(Ben) wrote in message om...

Sorry, it was late, I typed it wrong, and messed up the +-'s. I wrote:

And what you need for bridging is:

(AMP CHANNEL 1+) --- PAIR 1+ ---- (SPEKAER +)
(AMP CHANNEL 1-) --- PAIR 2+ (not used)
(AMP CHANNEL 2+) --- PAIR 1- ---- (SPEAKER -)
(AMP CHANNEL 2-) --- PAIR 2- (not used)


Which is wrong.
The correct wiring is:


(AMP CHANNEL 1+) --- PAIR 1+ ---- (SPEKAER +)
(AMP CHANNEL 1-) --- PAIR 2- (not used)
(AMP CHANNEL 2+) --- PAIR 1+ ---- (SPEAKER -)
(AMP CHANNEL 2-) --- PAIR 2- (not used)

-Ben

Thanks, I wasn't aware that Speakon connectors have more than one channel. I
have read the manual top to bottom (repeatedly) and must have missed that(I
re-read it after your post, its on page 9). The cables were provided with
the speakers, and go from Speakon to 1/4in.

Amp 1 (SpeakonCh1)--------(1/4) Peavey L
Amp 1 (SpeakonCh2)--------(1/4) Peavey R
Amp 2 (SpeakonCh1)--------(1/4) Subwoofer

All of the cables are a direct link, no splits anywhere. My guess is that
the wire is soldered inside the Speakon to +1 and -1, with no wires in the
cable for +2 and -2. I'll have to check tomorow, and will probably strip off
the Speakon and use the binding posts.

The 30hz filter is on for both channels.


Thanks for the help, this is the first time I've ever dealt with this type
of installation.




"Ben" wrote in message
m...
sk8rteck-

Ok you do not have the speaker connected to the amp correctly. That
is why bridging mode isn't working. Here is what I am talking about
with the speakon connector: Each speakon connector has room for FOUR
contacts in it. Now, in a normal single channel amp output, there are
only two contacts used, and the other two contacts do nothing.
Sometimes the metal pins for the other two contacts are not even there
in the connector. But sometimes they are used. For example, many pro
full-range cabinets are set up for bi-amping.. that is, the woofer in
the cabinet is driven from a different amp channel than mid (if there
is one) and tweeter). Then of course your sub cabinets get their own
amp which makes a tri-amped system. But it is a pain to run two
cables to every cabinet. So what they do is use a four conductor
cable with all 4 contacts used on the connector. Wire pair #1 carries
amp channel #1, wire pair #2 carries channel #2. So in a setup like
that, each cabinet would get it's own amp, but not bridged, just two
independent amp channels bundled through one cable.

Makes sense?

Ok so that is how the Behringer Channel #1 speaker jack is wired.
BOTH channels are available on that one connector. On the channel #2
connector, just channel #2 is available.

So how does this matter for bridging? Well, each channel output has a
(+) wire and a (-) wire. When you are bridging, you don't use the (-)
outputs at all, and the speaker is connected between the (+) of
channel #1 and the (+) of channel #2.. that is how it uses BOTH
channels of the amp.

So that is the deal. Your amp speaker jack #1, connector, and speaker
is
wired like this:

(AMP CHANNEL 1+) --- PAIR 1+ ---- (SPEKAER +)
(AMP CHANNEL 1-) --- PAIR 1- ---- (SPEAKER -)
(AMP CHANNEL 2+) --- PAIR 2+ (not used)
(AMP CHANNEL 2-) --- PAIR 2- (not used)

And what you need for bridging is:

(AMP CHANNEL 1+) --- PAIR 1+ ---- (SPEKAER +)
(AMP CHANNEL 1-) --- PAIR 2+ (not used)
(AMP CHANNEL 2+) --- PAIR 1- ---- (SPEAKER -)
(AMP CHANNEL 2-) --- PAIR 2- (not used)

That is why you need different connector wiring. If you don't want to
rewire the cable, you can get a speakon jack and solder two wires to
the right pins on that, put some tape on it.. then connect to the two
red binding posts on the amp. That will do the same thing. Use
decent wire for that...

The other thing you have set up wrong is that you should have both the
filter and the limiter on channel 2 turned OFF for bridge mode. Turn
the limiter on for channel 1 only. And the 30hz filter, probably
should be on for channel 1 also (I doubt you will get much bass down
there anyway).

So that's the deal. Hope it makes sense. BTW although I already know
how bridging works on other amps, ALL the info I just gave you about
this amp, the connector arrangements on it and the switch settings..
ALL that came from the PDF manual I got from looking on the Behringer
web site. You should read that manual.

Good luck.

-Ben




-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Thanks, I wasn't aware that Speakon connectors have more than one channel. I
have read the manual top to bottom (repeatedly) and must have missed that(I
re-read it after your post, its on page 9). The cables were provided with
the speakers, and go from Speakon to 1/4in.

Amp 1 (SpeakonCh1)--------(1/4) Peavey L
Amp 1 (SpeakonCh2)--------(1/4) Peavey R
Amp 2 (SpeakonCh1)--------(1/4) Subwoofer

All of the cables are a direct link, no splits anywhere. My guess is that
the wire is soldered inside the Speakon to +1 and -1, with no wires in the
cable for +2 and -2. I'll have to check tomorow, and will probably strip off
the Speakon and use the binding posts.

The 30hz filter is on for both channels.


Thanks for the help, this is the first time I've ever dealt with this type
of installation.




"Ben" wrote in message
m...
sk8rteck-

Ok you do not have the speaker connected to the amp correctly. That
is why bridging mode isn't working. Here is what I am talking about
with the speakon connector: Each speakon connector has room for FOUR
contacts in it. Now, in a normal single channel amp output, there are
only two contacts used, and the other two contacts do nothing.
Sometimes the metal pins for the other two contacts are not even there
in the connector. But sometimes they are used. For example, many pro
full-range cabinets are set up for bi-amping.. that is, the woofer in
the cabinet is driven from a different amp channel than mid (if there
is one) and tweeter). Then of course your sub cabinets get their own
amp which makes a tri-amped system. But it is a pain to run two
cables to every cabinet. So what they do is use a four conductor
cable with all 4 contacts used on the connector. Wire pair #1 carries
amp channel #1, wire pair #2 carries channel #2. So in a setup like
that, each cabinet would get it's own amp, but not bridged, just two
independent amp channels bundled through one cable.

