Hi,

as a hi-fi owner with some background knowledge of physics but no idea of
speaker or crossover design, I wonder if someone could help me out with this
simple question.

in a typical 2 way speaker, the bass driver dissipates many times as much
power as the tweeter. however, the nominal impedance of the two drivers is
often similar. if the voltage across the two is the same and the impedance
is the same, how does the bass driver draw so much more power than the
tweeter?

I realise this is a gross simplification of the actual design of a speaker.
Is it the crossover that increases the impedance as seen by the amplifier?
or am I totally missing something

Andy

Andy Fish wrote:
Hi,

as a hi-fi owner with some background knowledge of physics but no idea of
speaker or crossover design, I wonder if someone could help me out with this
simple question.

in a typical 2 way speaker, the bass driver dissipates many times as much
power as the tweeter. however, the nominal impedance of the two drivers is
often similar. if the voltage across the two is the same and the impedance
is the same, how does the bass driver draw so much more power than the
tweeter?

I realise this is a gross simplification of the actual design of a speaker.
Is it the crossover that increases the impedance as seen by the amplifier?
or am I totally missing something

Andy


The tweeter crossover network contains an attenuator.

A

"Andy Dee" wrote in message
...
Andy Fish wrote:
Hi,

as a hi-fi owner with some background knowledge of physics but no idea
of
speaker or crossover design, I wonder if someone could help me out with
this
simple question.

in a typical 2 way speaker, the bass driver dissipates many times as
much
power as the tweeter. however, the nominal impedance of the two drivers
is
often similar. if the voltage across the two is the same and the
impedance
is the same, how does the bass driver draw so much more power than the
tweeter?

I realise this is a gross simplification of the actual design of a
speaker.
Is it the crossover that increases the impedance as seen by the
amplifier?
or am I totally missing something

Andy


The tweeter crossover network contains an attenuator.

A


OK, thanks. This is obviously more difficult than I originally thought.

Here's my actual problem: I have a pair of bi-wireable speakers and I would
like to attenuate the top end by 2-3db. Is it possible to do this without
taking them apart - i.e. by connecting a series resistor and/or l-pad onto
the tweeter's speaker terminals?

On Sun, 05 Oct 2003 11:02:22 GMT, "Andy Fish"
wrote:

"Andy Dee" wrote in message
...
Andy Fish wrote:
Hi,

as a hi-fi owner with some background knowledge of physics but no idea
of
speaker or crossover design, I wonder if someone could help me out with
this
simple question.

in a typical 2 way speaker, the bass driver dissipates many times as
much
power as the tweeter. however, the nominal impedance of the two drivers
is
often similar. if the voltage across the two is the same and the
impedance
is the same, how does the bass driver draw so much more power than the
tweeter?

I realise this is a gross simplification of the actual design of a
speaker.
Is it the crossover that increases the impedance as seen by the
amplifier?
or am I totally missing something

Andy


The tweeter crossover network contains an attenuator.

A


OK, thanks. This is obviously more difficult than I originally thought.

Here's my actual problem: I have a pair of bi-wireable speakers and I would
like to attenuate the top end by 2-3db. Is it possible to do this without
taking them apart - i.e. by connecting a series resistor and/or l-pad onto
the tweeter's speaker terminals?


Put a wirewound pot (reasonably good power handling) in series with
the tweeter, and tweak to taste. Something around 20 ohms should do
it.

d

_____________________________

http://www.pearce.uk.com

Put a wirewound pot (reasonably good power handling) in series with
the tweeter, and tweak to taste. Something around 20 ohms should do
it.

d
Should have added - once you are happy, replace the pot with a power
resistor of the same value.

d

_____________________________

http://www.pearce.uk.com

Andy Fish wrote:
"Andy Dee" wrote in message
...

Andy Fish wrote:

Hi,

as a hi-fi owner with some background knowledge of physics but no idea

of

speaker or crossover design, I wonder if someone could help me out with

this

simple question.

in a typical 2 way speaker, the bass driver dissipates many times as

much

power as the tweeter. however, the nominal impedance of the two drivers

is

often similar. if the voltage across the two is the same and the

impedance

is the same, how does the bass driver draw so much more power than the
tweeter?

I realise this is a gross simplification of the actual design of a

speaker.

Is it the crossover that increases the impedance as seen by the

amplifier?

or am I totally missing something

Andy



The tweeter crossover network contains an attenuator.

A



OK, thanks. This is obviously more difficult than I originally thought.

Here's my actual problem: I have a pair of bi-wireable speakers and I would
like to attenuate the top end by 2-3db. Is it possible to do this without
taking them apart - i.e. by connecting a series resistor and/or l-pad onto
the tweeter's speaker terminals?


