"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

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".

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".

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".

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".

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.

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.

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.

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.

"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

"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

"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

"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

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.

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.

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.

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.

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?

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?

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?

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?

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.

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.

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.

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.

"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.

"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.

"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.

"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.

"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...

"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...

"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...

"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...

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...

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...

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...

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...

"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

"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

"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