I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel). However I
don't know what the power handling capability of the resistor and
inductor should be, as this is not a crossover, all the power passes
through this network.

The JX92s can handle 100W (50W continous) of power.
Does this mean my resistor and inductor must also be rated at 100W?
For wire wound components could I use the gauge of the wire as an
indication of what power they can handle?
Is there any advantage of an air cored inductor over a ferrite cored
one?

I see Falcon Acoustics http://www.falcon-acoustics.co.uk/pl28p6.htm
sell 50W resistors but most UK retailers don't indicate the power
handling of their components.

Thanks
Ivan

(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

However I don't know what the power handling capability of the
resistor and inductor should be, as this is not a crossover, all the
power passes through this network.

The JX92s can handle 100W (50W continous) of power. Does this mean
my resistor and inductor must also be rated at 100W?

It depends what you want to melt first in the event of overpowering,
but typically you'd want the speaker to be the weak link.
Consequently, if the inductor is going in series with the speaker, it
too should handle at least 100W. Yes, wire gauge can help you
approximate the power handling rather nicely on a wound component.

Curious though...how exactly are you going to connect this resistor in
this network? And what's the nominal impdance of the driver? These
will lend insight into your resistor power requirements.

I see Falcon Acoustics http://www.falcon-acoustics.co.uk/pl28p6.htm
sell 50W resistors but most UK retailers don't indicate the power
handling of their components.

That's rather irresponsible of them. Resistors are simple things, but
at the bare minimum they need to tell you its resistance, the
tolerance of that resistance, and the power handling.

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

However I don't know what the power handling capability of the
resistor and inductor should be, as this is not a crossover, all the
power passes through this network.

The JX92s can handle 100W (50W continous) of power. Does this mean
my resistor and inductor must also be rated at 100W?

It depends what you want to melt first in the event of overpowering,
but typically you'd want the speaker to be the weak link.
Consequently, if the inductor is going in series with the speaker, it
too should handle at least 100W. Yes, wire gauge can help you
approximate the power handling rather nicely on a wound component.

Curious though...how exactly are you going to connect this resistor in
this network? And what's the nominal impdance of the driver? These
will lend insight into your resistor power requirements.

I see Falcon Acoustics http://www.falcon-acoustics.co.uk/pl28p6.htm
sell 50W resistors but most UK retailers don't indicate the power
handling of their components.

That's rather irresponsible of them. Resistors are simple things, but
at the bare minimum they need to tell you its resistance, the
tolerance of that resistance, and the power handling.

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

However I don't know what the power handling capability of the
resistor and inductor should be, as this is not a crossover, all the
power passes through this network.

The JX92s can handle 100W (50W continous) of power. Does this mean
my resistor and inductor must also be rated at 100W?

It depends what you want to melt first in the event of overpowering,
but typically you'd want the speaker to be the weak link.
Consequently, if the inductor is going in series with the speaker, it
too should handle at least 100W. Yes, wire gauge can help you
approximate the power handling rather nicely on a wound component.

Curious though...how exactly are you going to connect this resistor in
this network? And what's the nominal impdance of the driver? These
will lend insight into your resistor power requirements.

I see Falcon Acoustics http://www.falcon-acoustics.co.uk/pl28p6.htm
sell 50W resistors but most UK retailers don't indicate the power
handling of their components.

That's rather irresponsible of them. Resistors are simple things, but
at the bare minimum they need to tell you its resistance, the
tolerance of that resistance, and the power handling.

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

(Ivan) wrote in message . com...
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel). However I
don't know what the power handling capability of the resistor and
inductor should be, as this is not a crossover, all the power passes
through this network.

The JX92s can handle 100W (50W continous) of power.
Does this mean my resistor and inductor must also be rated at 100W?

Nope, not at all. It depends upon the value of the components.

Let's, for the purpose of illustration, assume that the driver
has a nominal impedance of 8 ohms. And you say the power handling
is 50 watts, and we'll assume that's the thermal rating of the
driver. Since:

P = I^2 R

where I is current, R is resistance and P is power, then since:

50 W = I^2 8 ohms

then

I^2 = 50 W / 8 Ohms

I = 2.5 amps

This means that when the driver isdissipating 50 watts, 2.5 amps
are passing through it.