Makes sense?

Ok so that is how the Behringer Channel #1 speaker jack is wired.
BOTH channels are available on that one connector. On the channel #2
connector, just channel #2 is available.

So how does this matter for bridging? Well, each channel output has a
(+) wire and a (-) wire. When you are bridging, you don't use the (-)
outputs at all, and the speaker is connected between the (+) of
channel #1 and the (+) of channel #2.. that is how it uses BOTH
channels of the amp.

So that is the deal. Your amp speaker jack #1, connector, and speaker
is
wired like this:

(AMP CHANNEL 1+) --- PAIR 1+ ---- (SPEKAER +)
(AMP CHANNEL 1-) --- PAIR 1- ---- (SPEAKER -)
(AMP CHANNEL 2+) --- PAIR 2+ (not used)
(AMP CHANNEL 2-) --- PAIR 2- (not used)

And what you need for bridging is:

(AMP CHANNEL 1+) --- PAIR 1+ ---- (SPEKAER +)
(AMP CHANNEL 1-) --- PAIR 2+ (not used)
(AMP CHANNEL 2+) --- PAIR 1- ---- (SPEAKER -)
(AMP CHANNEL 2-) --- PAIR 2- (not used)

That is why you need different connector wiring. If you don't want to
rewire the cable, you can get a speakon jack and solder two wires to
the right pins on that, put some tape on it.. then connect to the two
red binding posts on the amp. That will do the same thing. Use
decent wire for that...

The other thing you have set up wrong is that you should have both the
filter and the limiter on channel 2 turned OFF for bridge mode. Turn
the limiter on for channel 1 only. And the 30hz filter, probably
should be on for channel 1 also (I doubt you will get much bass down
there anyway).

So that's the deal. Hope it makes sense. BTW although I already know
how bridging works on other amps, ALL the info I just gave you about
this amp, the connector arrangements on it and the switch settings..
ALL that came from the PDF manual I got from looking on the Behringer
web site. You should read that manual.

Good luck.

-Ben




-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Thanks, I wasn't aware that Speakon connectors have more than one channel. I
have read the manual top to bottom (repeatedly) and must have missed that(I
re-read it after your post, its on page 9). The cables were provided with
the speakers, and go from Speakon to 1/4in.

Amp 1 (SpeakonCh1)--------(1/4) Peavey L
Amp 1 (SpeakonCh2)--------(1/4) Peavey R
Amp 2 (SpeakonCh1)--------(1/4) Subwoofer

All of the cables are a direct link, no splits anywhere. My guess is that
the wire is soldered inside the Speakon to +1 and -1, with no wires in the
cable for +2 and -2. I'll have to check tomorow, and will probably strip off
the Speakon and use the binding posts.

The 30hz filter is on for both channels.


Thanks for the help, this is the first time I've ever dealt with this type
of installation.




"Ben" wrote in message
m...
sk8rteck-

Ok you do not have the speaker connected to the amp correctly. That
is why bridging mode isn't working. Here is what I am talking about
with the speakon connector: Each speakon connector has room for FOUR
contacts in it. Now, in a normal single channel amp output, there are
only two contacts used, and the other two contacts do nothing.
Sometimes the metal pins for the other two contacts are not even there
in the connector. But sometimes they are used. For example, many pro
full-range cabinets are set up for bi-amping.. that is, the woofer in
the cabinet is driven from a different amp channel than mid (if there
is one) and tweeter). Then of course your sub cabinets get their own
amp which makes a tri-amped system. But it is a pain to run two
cables to every cabinet. So what they do is use a four conductor
cable with all 4 contacts used on the connector. Wire pair #1 carries
amp channel #1, wire pair #2 carries channel #2. So in a setup like
that, each cabinet would get it's own amp, but not bridged, just two
independent amp channels bundled through one cable.

Makes sense?

Ok so that is how the Behringer Channel #1 speaker jack is wired.
BOTH channels are available on that one connector. On the channel #2
connector, just channel #2 is available.

So how does this matter for bridging? Well, each channel output has a
(+) wire and a (-) wire. When you are bridging, you don't use the (-)
outputs at all, and the speaker is connected between the (+) of
channel #1 and the (+) of channel #2.. that is how it uses BOTH
channels of the amp.

So that is the deal. Your amp speaker jack #1, connector, and speaker
is
wired like this:

(AMP CHANNEL 1+) --- PAIR 1+ ---- (SPEKAER +)
(AMP CHANNEL 1-) --- PAIR 1- ---- (SPEAKER -)
(AMP CHANNEL 2+) --- PAIR 2+ (not used)
(AMP CHANNEL 2-) --- PAIR 2- (not used)

And what you need for bridging is:

(AMP CHANNEL 1+) --- PAIR 1+ ---- (SPEKAER +)
(AMP CHANNEL 1-) --- PAIR 2+ (not used)
(AMP CHANNEL 2+) --- PAIR 1- ---- (SPEAKER -)
(AMP CHANNEL 2-) --- PAIR 2- (not used)

That is why you need different connector wiring. If you don't want to
rewire the cable, you can get a speakon jack and solder two wires to
the right pins on that, put some tape on it.. then connect to the two
red binding posts on the amp. That will do the same thing. Use
decent wire for that...

The other thing you have set up wrong is that you should have both the
filter and the limiter on channel 2 turned OFF for bridge mode. Turn
the limiter on for channel 1 only. And the 30hz filter, probably
should be on for channel 1 also (I doubt you will get much bass down
there anyway).