Absolutely no problem if the unit is bi-wireable, you can add an L-pad
to the HF terminals.

This may slightly affect the resposne of the HF crossover section as it
will increase the source impeadance seen on the input.

Most terminals are on a plastic panel that can be removed. If your is,
you could solder the extra resistors on the inside after you are happy
with the sound.

A

In article 2wSfb.9398$fR7.78736602@news-
text.cableinet.net, says...
Hi,

as a hi-fi owner with some background knowledge of physics but no idea of
speaker or crossover design, I wonder if someone could help me out with this
simple question.

in a typical 2 way speaker, the bass driver dissipates many times as much
power as the tweeter. however, the nominal impedance of the two drivers is
often similar. if the voltage across the two is the same and the impedance
is the same, how does the bass driver draw so much more power than the
tweeter?

The answer to your question is this.

The bass driver consumes more power simply because there is
more power made available in the lower end of the spectrum
than the upper end. It is _not_ an issue of driver design,
_nor_ does it have Anything to do with the crossover.
There's simply more energy in sound as you go down in the
audible spectrum.

Now, given your follow-up post, it appears that this really
wasn't the question you wanted to ask.

If your objective is to reduce the high end response, and
you're using these speakers in a biamp (bi-wired)
configuration, the most expedient correct way to do it is to
use an L-pad, wired between the amplifier and the tweeter.

You could also design a resister pad of the correct
impedance, but it's certainly more hassle. Simply putting a
resister is series will affect the load impedance, which may
introduce other issues of amp stability and frequency
response.
--
Mark

The truth as I perceive it to be.
Your perception may be different.

Triple Z is spam control.

On Sun, 05 Oct 2003 10:37:50 GMT, "Andy Fish"
wrote:

Hi,

as a hi-fi owner with some background knowledge of physics but no idea of
speaker or crossover design, I wonder if someone could help me out with this
simple question.

in a typical 2 way speaker, the bass driver dissipates many times as much
power as the tweeter. however, the nominal impedance of the two drivers is
often similar. if the voltage across the two is the same and the impedance
is the same, how does the bass driver draw so much more power than the
tweeter?

I realise this is a gross simplification of the actual design of a speaker.
Is it the crossover that increases the impedance as seen by the amplifier?
or am I totally missing something

The crossover splits the input voltage according to frequency. The
power distribution spectrum of most music is such that the vast
majority of the power occurs below 2kHz, hence the bass driver is
required to handle more power. It follows that bass/mid drivers
typically have 5-10 times the power handling capability of tweeters,
because that's pretty much how the power requirement distributes.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Sun, 05 Oct 2003 11:02:22 GMT, "Andy Fish"
wrote:

Here's my actual problem: I have a pair of bi-wireable speakers and I would
like to attenuate the top end by 2-3db.

Why? Do you find that the speakers have excessive treble?

Is it possible to do this without
taking them apart - i.e. by connecting a series resistor and/or l-pad onto
the tweeter's speaker terminals?

Yes. If they're bi-wireable, you just need to add a 3-5 ohm resistor
in series with the treble terminals.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Sun, 05 Oct 2003 12:06:49 +0100, Don Pearce
wrote:

On Sun, 05 Oct 2003 11:02:22 GMT, "Andy Fish"
wrote:

"Andy Dee" wrote in message
...

Here's my actual problem: I have a pair of bi-wireable speakers and I would
like to attenuate the top end by 2-3db. Is it possible to do this without
taking them apart - i.e. by connecting a series resistor and/or l-pad onto
the tweeter's speaker terminals?


Put a wirewound pot (reasonably good power handling) in series with
the tweeter, and tweak to taste. Something around 20 ohms should do
it.

WHAT? A 20 ohm resistor will drop the treble by around 10dB. This is a
*lot* more than the original poster desires.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Sun, 05 Oct 2003 20:12:05 GMT, (Stewart
Pinkerton) wrote:

On Sun, 05 Oct 2003 12:06:49 +0100, Don Pearce
wrote:

On Sun, 05 Oct 2003 11:02:22 GMT, "Andy Fish"
wrote:

"Andy Dee" wrote in message
.. .

Here's my actual problem: I have a pair of bi-wireable speakers and I would
like to attenuate the top end by 2-3db. Is it possible to do this without
taking them apart - i.e. by connecting a series resistor and/or l-pad onto
the tweeter's speaker terminals?


Put a wirewound pot (reasonably good power handling) in series with
the tweeter, and tweak to taste. Something around 20 ohms should do
it.

WHAT? A 20 ohm resistor will drop the treble by around 10dB. This is a
*lot* more than the original poster desires.