Now, assume the resistor has a value of 2 ohms. Going right back
to our equation above:

P = I^2 R

then

P = 2.5^2 * 2

P = 6.25 * 2

P = 12.5 watts

So even though the driver is dissipating 50 watts, the resistor
only dissipates 12.5.

So, without knowing the value of the components and such, you
can't answer the question.

And your question is further complicated by the fact that you also
have an inductor, and you give us no idea of the value of the
inductor nor the topology of the circuit, both crucial to determining
the power dissipation of the resistor (the power dissipation of the
inductor is, rest assured, almost totally irrelevant. It's DC
resistance is very low compared to the driver, and so it will be
dissipating very little power. For example, if its resistance is
0.25 ohms, it will be dissipating less than 2 watts).

The topology is important becuase your description implies some sort
of frequency ocrrection network, which means that at some frequencies,
most of the current may be going through the resistor, while at other
frequencies, mos6t may be going through the inductor, so that complicates
things even further.

(Ivan) wrote in message . com...
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel). However I
don't know what the power handling capability of the resistor and
inductor should be, as this is not a crossover, all the power passes
through this network.

The JX92s can handle 100W (50W continous) of power.
Does this mean my resistor and inductor must also be rated at 100W?

Nope, not at all. It depends upon the value of the components.

Let's, for the purpose of illustration, assume that the driver
has a nominal impedance of 8 ohms. And you say the power handling
is 50 watts, and we'll assume that's the thermal rating of the
driver. Since:

P = I^2 R

where I is current, R is resistance and P is power, then since:

50 W = I^2 8 ohms

then

I^2 = 50 W / 8 Ohms

I = 2.5 amps

This means that when the driver isdissipating 50 watts, 2.5 amps
are passing through it.

Now, assume the resistor has a value of 2 ohms. Going right back
to our equation above:

P = I^2 R

then

P = 2.5^2 * 2

P = 6.25 * 2

P = 12.5 watts

So even though the driver is dissipating 50 watts, the resistor
only dissipates 12.5.

So, without knowing the value of the components and such, you
can't answer the question.

And your question is further complicated by the fact that you also
have an inductor, and you give us no idea of the value of the
inductor nor the topology of the circuit, both crucial to determining
the power dissipation of the resistor (the power dissipation of the
inductor is, rest assured, almost totally irrelevant. It's DC
resistance is very low compared to the driver, and so it will be
dissipating very little power. For example, if its resistance is
0.25 ohms, it will be dissipating less than 2 watts).

The topology is important becuase your description implies some sort
of frequency ocrrection network, which means that at some frequencies,
most of the current may be going through the resistor, while at other
frequencies, mos6t may be going through the inductor, so that complicates
things even further.

(Ivan) wrote in message . com...
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel). However I
don't know what the power handling capability of the resistor and
inductor should be, as this is not a crossover, all the power passes
through this network.

The JX92s can handle 100W (50W continous) of power.
Does this mean my resistor and inductor must also be rated at 100W?

Nope, not at all. It depends upon the value of the components.

Let's, for the purpose of illustration, assume that the driver
has a nominal impedance of 8 ohms. And you say the power handling
is 50 watts, and we'll assume that's the thermal rating of the
driver. Since:

P = I^2 R

where I is current, R is resistance and P is power, then since:

50 W = I^2 8 ohms

then

I^2 = 50 W / 8 Ohms

I = 2.5 amps

This means that when the driver isdissipating 50 watts, 2.5 amps
are passing through it.

Now, assume the resistor has a value of 2 ohms. Going right back
to our equation above:

P = I^2 R

then

P = 2.5^2 * 2

P = 6.25 * 2

P = 12.5 watts

So even though the driver is dissipating 50 watts, the resistor
only dissipates 12.5.

So, without knowing the value of the components and such, you
can't answer the question.

And your question is further complicated by the fact that you also
have an inductor, and you give us no idea of the value of the
inductor nor the topology of the circuit, both crucial to determining
the power dissipation of the resistor (the power dissipation of the
inductor is, rest assured, almost totally irrelevant. It's DC
resistance is very low compared to the driver, and so it will be
dissipating very little power. For example, if its resistance is
0.25 ohms, it will be dissipating less than 2 watts).