So that's the deal. Hope it makes sense. BTW although I already know
how bridging works on other amps, ALL the info I just gave you about
this amp, the connector arrangements on it and the switch settings..
ALL that came from the PDF manual I got from looking on the Behringer
web site. You should read that manual.

Good luck.

-Ben




-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Thanks, I wasn't aware that Speakon connectors have more than one channel. I
have read the manual top to bottom (repeatedly) and must have missed that(I
re-read it after your post, its on page 9). The cables were provided with
the speakers, and go from Speakon to 1/4in.

Amp 1 (SpeakonCh1)--------(1/4) Peavey L
Amp 1 (SpeakonCh2)--------(1/4) Peavey R
Amp 2 (SpeakonCh1)--------(1/4) Subwoofer

All of the cables are a direct link, no splits anywhere. My guess is that
the wire is soldered inside the Speakon to +1 and -1, with no wires in the
cable for +2 and -2. I'll have to check tomorow, and will probably strip off
the Speakon and use the binding posts.

The 30hz filter is on for both channels.


Thanks for the help, this is the first time I've ever dealt with this type
of installation.




"Ben" wrote in message
m...
sk8rteck-

Ok you do not have the speaker connected to the amp correctly. That
is why bridging mode isn't working. Here is what I am talking about
with the speakon connector: Each speakon connector has room for FOUR
contacts in it. Now, in a normal single channel amp output, there are
only two contacts used, and the other two contacts do nothing.
Sometimes the metal pins for the other two contacts are not even there
in the connector. But sometimes they are used. For example, many pro
full-range cabinets are set up for bi-amping.. that is, the woofer in
the cabinet is driven from a different amp channel than mid (if there
is one) and tweeter). Then of course your sub cabinets get their own
amp which makes a tri-amped system. But it is a pain to run two
cables to every cabinet. So what they do is use a four conductor
cable with all 4 contacts used on the connector. Wire pair #1 carries
amp channel #1, wire pair #2 carries channel #2. So in a setup like
that, each cabinet would get it's own amp, but not bridged, just two
independent amp channels bundled through one cable.

Makes sense?

Ok so that is how the Behringer Channel #1 speaker jack is wired.
BOTH channels are available on that one connector. On the channel #2
connector, just channel #2 is available.

So how does this matter for bridging? Well, each channel output has a
(+) wire and a (-) wire. When you are bridging, you don't use the (-)
outputs at all, and the speaker is connected between the (+) of
channel #1 and the (+) of channel #2.. that is how it uses BOTH
channels of the amp.

So that is the deal. Your amp speaker jack #1, connector, and speaker
is
wired like this:

(AMP CHANNEL 1+) --- PAIR 1+ ---- (SPEKAER +)
(AMP CHANNEL 1-) --- PAIR 1- ---- (SPEAKER -)
(AMP CHANNEL 2+) --- PAIR 2+ (not used)
(AMP CHANNEL 2-) --- PAIR 2- (not used)

And what you need for bridging is:

(AMP CHANNEL 1+) --- PAIR 1+ ---- (SPEKAER +)
(AMP CHANNEL 1-) --- PAIR 2+ (not used)
(AMP CHANNEL 2+) --- PAIR 1- ---- (SPEAKER -)
(AMP CHANNEL 2-) --- PAIR 2- (not used)

That is why you need different connector wiring. If you don't want to
rewire the cable, you can get a speakon jack and solder two wires to
the right pins on that, put some tape on it.. then connect to the two
red binding posts on the amp. That will do the same thing. Use
decent wire for that...

The other thing you have set up wrong is that you should have both the
filter and the limiter on channel 2 turned OFF for bridge mode. Turn
the limiter on for channel 1 only. And the 30hz filter, probably
should be on for channel 1 also (I doubt you will get much bass down
there anyway).

So that's the deal. Hope it makes sense. BTW although I already know
how bridging works on other amps, ALL the info I just gave you about
this amp, the connector arrangements on it and the switch settings..
ALL that came from the PDF manual I got from looking on the Behringer
web site. You should read that manual.

Good luck.

-Ben




-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Svante wrote:

"G M" wrote in message ...
A voice coil is nothing but a coil of wire with almost zero ohms of
resistance. While the voice coil is moving through a magnetic field,
counter electro-motive force generated creates the impedence.

Nope. Typical resistance is about 3 ohms for a 4 ohm speaker and 6
ohms for a 8 ohm speaker. So 75% of its nominal impedance is the DC
resistance. Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in the magnet?

If you clip the signal going into the speaker, you will end up with the
speaker not smoothly moving in and out, but a longer than usual standstill
at the most outward and inward positions. It will then lose impedence,
creating much higher current, and heat generated. If the problem is not
corrected, you will melt the fine wire in the voice coil.

Wrong. It is not the position of the cone that is proportional to the
input voltage (above the system resonance) but the cone
*acceleration*. So the cone will not "stop" because the input voltage
remains fixed. Not that it would matter, though.

I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then stays there.

That is why, as a general rule, you can over-power a lower rated speaker
with a larger clean running amp, than run a clipping amp into a higher rated
speaker.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Svante wrote:

"G M" wrote in message ...
A voice coil is nothing but a coil of wire with almost zero ohms of
resistance. While the voice coil is moving through a magnetic field,
counter electro-motive force generated creates the impedence.

Nope. Typical resistance is about 3 ohms for a 4 ohm speaker and 6
ohms for a 8 ohm speaker. So 75% of its nominal impedance is the DC
resistance. Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in the magnet?

If you clip the signal going into the speaker, you will end up with the
speaker not smoothly moving in and out, but a longer than usual standstill
at the most outward and inward positions. It will then lose impedence,
creating much higher current, and heat generated. If the problem is not
corrected, you will melt the fine wire in the voice coil.

Wrong. It is not the position of the cone that is proportional to the
input voltage (above the system resonance) but the cone
*acceleration*. So the cone will not "stop" because the input voltage
remains fixed. Not that it would matter, though.