A 20 ohm pot, Stewart, not a 20 ohm resistor. I'm not expecting the
wanted value to be anything like that much.

d

_____________________________

http://www.pearce.uk.com

The best way is to use a graphic equalizer between the preamp and the power
amp. If you have a tape monitor section of your receiver or amplifier you
are not using, you can insert a graphic equalizer unit between them.

Or, you can get inside of the speaker box, and wire a standard L-Pad (pot
designed to adjust speaker level) between the tweeter and the crossover
output. You can then mount the L-Pad on the rear of the box. This will
allow you to control the output to the tweeter to match your taste. There
are speaker boxes on the market that have built in EQ pots to adjust the
highs, and even the mids.

--

Greetings,

Jerry Greenberg GLG Technologies GLG
=========================================
WebPage http://www.zoom-one.com
Electronics http://www.zoom-one.com/electron.htm
Instruments http://www.zoom-one.com/glgtech.htm
=========================================


"Andy Fish" wrote in message
...
Hi,

as a hi-fi owner with some background knowledge of physics but no idea of
speaker or crossover design, I wonder if someone could help me out with this
simple question.

in a typical 2 way speaker, the bass driver dissipates many times as much
power as the tweeter. however, the nominal impedance of the two drivers is
often similar. if the voltage across the two is the same and the impedance
is the same, how does the bass driver draw so much more power than the
tweeter?

I realise this is a gross simplification of the actual design of a speaker.
Is it the crossover that increases the impedance as seen by the amplifier?
or am I totally missing something

Andy

thanks for all the replies

I couldn't easily locate a pot less than 1k ohms or with power handling in
excess of 0.5 watt, so I just bought a few fixed wirewound resistors to try
out.

Ended up with 2.2 ohm in series with the tweeter which subjectively is
giving me the sound I was looking for. I'm not sure whether I should have
gone for an L-pad, but the simple series resistor sounds OK to me.

Here's the reason I was concerned about using an L-pad, and it's the reason
for the original question:

The crossover network on the tweeter has high resistance at low frequencies
and this is why it doesn't have to handle the bass energy. however, an l-pad
situated between the amp and the crossover would have a relatively low
resistance (say 20 ohm) at all frequencies. this would mean the l-pad would
potentially have to dissipate a lot of power were I to run the system at
high volume. With a simple resistor in series with the crossover, it only
ever gets get high frequency (i.e. lower power) energy.

Andy


"Don Pearce" wrote in message
...
On Sun, 05 Oct 2003 20:12:05 GMT, (Stewart
Pinkerton) wrote:

On Sun, 05 Oct 2003 12:06:49 +0100, Don Pearce
wrote:

On Sun, 05 Oct 2003 11:02:22 GMT, "Andy Fish"
wrote:

"Andy Dee" wrote in message
.. .

Here's my actual problem: I have a pair of bi-wireable speakers and I
would
like to attenuate the top end by 2-3db. Is it possible to do this
without
taking them apart - i.e. by connecting a series resistor and/or l-pad
onto
the tweeter's speaker terminals?


Put a wirewound pot (reasonably good power handling) in series with
the tweeter, and tweak to taste. Something around 20 ohms should do
it.

WHAT? A 20 ohm resistor will drop the treble by around 10dB. This is a
*lot* more than the original poster desires.

A 20 ohm pot, Stewart, not a 20 ohm resistor. I'm not expecting the
wanted value to be anything like that much.

d

_____________________________

http://www.pearce.uk.com

"Andy Fish" wrote in message ...
thanks for all the replies

I couldn't easily locate a pot less than 1k ohms or with power handling in
excess of 0.5 watt, so I just bought a few fixed wirewound resistors to try
out.

Ended up with 2.2 ohm in series with the tweeter which subjectively is
giving me the sound I was looking for. I'm not sure whether I should have
gone for an L-pad, but the simple series resistor sounds OK to me.

Here's the reason I was concerned about using an L-pad, and it's the reason
for the original question:

The crossover network on the tweeter has high resistance at low frequencies
and this is why it doesn't have to handle the bass energy. however, an l-pad
situated between the amp and the crossover would have a relatively low
resistance (say 20 ohm) at all frequencies. this would mean the l-pad would
potentially have to dissipate a lot of power were I to run the system at
high volume.

That's because you're putting it in the wrong place. The L-pad
is properly placed on the output of the crossover, NOT the input,
between the rossover and the tweeter. In doing so, the L-pad
only needs to deal with the passband energy of the tweeter AND
provides a reasonably constant load for the tweeter crossover
network.