The topology is important becuase your description implies some sort
of frequency ocrrection network, which means that at some frequencies,
most of the current may be going through the resistor, while at other
frequencies, mos6t may be going through the inductor, so that complicates
things even further.

(Todd H.) writes:

(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

However I don't know what the power handling capability of the
resistor and inductor should be, as this is not a crossover, all the
power passes through this network.

The JX92s can handle 100W (50W continous) of power. Does this mean
my resistor and inductor must also be rated at 100W?

It depends what you want to melt first in the event of overpowering,
but typically you'd want the speaker to be the weak link.
Consequently, if the inductor is going in series with the speaker, it
too should handle at least 100W.

Oy. Strike that. That inductor needs to handle all the series
_current_ is what I meant to say (duh). The inductor needn't be able
to dissipate 100W.

Yes, wire gauge can help you approximate the power handling rather
nicely on a wound component.

"Current Handling" would be a far better statement.

Curious though...how exactly are you going to connect this resistor in
this network? And what's the nominal impdance of the driver? These
will lend insight into your resistor power requirements.

I see Falcon Acoustics http://www.falcon-acoustics.co.uk/pl28p6.htm
sell 50W resistors but most UK retailers don't indicate the power
handling of their components.

That's rather irresponsible of them. Resistors are simple things, but
at the bare minimum they need to tell you its resistance, the
tolerance of that resistance, and the power handling.

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

(Todd H.) writes:

(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

However I don't know what the power handling capability of the
resistor and inductor should be, as this is not a crossover, all the
power passes through this network.

The JX92s can handle 100W (50W continous) of power. Does this mean
my resistor and inductor must also be rated at 100W?

It depends what you want to melt first in the event of overpowering,
but typically you'd want the speaker to be the weak link.
Consequently, if the inductor is going in series with the speaker, it
too should handle at least 100W.

Oy. Strike that. That inductor needs to handle all the series
_current_ is what I meant to say (duh). The inductor needn't be able
to dissipate 100W.

Yes, wire gauge can help you approximate the power handling rather
nicely on a wound component.

"Current Handling" would be a far better statement.

Curious though...how exactly are you going to connect this resistor in
this network? And what's the nominal impdance of the driver? These
will lend insight into your resistor power requirements.

I see Falcon Acoustics http://www.falcon-acoustics.co.uk/pl28p6.htm
sell 50W resistors but most UK retailers don't indicate the power
handling of their components.

That's rather irresponsible of them. Resistors are simple things, but
at the bare minimum they need to tell you its resistance, the
tolerance of that resistance, and the power handling.

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

(Todd H.) writes:

(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

However I don't know what the power handling capability of the
resistor and inductor should be, as this is not a crossover, all the
power passes through this network.

The JX92s can handle 100W (50W continous) of power. Does this mean
my resistor and inductor must also be rated at 100W?

It depends what you want to melt first in the event of overpowering,
but typically you'd want the speaker to be the weak link.
Consequently, if the inductor is going in series with the speaker, it
too should handle at least 100W.

Oy. Strike that. That inductor needs to handle all the series
_current_ is what I meant to say (duh). The inductor needn't be able
to dissipate 100W.

Yes, wire gauge can help you approximate the power handling rather
nicely on a wound component.

"Current Handling" would be a far better statement.

Curious though...how exactly are you going to connect this resistor in
this network? And what's the nominal impdance of the driver? These
will lend insight into your resistor power requirements.

I see Falcon Acoustics http://www.falcon-acoustics.co.uk/pl28p6.htm
sell 50W resistors but most UK retailers don't indicate the power
handling of their components.

That's rather irresponsible of them. Resistors are simple things, but
at the bare minimum they need to tell you its resistance, the
tolerance of that resistance, and the power handling.

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

Ivan wrote:

I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Don't!