I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then stays there.

That is why, as a general rule, you can over-power a lower rated speaker
with a larger clean running amp, than run a clipping amp into a higher rated
speaker.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Svante wrote:

"G M" wrote in message ...
A voice coil is nothing but a coil of wire with almost zero ohms of
resistance. While the voice coil is moving through a magnetic field,
counter electro-motive force generated creates the impedence.

Nope. Typical resistance is about 3 ohms for a 4 ohm speaker and 6
ohms for a 8 ohm speaker. So 75% of its nominal impedance is the DC
resistance. Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in the magnet?

If you clip the signal going into the speaker, you will end up with the
speaker not smoothly moving in and out, but a longer than usual standstill
at the most outward and inward positions. It will then lose impedence,
creating much higher current, and heat generated. If the problem is not
corrected, you will melt the fine wire in the voice coil.

Wrong. It is not the position of the cone that is proportional to the
input voltage (above the system resonance) but the cone
*acceleration*. So the cone will not "stop" because the input voltage
remains fixed. Not that it would matter, though.

I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then stays there.

That is why, as a general rule, you can over-power a lower rated speaker
with a larger clean running amp, than run a clipping amp into a higher rated
speaker.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Svante wrote:

"G M" wrote in message ...
A voice coil is nothing but a coil of wire with almost zero ohms of
resistance. While the voice coil is moving through a magnetic field,
counter electro-motive force generated creates the impedence.

Nope. Typical resistance is about 3 ohms for a 4 ohm speaker and 6
ohms for a 8 ohm speaker. So 75% of its nominal impedance is the DC
resistance. Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in the magnet?

If you clip the signal going into the speaker, you will end up with the
speaker not smoothly moving in and out, but a longer than usual standstill
at the most outward and inward positions. It will then lose impedence,
creating much higher current, and heat generated. If the problem is not
corrected, you will melt the fine wire in the voice coil.

Wrong. It is not the position of the cone that is proportional to the
input voltage (above the system resonance) but the cone
*acceleration*. So the cone will not "stop" because the input voltage
remains fixed. Not that it would matter, though.

I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then stays there.

That is why, as a general rule, you can over-power a lower rated speaker
with a larger clean running amp, than run a clipping amp into a higher rated
speaker.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

wrote in message
...
On Thu, 12 Feb 2004 23:19:30 -0800, gwhite
wrote:



Svante wrote:

snip

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Which is fine so far as it goes, but the OP's quaestion was about
sub-woofers and the URL above is about tweeter burn-out and how Rane's
limiters can save them.

How is that anything to do with the question?

It explains one plausible source for the 'clipping hurts speakers' myth.

wrote in message
...
On Thu, 12 Feb 2004 23:19:30 -0800, gwhite
wrote:



Svante wrote:

snip

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Which is fine so far as it goes, but the OP's quaestion was about
sub-woofers and the URL above is about tweeter burn-out and how Rane's
limiters can save them.

How is that anything to do with the question?

It explains one plausible source for the 'clipping hurts speakers' myth.

wrote in message
...
On Thu, 12 Feb 2004 23:19:30 -0800, gwhite
wrote:



Svante wrote:

snip

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Which is fine so far as it goes, but the OP's quaestion was about
sub-woofers and the URL above is about tweeter burn-out and how Rane's
limiters can save them.

How is that anything to do with the question?

It explains one plausible source for the 'clipping hurts speakers' myth.

wrote in message
...
On Thu, 12 Feb 2004 23:19:30 -0800, gwhite
wrote:



Svante wrote:

snip

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Which is fine so far as it goes, but the OP's quaestion was about
sub-woofers and the URL above is about tweeter burn-out and how Rane's
limiters can save them.

How is that anything to do with the question?

It explains one plausible source for the 'clipping hurts speakers' myth.

gwhite wrote in message ...
Svante wrote:

"G M" wrote in message ...
A voice coil is nothing but a coil of wire with almost zero ohms of
resistance. While the voice coil is moving through a magnetic field,
counter electro-motive force generated creates the impedence.

Nope. Typical resistance is about 3 ohms for a 4 ohm speaker and 6
ohms for a 8 ohm speaker. So 75% of its nominal impedance is the DC
resistance. Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in
the magnet?

That would not heat the coil, except indirectly. This power is not
dissipated in the coil, at least not directly.


If you clip the signal going into the speaker, you will end up with the
speaker not smoothly moving in and out, but a longer than usual standstill
at the most outward and inward positions. It will then lose impedence,
creating much higher current, and heat generated. If the problem is not
corrected, you will melt the fine wire in the voice coil.

Wrong. It is not the position of the cone that is proportional to the
input voltage (above the system resonance) but the cone
*acceleration*. So the cone will not "stop" because the input voltage
remains fixed. Not that it would matter, though.

I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then
stays there.

Yes, that is DC, which is hardly "above system resonance" as I wrote.
The system resonance is usually in the low end of the usable frequency
range, so most signals will appear above the system resonance. DC will
not, obviously.


That is why, as a general rule, you can over-power a lower rated speaker
with a larger clean running amp, than run a clipping amp into a
higher rated
speaker.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Interesting link, but remember that it is produced by someone who
wants to sell a limiter... :-) I actually did similar power
calculations after my post yesterday and was a bit surprised by the
low HF power levels during clipping, but I do not agree with their
explanation to why tweeters burn. I am not sure I have an alternative
though.
Still, the ONLY thing that can melt the voice coil is the power
delivered into the DC resistance.

gwhite wrote in message ...
Svante wrote:

"G M" wrote in message ...
A voice coil is nothing but a coil of wire with almost zero ohms of
resistance. While the voice coil is moving through a magnetic field,
counter electro-motive force generated creates the impedence.

Nope. Typical resistance is about 3 ohms for a 4 ohm speaker and 6
ohms for a 8 ohm speaker. So 75% of its nominal impedance is the DC
resistance. Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in
the magnet?