You understand that, depending upon the topology of the tweeter
crossover, not only does you 2.2 ohm series resistor change the
level, it also changes the shape of the frequency response of
the tweeter crossover.

The best place for any tweeter attenuation, be it an L-pad or
comething else, is AFTER the tweeter crossover, BEFORE the tweeter
itself.

With a simple resistor in series with the crossover, it only
ever gets get high frequency (i.e. lower power) energy.

Thanks, I think I get it all now.

"Dick Pierce" wrote in message
om...
"Andy Fish" wrote in message
...
thanks for all the replies

I couldn't easily locate a pot less than 1k ohms or with power handling
in
excess of 0.5 watt, so I just bought a few fixed wirewound resistors to
try
out.

Ended up with 2.2 ohm in series with the tweeter which subjectively is
giving me the sound I was looking for. I'm not sure whether I should
have
gone for an L-pad, but the simple series resistor sounds OK to me.

Here's the reason I was concerned about using an L-pad, and it's the
reason
for the original question:

The crossover network on the tweeter has high resistance at low
frequencies
and this is why it doesn't have to handle the bass energy. however, an
l-pad
situated between the amp and the crossover would have a relatively low
resistance (say 20 ohm) at all frequencies. this would mean the l-pad
would
potentially have to dissipate a lot of power were I to run the system at
high volume.

That's because you're putting it in the wrong place. The L-pad
is properly placed on the output of the crossover, NOT the input,
between the rossover and the tweeter. In doing so, the L-pad
only needs to deal with the passband energy of the tweeter AND
provides a reasonably constant load for the tweeter crossover
network.

You understand that, depending upon the topology of the tweeter
crossover, not only does you 2.2 ohm series resistor change the
level, it also changes the shape of the frequency response of
the tweeter crossover.

The best place for any tweeter attenuation, be it an L-pad or
comething else, is AFTER the tweeter crossover, BEFORE the tweeter
itself.

With a simple resistor in series with the crossover, it only
ever gets get high frequency (i.e. lower power) energy.

"Dick Pierce" wrote in message
om

You understand that, depending upon the topology of the tweeter
crossover, not only does you 2.2 ohm series resistor change the
level, it also changes the shape of the frequency response of
the tweeter crossover.

Extrememely worthwhile point. As a rule series resistors move the tweeter
crossover point down, which potentially decreases the power handling
capacity of the tweeter, not to mention other likely audible changes in
frequency response.

The best place for any tweeter attenuation, be it an L-pad or
something else, is AFTER the tweeter crossover, BEFORE the tweeter
itself.

With a simple resistor in series with the crossover, it only
ever gets get high frequency (i.e. lower power) energy.

Agreed.

"Arny Krueger" wrote in message ...
"Dick Pierce" wrote in message
om

You understand that, depending upon the topology of the tweeter
crossover, not only does you 2.2 ohm series resistor change the
level, it also changes the shape of the frequency response of
the tweeter crossover.

Extrememely worthwhile point. As a rule series resistors move the tweeter
crossover point down, which potentially decreases the power handling
capacity of the tweeter, not to mention other likely audible changes in
frequency response.

Not necessarily true for anything other than first-order networks.
For higher order networks, the crossover, i.e., the -3dB point, may
change only small amount, but the passband response of the network
may change dramatically. As an example, I had a client for whom I
designed a network. I also provided him with a set of L-pad
attenuator values should he feel the need to tune things to his
liking. To save cost, he tried it with the single series resistor.
The result was that the -3 dB point of the tweeter pass function
moved barely 1/4 octave, but had a significant (about 4 dB) peak
right above the crossover, a dip and then a significant rise above
that.

"Dick Pierce" wrote in message
om
"Arny Krueger" wrote in message
...
"Dick Pierce" wrote in message
om

You understand that, depending upon the topology of the tweeter
crossover, not only does you 2.2 ohm series resistor change the
level, it also changes the shape of the frequency response of
the tweeter crossover.

Extrememely worthwhile point. As a rule series resistors move the
tweeter crossover point down, which potentially decreases the power
handling capacity of the tweeter, not to mention other likely
audible changes in frequency response.

Not necessarily true for anything other than first-order networks.
For higher order networks, the crossover, i.e., the -3dB point, may
change only small amount, but the passband response of the network
may change dramatically. As an example, I had a client for whom I
designed a network. I also provided him with a set of L-pad
attenuator values should he feel the need to tune things to his
liking. To save cost, he tried it with the single series resistor.
The result was that the -3 dB point of the tweeter pass function
moved barely 1/4 octave, but had a significant (about 4 dB) peak
right above the crossover, a dip and then a significant rise above
that.

Interesting. Noted.