Put in on the input of the amp instead, it is fairly simple to design a
low end shelving boost (actually a low hf shelving cut ... ) that can be
implemented with passive components and work well if fed via a sensibly
low impedance source, most modern preamp outputs are just that. You
could design it for use in a tape monitor loop if you are constrained by
having an integrated amp with no pre-main separation option.

Ivan


Kind regards

Peter Larsen

--
************************************************** ***********
* My site is at: http://www.muyiovatki.dk *
************************************************** ***********

Ivan wrote:

I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Don't!

Put in on the input of the amp instead, it is fairly simple to design a
low end shelving boost (actually a low hf shelving cut ... ) that can be
implemented with passive components and work well if fed via a sensibly
low impedance source, most modern preamp outputs are just that. You
could design it for use in a tape monitor loop if you are constrained by
having an integrated amp with no pre-main separation option.

Ivan


Kind regards

Peter Larsen

--
************************************************** ***********
* My site is at: http://www.muyiovatki.dk *
************************************************** ***********

Ivan wrote:

I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Don't!

Put in on the input of the amp instead, it is fairly simple to design a
low end shelving boost (actually a low hf shelving cut ... ) that can be
implemented with passive components and work well if fed via a sensibly
low impedance source, most modern preamp outputs are just that. You
could design it for use in a tape monitor loop if you are constrained by
having an integrated amp with no pre-main separation option.

Ivan


Kind regards

Peter Larsen

--
************************************************** ***********
* My site is at: http://www.muyiovatki.dk *
************************************************** ***********

(Todd H.) wrote in message ...
(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

Hmm... Strictly, no. Consider this network:

U1 U0
----------Inductor----------------
| |
Capacitor Speaker
| |
----------------------------------

Let us for the sake of argument assume that the speaker is resistive.
The transfer function will be:


U0 s^2*LC
-- = -------------------
U1 1+s*L/R+s^2*LC

set s=jw and take its absolute value:


|U0| w^2*LC
|--| = -----------------------------
|U1| sqrt((w*L/R)^2+(1-s^2*LC)^2)

(Hope I got it right)

This is a standard second order high-pass circuit. At resonance the
output is greater than at higher frequencies if Q1.

Q=R*sqrt(C/L)

So, given that the L and C is selected such that the Q is greater than
1 the power delivered to the speaker will increase around the cut-off
frequency of the filter.

But: There is a big but here. At those frequencies the impedance that
the amplifier sees will become less than the impedance of the speaker.
This is what causes the amplifier to deliver more power. Since
amplifiers are designed for a certain load, this is potentially bad!
Possibly, the technique can be used to raise the level at low
frequencies slightly, and possibly it would work well around the
speaker's mechanical resonance (where the electrical impedance is high
anyway).

I think that Audio Pro had a closed box model in the 80's that did
this (model name was "2-25") but i am not sure. Anyway, if you do
this, be careful and measure the impedance as seen by the amplifier
before connecting it to the amplifier!

Also, the inductor values are likely to be huge...

(Todd H.) wrote in message ...
(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

Hmm... Strictly, no. Consider this network:

U1 U0
----------Inductor----------------
| |
Capacitor Speaker
| |
----------------------------------

Let us for the sake of argument assume that the speaker is resistive.
The transfer function will be:


U0 s^2*LC
-- = -------------------
U1 1+s*L/R+s^2*LC

set s=jw and take its absolute value:


|U0| w^2*LC
|--| = -----------------------------
|U1| sqrt((w*L/R)^2+(1-s^2*LC)^2)

(Hope I got it right)

This is a standard second order high-pass circuit. At resonance the
output is greater than at higher frequencies if Q1.

Q=R*sqrt(C/L)

So, given that the L and C is selected such that the Q is greater than
1 the power delivered to the speaker will increase around the cut-off
frequency of the filter.

But: There is a big but here. At those frequencies the impedance that
the amplifier sees will become less than the impedance of the speaker.
This is what causes the amplifier to deliver more power. Since
amplifiers are designed for a certain load, this is potentially bad!
Possibly, the technique can be used to raise the level at low
frequencies slightly, and possibly it would work well around the
speaker's mechanical resonance (where the electrical impedance is high
anyway).