That would not heat the coil, except indirectly. This power is not
dissipated in the coil, at least not directly.


If you clip the signal going into the speaker, you will end up with the
speaker not smoothly moving in and out, but a longer than usual standstill
at the most outward and inward positions. It will then lose impedence,
creating much higher current, and heat generated. If the problem is not
corrected, you will melt the fine wire in the voice coil.

Wrong. It is not the position of the cone that is proportional to the
input voltage (above the system resonance) but the cone
*acceleration*. So the cone will not "stop" because the input voltage
remains fixed. Not that it would matter, though.

I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then
stays there.

Yes, that is DC, which is hardly "above system resonance" as I wrote.
The system resonance is usually in the low end of the usable frequency
range, so most signals will appear above the system resonance. DC will
not, obviously.


That is why, as a general rule, you can over-power a lower rated speaker
with a larger clean running amp, than run a clipping amp into a
higher rated
speaker.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Interesting link, but remember that it is produced by someone who
wants to sell a limiter... :-) I actually did similar power
calculations after my post yesterday and was a bit surprised by the
low HF power levels during clipping, but I do not agree with their
explanation to why tweeters burn. I am not sure I have an alternative
though.
Still, the ONLY thing that can melt the voice coil is the power
delivered into the DC resistance.

gwhite wrote in message ...
Svante wrote:

"G M" wrote in message ...
A voice coil is nothing but a coil of wire with almost zero ohms of
resistance. While the voice coil is moving through a magnetic field,
counter electro-motive force generated creates the impedence.

Nope. Typical resistance is about 3 ohms for a 4 ohm speaker and 6
ohms for a 8 ohm speaker. So 75% of its nominal impedance is the DC
resistance. Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in
the magnet?

That would not heat the coil, except indirectly. This power is not
dissipated in the coil, at least not directly.


If you clip the signal going into the speaker, you will end up with the
speaker not smoothly moving in and out, but a longer than usual standstill
at the most outward and inward positions. It will then lose impedence,
creating much higher current, and heat generated. If the problem is not
corrected, you will melt the fine wire in the voice coil.

Wrong. It is not the position of the cone that is proportional to the
input voltage (above the system resonance) but the cone
*acceleration*. So the cone will not "stop" because the input voltage
remains fixed. Not that it would matter, though.

I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then
stays there.

Yes, that is DC, which is hardly "above system resonance" as I wrote.
The system resonance is usually in the low end of the usable frequency
range, so most signals will appear above the system resonance. DC will
not, obviously.


That is why, as a general rule, you can over-power a lower rated speaker
with a larger clean running amp, than run a clipping amp into a
higher rated
speaker.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Interesting link, but remember that it is produced by someone who
wants to sell a limiter... :-) I actually did similar power
calculations after my post yesterday and was a bit surprised by the
low HF power levels during clipping, but I do not agree with their
explanation to why tweeters burn. I am not sure I have an alternative
though.
Still, the ONLY thing that can melt the voice coil is the power
delivered into the DC resistance.

gwhite wrote in message ...
Svante wrote:

"G M" wrote in message ...
A voice coil is nothing but a coil of wire with almost zero ohms of
resistance. While the voice coil is moving through a magnetic field,
counter electro-motive force generated creates the impedence.

Nope. Typical resistance is about 3 ohms for a 4 ohm speaker and 6
ohms for a 8 ohm speaker. So 75% of its nominal impedance is the DC
resistance. Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in
the magnet?

That would not heat the coil, except indirectly. This power is not
dissipated in the coil, at least not directly.


If you clip the signal going into the speaker, you will end up with the
speaker not smoothly moving in and out, but a longer than usual standstill
at the most outward and inward positions. It will then lose impedence,
creating much higher current, and heat generated. If the problem is not
corrected, you will melt the fine wire in the voice coil.

Wrong. It is not the position of the cone that is proportional to the
input voltage (above the system resonance) but the cone
*acceleration*. So the cone will not "stop" because the input voltage
remains fixed. Not that it would matter, though.

I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then
stays there.

Yes, that is DC, which is hardly "above system resonance" as I wrote.
The system resonance is usually in the low end of the usable frequency
range, so most signals will appear above the system resonance. DC will
not, obviously.


That is why, as a general rule, you can over-power a lower rated speaker
with a larger clean running amp, than run a clipping amp into a
higher rated
speaker.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Interesting link, but remember that it is produced by someone who
wants to sell a limiter... :-) I actually did similar power
calculations after my post yesterday and was a bit surprised by the
low HF power levels during clipping, but I do not agree with their
explanation to why tweeters burn. I am not sure I have an alternative
though.
Still, the ONLY thing that can melt the voice coil is the power
delivered into the DC resistance.

rols wrote:

On Thu, 12 Feb 2004 23:19:30 -0800, gwhite
wrote:



Svante wrote:

snip

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Which is fine so far as it goes, but the OP's quaestion was about
sub-woofers and the URL above is about tweeter burn-out and how Rane's
limiters can save them.

Who said I was trying to answer the OP's question?

How is that anything to do with the question?

Why does it have to be? Since when does every post in a usenet thread deal
precisely with the original concern?

rols wrote:

On Thu, 12 Feb 2004 23:19:30 -0800, gwhite
wrote:



Svante wrote:

snip

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Which is fine so far as it goes, but the OP's quaestion was about
sub-woofers and the URL above is about tweeter burn-out and how Rane's
limiters can save them.

Who said I was trying to answer the OP's question?

How is that anything to do with the question?

Why does it have to be? Since when does every post in a usenet thread deal
precisely with the original concern?

rols wrote:

On Thu, 12 Feb 2004 23:19:30 -0800, gwhite
wrote:



Svante wrote:

snip

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Which is fine so far as it goes, but the OP's quaestion was about
sub-woofers and the URL above is about tweeter burn-out and how Rane's
limiters can save them.

