers. I neither stated nor implied that one required the other -- simply that a good power amplifier design would have both. Regarding the SET amp not being popular, try rec.audio.tubes with that line... And I'm sure somewhere there is a forum or newsgroup where the flat-earth theory is very popular... I'm no tube bigot, but there are certainly some very fine Class A SET amps out there that offer pretty great performance. There are those who will have nothing else (negative feedback being completely evil and all) because of the softer clipping and numerous other reasons. I still prefer solid state, but even among those you have to spend quite a bit of money to get output impedance values below or approaching .1 ohms. While Krells, Meridians, CJs and other quality solid-state amps can handily beat even that value, the ability for this to provide audible "isolation" in a biwired configuration is questionable at best. So, I call biwiring "Hogwash". I don't do it. Do you? -afh3 |
#280
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Bi-wiring - Hogwash?
"chung" wrote in message
vers.com... afh3 wrote: "chung" wrote in message vers.com... You can have an extremely low impedance, and yet not be able to source a lot of current. An op amp is a classical example. Yes. But without the ability to source a lot of current, I'm sure you can see why we don't use 741's for output stages of amplifiers. I neither stated nor implied that one required the other -- simply that a good power amplifier design would have both. Regarding the SET amp not being popular, try rec.audio.tubes with that line... And I'm sure somewhere there is a forum or newsgroup where the flat-earth theory is very popular... I'm no tube bigot, but there are certainly some very fine Class A SET amps out there that offer pretty great performance. There are those who will have nothing else (negative feedback being completely evil and all) because of the softer clipping and numerous other reasons. I still prefer solid state, but even among those you have to spend quite a bit of money to get output impedance values below or approaching .1 ohms. While Krells, Meridians, CJs and other quality solid-state amps can handily beat even that value, the ability for this to provide audible "isolation" in a biwired configuration is questionable at best. So, I call biwiring "Hogwash". I don't do it. Do you? -afh3 |
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