I think that Audio Pro had a closed box model in the 80's that did
this (model name was "2-25") but i am not sure. Anyway, if you do
this, be careful and measure the impedance as seen by the amplifier
before connecting it to the amplifier!

Also, the inductor values are likely to be huge...

(Todd H.) wrote in message ...
(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

Hmm... Strictly, no. Consider this network:

U1 U0
----------Inductor----------------
| |
Capacitor Speaker
| |
----------------------------------

Let us for the sake of argument assume that the speaker is resistive.
The transfer function will be:


U0 s^2*LC
-- = -------------------
U1 1+s*L/R+s^2*LC

set s=jw and take its absolute value:


|U0| w^2*LC
|--| = -----------------------------
|U1| sqrt((w*L/R)^2+(1-s^2*LC)^2)

(Hope I got it right)

This is a standard second order high-pass circuit. At resonance the
output is greater than at higher frequencies if Q1.

Q=R*sqrt(C/L)

So, given that the L and C is selected such that the Q is greater than
1 the power delivered to the speaker will increase around the cut-off
frequency of the filter.

But: There is a big but here. At those frequencies the impedance that
the amplifier sees will become less than the impedance of the speaker.
This is what causes the amplifier to deliver more power. Since
amplifiers are designed for a certain load, this is potentially bad!
Possibly, the technique can be used to raise the level at low
frequencies slightly, and possibly it would work well around the
speaker's mechanical resonance (where the electrical impedance is high
anyway).

I think that Audio Pro had a closed box model in the 80's that did
this (model name was "2-25") but i am not sure. Anyway, if you do
this, be careful and measure the impedance as seen by the amplifier
before connecting it to the amplifier!

Also, the inductor values are likely to be huge...

"Ivan" wrote in message
om

I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel). However I
don't know what the power handling capability of the resistor and
inductor should be, as this is not a crossover, all the power passes
through this network.

The JX92s can handle 100W (50W continuous) of power.
Does this mean my resistor and inductor must also be rated at 100W?

Remember that your project, as inadvisable as it may be, involves music
signals which have far less average (heating) power than normal steady-state
AC or DC signals.

This suggests that your resistors may not need to be rated at 100 watts if a
complex audio signal of 100 watts is being handled. Furthermore, the entire
power applied to this system is not dissipated in the resistor. The minimum
peak-to-average ratio of music even when highly compressed is 8-10 dB. This
means that if you apply 10 volts peak audio to a resistor, the heating of
the resistor will be equivalent to only about 3-4 volts of steady-state DC.
That's the difference between 100 watts and 16 watts, for example.

However, this project also takes you down a the road of component power
ratings. For example, most resistor power ratings relate to fairly high
operating temperatures.

I recently tested some high-quality non-inductive 8 ohm 250 watt resistors,
and found that when they are dissipating 250 watts in air, their surface
temperature was over 300 degrees F.

300+ degree F operation can be problematical in a number of practical ways.
For example it would soften or melt vinyl wire insulation, and it would char
wood and other organic materials nearby. Charred wood loses its structural
capabilities and it causes odors to be released (i.e., smoke) that raise
listener concerns. Open flames are possible.

I've seen power resistors get so hot they unsoldered themselves, but were
not permanently damaged! Many resistors dramatically change their
resistance when they get this hot. The good ones restore normal values when
cooled, but this is not a global rule.

Depending on how these resistors are made, you may want to overrate them
dramatically in the interest of keeping operational temperatures low enough
so that equipment and parts in their near vicinity are not damaged.

For wire wound components could I use the gauge of the wire as an
indication of what power they can handle?

You'll have to further derate the power because the wire is in a coil that
traps heat.

Is there any advantage of an air cored inductor over a ferrite cored one?

I've seen both kinds of inductors melted down, and/or melting components
near them.

I see Falcon Acoustics http://www.falcon-acoustics.co.uk/pl28p6.htm
sell 50W resistors but most UK retailers don't indicate the power
handling of their components.

I can't believe this. In US catalogs like Digi-Key and Mouser, resistors are
clearly specd by power rating. However you may have to read manufacturer
spec sheets to see what the power ratings mean in practical terms.