Who said I was trying to answer the OP's question?

How is that anything to do with the question?

Why does it have to be? Since when does every post in a usenet thread deal
precisely with the original concern?

rols wrote:

On Thu, 12 Feb 2004 23:19:30 -0800, gwhite
wrote:



Svante wrote:

snip

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Which is fine so far as it goes, but the OP's quaestion was about
sub-woofers and the URL above is about tweeter burn-out and how Rane's
limiters can save them.

Who said I was trying to answer the OP's question?

How is that anything to do with the question?

Why does it have to be? Since when does every post in a usenet thread deal
precisely with the original concern?

Svante wrote:

gwhite wrote in message ...
Svante wrote:

Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in
the magnet?

That would not heat the coil, except indirectly. This power is not
dissipated in the coil, at least not directly.

I agree, I was only pointing out that heat (essentially loss) is produced by
other mechanisms too. You said "only."


I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then
stays there.

Yes, that is DC, which is hardly "above system resonance" as I wrote.
The system resonance is usually in the low end of the usable frequency
range, so most signals will appear above the system resonance. DC will
not, obviously.

It doesn't matter if it is below resonance. At any given frequency, the peak
excursion will be proportional to the voltage across the terminals, mass
(inertia storage) and elasticity (the other storage) included.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Interesting link, but remember that it is produced by someone who
wants to sell a limiter... :-) I actually did similar power
calculations after my post yesterday and was a bit surprised by the
low HF power levels during clipping, but I do not agree with their
explanation to why tweeters burn. I am not sure I have an alternative
though.

Regardless if they want to sell a limiter, the fact is they are quite likely
right about the tweeter harmonic burnout myth. Just like you say below, it is
*power* that will wreck a driver, and if it isn't from harmonic energy, it then
has to be from something else. They burn out because the volume is turned up.


Still, the ONLY thing that can melt the voice coil is the power
delivered into the DC resistance.

Agreed.

Svante wrote:

gwhite wrote in message ...
Svante wrote:

Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in
the magnet?

That would not heat the coil, except indirectly. This power is not
dissipated in the coil, at least not directly.

I agree, I was only pointing out that heat (essentially loss) is produced by
other mechanisms too. You said "only."


I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then
stays there.

Yes, that is DC, which is hardly "above system resonance" as I wrote.
The system resonance is usually in the low end of the usable frequency
range, so most signals will appear above the system resonance. DC will
not, obviously.

It doesn't matter if it is below resonance. At any given frequency, the peak
excursion will be proportional to the voltage across the terminals, mass
(inertia storage) and elasticity (the other storage) included.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Interesting link, but remember that it is produced by someone who
wants to sell a limiter... :-) I actually did similar power
calculations after my post yesterday and was a bit surprised by the
low HF power levels during clipping, but I do not agree with their
explanation to why tweeters burn. I am not sure I have an alternative
though.

Regardless if they want to sell a limiter, the fact is they are quite likely
right about the tweeter harmonic burnout myth. Just like you say below, it is
*power* that will wreck a driver, and if it isn't from harmonic energy, it then
has to be from something else. They burn out because the volume is turned up.


Still, the ONLY thing that can melt the voice coil is the power
delivered into the DC resistance.

Agreed.

Svante wrote:

gwhite wrote in message ...
Svante wrote:

Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in
the magnet?

That would not heat the coil, except indirectly. This power is not
dissipated in the coil, at least not directly.

I agree, I was only pointing out that heat (essentially loss) is produced by
other mechanisms too. You said "only."


I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then
stays there.

Yes, that is DC, which is hardly "above system resonance" as I wrote.
The system resonance is usually in the low end of the usable frequency
range, so most signals will appear above the system resonance. DC will
not, obviously.

It doesn't matter if it is below resonance. At any given frequency, the peak
excursion will be proportional to the voltage across the terminals, mass
(inertia storage) and elasticity (the other storage) included.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Interesting link, but remember that it is produced by someone who
wants to sell a limiter... :-) I actually did similar power
calculations after my post yesterday and was a bit surprised by the
low HF power levels during clipping, but I do not agree with their
explanation to why tweeters burn. I am not sure I have an alternative
though.

Regardless if they want to sell a limiter, the fact is they are quite likely
right about the tweeter harmonic burnout myth. Just like you say below, it is
*power* that will wreck a driver, and if it isn't from harmonic energy, it then
has to be from something else. They burn out because the volume is turned up.


Still, the ONLY thing that can melt the voice coil is the power
delivered into the DC resistance.

Agreed.

Svante wrote:

gwhite wrote in message ...
Svante wrote:

Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in
the magnet?

That would not heat the coil, except indirectly. This power is not
dissipated in the coil, at least not directly.

I agree, I was only pointing out that heat (essentially loss) is produced by
other mechanisms too. You said "only."


I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then
stays there.

Yes, that is DC, which is hardly "above system resonance" as I wrote.
The system resonance is usually in the low end of the usable frequency
range, so most signals will appear above the system resonance. DC will
not, obviously.

It doesn't matter if it is below resonance. At any given frequency, the peak
excursion will be proportional to the voltage across the terminals, mass
(inertia storage) and elasticity (the other storage) included.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Interesting link, but remember that it is produced by someone who
wants to sell a limiter... :-) I actually did similar power
calculations after my post yesterday and was a bit surprised by the
low HF power levels during clipping, but I do not agree with their
explanation to why tweeters burn. I am not sure I have an alternative
though.

Regardless if they want to sell a limiter, the fact is they are quite likely
right about the tweeter harmonic burnout myth. Just like you say below, it is
*power* that will wreck a driver, and if it isn't from harmonic energy, it then
has to be from something else. They burn out because the volume is turned up.


Still, the ONLY thing that can melt the voice coil is the power
delivered into the DC resistance.

Agreed.

wrote in message
...
"How is that anything to do with the question?"

Why not answer it then dickhead?

wrote in message
...
"How is that anything to do with the question?"

Why not answer it then dickhead?

wrote in message
...
"How is that anything to do with the question?"

Why not answer it then dickhead?

wrote in message
...
"How is that anything to do with the question?"

Why not answer it then dickhead?

"Ben Down" wrote in message
...

wrote in message
...
"How is that anything to do with the question?"

Why not answer it then dickhead?



Why complain about someone's post if you are just posting a complaint?

"Ben Down" wrote in message
...

wrote in message
...
"How is that anything to do with the question?"

Why not answer it then dickhead?



Why complain about someone's post if you are just posting a complaint?

"Ben Down" wrote in message
...

wrote in message
...
"How is that anything to do with the question?"

Why not answer it then dickhead?



Why complain about someone's post if you are just posting a complaint?

"Ben Down" wrote in message
...

wrote in message
...
"How is that anything to do with the question?"

Why not answer it then dickhead?



Why complain about someone's post if you are just posting a complaint?

gwhite wrote in message ...
Svante wrote:

gwhite wrote in message ...
Svante wrote:

Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in
the magnet?

That would not heat the coil, except indirectly. This power is not
dissipated in the coil, at least not directly.

I agree, I was only pointing out that heat (essentially loss) is produced by
other mechanisms too. You said "only."

Yes, you are right, I was thinking "in the coil", and "mainly" but
wrote "only. Sorry.

I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then
stays there.

Yes, that is DC, which is hardly "above system resonance" as I wrote.
The system resonance is usually in the low end of the usable frequency
range, so most signals will appear above the system resonance. DC will
not, obviously.

It doesn't matter if it is below resonance. At any given frequency, the peak
excursion will be proportional to the voltage across the terminals, mass
(inertia storage) and elasticity (the other storage) included.

Mmm... But the issue here was clipping, and the poster had a
time-domain approach. I got the impression that he thought that when
voltage is clipped, the cone suddenly stops at a certain position. It
doesn't.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Interesting link, but remember that it is produced by someone who
wants to sell a limiter... :-) I actually did similar power
calculations after my post yesterday and was a bit surprised by the
low HF power levels during clipping, but I do not agree with their
explanation to why tweeters burn. I am not sure I have an alternative
though.

Regardless if they want to sell a limiter, the fact is they are quite likely
right about the tweeter harmonic burnout myth. Just like you say below, it is
*power* that will wreck a driver, and if it isn't from harmonic energy, it then
has to be from something else. They burn out because the volume is turned up.

So then a limiter would lead to compression, and less audible
distorsion, and a *higher* risk of tweeter burnout... :-)

The swedish radio had a problem a few years ago. The midranges in
their Yamaha monitor speakers tended to burn. Finally they found the
cause: It was pretty common that a 1000Hz tone was played in the
loudspeakers for long times when nobody was in the room to turn it
down. This resulted in concentration of energy to the midrange
speaker, and meltdown.

Still, the ONLY thing that can melt the voice coil is the power
delivered into the DC resistance.

Agreed.

gwhite wrote in message ...
Svante wrote:

gwhite wrote in message ...
Svante wrote:

Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in
the magnet?

That would not heat the coil, except indirectly. This power is not
dissipated in the coil, at least not directly.

I agree, I was only pointing out that heat (essentially loss) is produced by
other mechanisms too. You said "only."

Yes, you are right, I was thinking "in the coil", and "mainly" but
wrote "only. Sorry.

I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then
stays there.

Yes, that is DC, which is hardly "above system resonance" as I wrote.
The system resonance is usually in the low end of the usable frequency
range, so most signals will appear above the system resonance. DC will
not, obviously.

It doesn't matter if it is below resonance. At any given frequency, the peak
excursion will be proportional to the voltage across the terminals, mass
(inertia storage) and elasticity (the other storage) included.

Mmm... But the issue here was clipping, and the poster had a
time-domain approach. I got the impression that he thought that when
voltage is clipped, the cone suddenly stops at a certain position. It
doesn't.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Interesting link, but remember that it is produced by someone who
wants to sell a limiter... :-) I actually did similar power
calculations after my post yesterday and was a bit surprised by the
low HF power levels during clipping, but I do not agree with their
explanation to why tweeters burn. I am not sure I have an alternative
though.

Regardless if they want to sell a limiter, the fact is they are quite likely
right about the tweeter harmonic burnout myth. Just like you say below, it is
*power* that will wreck a driver, and if it isn't from harmonic energy, it then
has to be from something else. They burn out because the volume is turned up.

So then a limiter would lead to compression, and less audible
distorsion, and a *higher* risk of tweeter burnout... :-)

The swedish radio had a problem a few years ago. The midranges in
their Yamaha monitor speakers tended to burn. Finally they found the
cause: It was pretty common that a 1000Hz tone was played in the
loudspeakers for long times when nobody was in the room to turn it
down. This resulted in concentration of energy to the midrange
speaker, and meltdown.

Still, the ONLY thing that can melt the voice coil is the power
delivered into the DC resistance.

Agreed.

gwhite wrote in message ...
Svante wrote:

gwhite wrote in message ...
Svante wrote:

Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in
the magnet?

That would not heat the coil, except indirectly. This power is not
dissipated in the coil, at least not directly.

I agree, I was only pointing out that heat (essentially loss) is produced by
other mechanisms too. You said "only."

Yes, you are right, I was thinking "in the coil", and "mainly" but
wrote "only. Sorry.

I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then
stays there.

Yes, that is DC, which is hardly "above system resonance" as I wrote.
The system resonance is usually in the low end of the usable frequency
range, so most signals will appear above the system resonance. DC will
not, obviously.

It doesn't matter if it is below resonance. At any given frequency, the peak
excursion will be proportional to the voltage across the terminals, mass
(inertia storage) and elasticity (the other storage) included.