"Ivan" wrote in message
om

I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel). However I
don't know what the power handling capability of the resistor and
inductor should be, as this is not a crossover, all the power passes
through this network.

The JX92s can handle 100W (50W continuous) of power.
Does this mean my resistor and inductor must also be rated at 100W?

Remember that your project, as inadvisable as it may be, involves music
signals which have far less average (heating) power than normal steady-state
AC or DC signals.

This suggests that your resistors may not need to be rated at 100 watts if a
complex audio signal of 100 watts is being handled. Furthermore, the entire
power applied to this system is not dissipated in the resistor. The minimum
peak-to-average ratio of music even when highly compressed is 8-10 dB. This
means that if you apply 10 volts peak audio to a resistor, the heating of
the resistor will be equivalent to only about 3-4 volts of steady-state DC.
That's the difference between 100 watts and 16 watts, for example.

However, this project also takes you down a the road of component power
ratings. For example, most resistor power ratings relate to fairly high
operating temperatures.

I recently tested some high-quality non-inductive 8 ohm 250 watt resistors,
and found that when they are dissipating 250 watts in air, their surface
temperature was over 300 degrees F.

300+ degree F operation can be problematical in a number of practical ways.
For example it would soften or melt vinyl wire insulation, and it would char
wood and other organic materials nearby. Charred wood loses its structural
capabilities and it causes odors to be released (i.e., smoke) that raise
listener concerns. Open flames are possible.

I've seen power resistors get so hot they unsoldered themselves, but were
not permanently damaged! Many resistors dramatically change their
resistance when they get this hot. The good ones restore normal values when
cooled, but this is not a global rule.

Depending on how these resistors are made, you may want to overrate them
dramatically in the interest of keeping operational temperatures low enough
so that equipment and parts in their near vicinity are not damaged.

For wire wound components could I use the gauge of the wire as an
indication of what power they can handle?

You'll have to further derate the power because the wire is in a coil that
traps heat.

Is there any advantage of an air cored inductor over a ferrite cored one?

I've seen both kinds of inductors melted down, and/or melting components
near them.

I see Falcon Acoustics http://www.falcon-acoustics.co.uk/pl28p6.htm
sell 50W resistors but most UK retailers don't indicate the power
handling of their components.

I can't believe this. In US catalogs like Digi-Key and Mouser, resistors are
clearly specd by power rating. However you may have to read manufacturer
spec sheets to see what the power ratings mean in practical terms.

"Ivan" wrote in message
om

I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel). However I
don't know what the power handling capability of the resistor and
inductor should be, as this is not a crossover, all the power passes
through this network.

The JX92s can handle 100W (50W continuous) of power.
Does this mean my resistor and inductor must also be rated at 100W?

Remember that your project, as inadvisable as it may be, involves music
signals which have far less average (heating) power than normal steady-state
AC or DC signals.

This suggests that your resistors may not need to be rated at 100 watts if a
complex audio signal of 100 watts is being handled. Furthermore, the entire
power applied to this system is not dissipated in the resistor. The minimum
peak-to-average ratio of music even when highly compressed is 8-10 dB. This
means that if you apply 10 volts peak audio to a resistor, the heating of
the resistor will be equivalent to only about 3-4 volts of steady-state DC.
That's the difference between 100 watts and 16 watts, for example.

However, this project also takes you down a the road of component power
ratings. For example, most resistor power ratings relate to fairly high
operating temperatures.

I recently tested some high-quality non-inductive 8 ohm 250 watt resistors,
and found that when they are dissipating 250 watts in air, their surface
temperature was over 300 degrees F.

300+ degree F operation can be problematical in a number of practical ways.
For example it would soften or melt vinyl wire insulation, and it would char
wood and other organic materials nearby. Charred wood loses its structural
capabilities and it causes odors to be released (i.e., smoke) that raise
listener concerns. Open flames are possible.

I've seen power resistors get so hot they unsoldered themselves, but were
not permanently damaged! Many resistors dramatically change their
resistance when they get this hot. The good ones restore normal values when
cooled, but this is not a global rule.

Depending on how these resistors are made, you may want to overrate them
dramatically in the interest of keeping operational temperatures low enough
so that equipment and parts in their near vicinity are not damaged.