Mmm... But the issue here was clipping, and the poster had a
time-domain approach. I got the impression that he thought that when
voltage is clipped, the cone suddenly stops at a certain position. It
doesn't.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Interesting link, but remember that it is produced by someone who
wants to sell a limiter... :-) I actually did similar power
calculations after my post yesterday and was a bit surprised by the
low HF power levels during clipping, but I do not agree with their
explanation to why tweeters burn. I am not sure I have an alternative
though.

Regardless if they want to sell a limiter, the fact is they are quite likely
right about the tweeter harmonic burnout myth. Just like you say below, it is
*power* that will wreck a driver, and if it isn't from harmonic energy, it then
has to be from something else. They burn out because the volume is turned up.

So then a limiter would lead to compression, and less audible
distorsion, and a *higher* risk of tweeter burnout... :-)

The swedish radio had a problem a few years ago. The midranges in
their Yamaha monitor speakers tended to burn. Finally they found the
cause: It was pretty common that a 1000Hz tone was played in the
loudspeakers for long times when nobody was in the room to turn it
down. This resulted in concentration of energy to the midrange
speaker, and meltdown.

Still, the ONLY thing that can melt the voice coil is the power
delivered into the DC resistance.

Agreed.

gwhite wrote in message ...
Svante wrote:

gwhite wrote in message ...
Svante wrote:

Furthermore, the only part of the power delivered to the
speaker that turns into heat, is that dissipated in this DC
resistance. If the coil had a zero ohm resistance you could not melt
it (P=I*I*R).

Why wouldn't there be eddy currents (or other loss mechanisms) in
the magnet?

That would not heat the coil, except indirectly. This power is not
dissipated in the coil, at least not directly.

I agree, I was only pointing out that heat (essentially loss) is produced by
other mechanisms too. You said "only."

Yes, you are right, I was thinking "in the coil", and "mainly" but
wrote "only. Sorry.

I can put a DC battery of a fixed voltage on my speaker terminals and I can
guarantee you that it moves to the same position everytime and then
stays there.

Yes, that is DC, which is hardly "above system resonance" as I wrote.
The system resonance is usually in the low end of the usable frequency
range, so most signals will appear above the system resonance. DC will
not, obviously.

It doesn't matter if it is below resonance. At any given frequency, the peak
excursion will be proportional to the voltage across the terminals, mass
(inertia storage) and elasticity (the other storage) included.

Mmm... But the issue here was clipping, and the poster had a
time-domain approach. I got the impression that he thought that when
voltage is clipped, the cone suddenly stops at a certain position. It
doesn't.

Nope. The problem with clipping amplifiers is mainly that the spectral
content of the signal is shifted towards higher frequencies as the
signal is clipped, and also that the dynamics of the signal is lost.
This will lead to a constantly high high frequency content of the
signal. This is fed to the tweeter, and the tweeter burns. In a clean
signal, the high frequency content is much lower, so the power
delivered to the tweeter is less, even if the total power is higher.
In a 1-way system, the ONLY thing that can melt the voice coil is the
power delivered into the DC resistance.

Myth and lore.

http://www.rane.com/pdf/note128.pdf

Interesting link, but remember that it is produced by someone who
wants to sell a limiter... :-) I actually did similar power
calculations after my post yesterday and was a bit surprised by the
low HF power levels during clipping, but I do not agree with their
explanation to why tweeters burn. I am not sure I have an alternative
though.

Regardless if they want to sell a limiter, the fact is they are quite likely
right about the tweeter harmonic burnout myth. Just like you say below, it is
*power* that will wreck a driver, and if it isn't from harmonic energy, it then
has to be from something else. They burn out because the volume is turned up.

So then a limiter would lead to compression, and less audible
distorsion, and a *higher* risk of tweeter burnout... :-)

The swedish radio had a problem a few years ago. The midranges in
their Yamaha monitor speakers tended to burn. Finally they found the
cause: It was pretty common that a 1000Hz tone was played in the
loudspeakers for long times when nobody was in the room to turn it
down. This resulted in concentration of energy to the midrange
speaker, and meltdown.

Still, the ONLY thing that can melt the voice coil is the power
delivered into the DC resistance.

Agreed.

" I can put a DC battery of a fixed voltage on my speaker terminals and I
can
guarantee you that it moves to the same position everytime and then stays
there."
Well YOU try it ....because you are wrong....

Three laws of electromagnetic induction, and two laws of electrolysis, all
proposed originally by English scientist Michael Faraday induction (1) a
changing magnetic field induces an electromagnetic force in a conductor (2)
the electromagnetic force is proportional to the rate of change of the field
(3) the direction of the induced electromagnetic force depends on the
orientation of the field. (Hutchinson Encyclopaedia)

It is the CHANGE in current that produces magnetic flux. Your battery will
induce current only at the moments of start up and stop.

" I can put a DC battery of a fixed voltage on my speaker terminals and I
can
guarantee you that it moves to the same position everytime and then stays
there."
Well YOU try it ....because you are wrong....

Three laws of electromagnetic induction, and two laws of electrolysis, all
proposed originally by English scientist Michael Faraday induction (1) a
changing magnetic field induces an electromagnetic force in a conductor (2)
the electromagnetic force is proportional to the rate of change of the field
(3) the direction of the induced electromagnetic force depends on the
orientation of the field. (Hutchinson Encyclopaedia)

It is the CHANGE in current that produces magnetic flux. Your battery will
induce current only at the moments of start up and stop.

" I can put a DC battery of a fixed voltage on my speaker terminals and I
can
guarantee you that it moves to the same position everytime and then stays
there."
Well YOU try it ....because you are wrong....

Three laws of electromagnetic induction, and two laws of electrolysis, all
proposed originally by English scientist Michael Faraday induction (1) a
changing magnetic field induces an electromagnetic force in a conductor (2)
the electromagnetic force is proportional to the rate of change of the field
(3) the direction of the induced electromagnetic force depends on the
orientation of the field. (Hutchinson Encyclopaedia)

It is the CHANGE in current that produces magnetic flux. Your battery will
induce current only at the moments of start up and stop.