For wire wound components could I use the gauge of the wire as an
indication of what power they can handle?

You'll have to further derate the power because the wire is in a coil that
traps heat.

Is there any advantage of an air cored inductor over a ferrite cored one?

I've seen both kinds of inductors melted down, and/or melting components
near them.

I see Falcon Acoustics http://www.falcon-acoustics.co.uk/pl28p6.htm
sell 50W resistors but most UK retailers don't indicate the power
handling of their components.

I can't believe this. In US catalogs like Digi-Key and Mouser, resistors are
clearly specd by power rating. However you may have to read manufacturer
spec sheets to see what the power ratings mean in practical terms.

(Svante) writes:

(Todd H.) wrote in message ...
(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

Hmm... Strictly, no. Consider this network:

U1 U0
----------Inductor----------------
| |
Capacitor Speaker
| |
----------------------------------

Let us for the sake of argument assume that the speaker is resistive.
The transfer function will be:


U0 s^2*LC
-- = -------------------
U1 1+s*L/R+s^2*LC

set s=jw and take its absolute value:


|U0| w^2*LC
|--| = -----------------------------
|U1| sqrt((w*L/R)^2+(1-s^2*LC)^2)

(Hope I got it right)

If you can provide me a value of L, C, R and omega for which that
transfer function is 1 and I'll concede that you've managed to
_boost_ anything without the benefit of an active component.

The network you've described is a 2nd order passive network that will
only attenuate high frequencies. It's not going to boost the lows
over the value they'd have in the absence of the network, unless you
rely on something really odd the amplifier is doing... but that'd be
cheating as the amplifier is a big box of active electronics. :-)

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

(Svante) writes:

(Todd H.) wrote in message ...
(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

Hmm... Strictly, no. Consider this network:

U1 U0
----------Inductor----------------
| |
Capacitor Speaker
| |
----------------------------------

Let us for the sake of argument assume that the speaker is resistive.
The transfer function will be:


U0 s^2*LC
-- = -------------------
U1 1+s*L/R+s^2*LC

set s=jw and take its absolute value:


|U0| w^2*LC
|--| = -----------------------------
|U1| sqrt((w*L/R)^2+(1-s^2*LC)^2)

(Hope I got it right)

If you can provide me a value of L, C, R and omega for which that
transfer function is 1 and I'll concede that you've managed to
_boost_ anything without the benefit of an active component.

The network you've described is a 2nd order passive network that will
only attenuate high frequencies. It's not going to boost the lows
over the value they'd have in the absence of the network, unless you
rely on something really odd the amplifier is doing... but that'd be
cheating as the amplifier is a big box of active electronics. :-)

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

(Svante) writes:

(Todd H.) wrote in message ...
(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

Hmm... Strictly, no. Consider this network:

U1 U0
----------Inductor----------------
| |
Capacitor Speaker
| |
----------------------------------

Let us for the sake of argument assume that the speaker is resistive.
The transfer function will be:


U0 s^2*LC
-- = -------------------
U1 1+s*L/R+s^2*LC

set s=jw and take its absolute value:


|U0| w^2*LC
|--| = -----------------------------
|U1| sqrt((w*L/R)^2+(1-s^2*LC)^2)

(Hope I got it right)

If you can provide me a value of L, C, R and omega for which that
transfer function is 1 and I'll concede that you've managed to
_boost_ anything without the benefit of an active component.

The network you've described is a 2nd order passive network that will
only attenuate high frequencies. It's not going to boost the lows
over the value they'd have in the absence of the network, unless you
rely on something really odd the amplifier is doing... but that'd be
cheating as the amplifier is a big box of active electronics. :-)

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

And your question is further complicated by the fact that you also
have an inductor, and you give us no idea of the value of the
inductor nor the topology of the circuit, both crucial to determining
the power dissipation of the resistor (the power dissipation of the
inductor is, rest assured, almost totally irrelevant. It's DC
resistance is very low compared to the driver, and so it will be
dissipating very little power. For example, if its resistance is
0.25 ohms, it will be dissipating less than 2 watts).

Okey Dokey, I'll make things a little clearer. My circuit does not
boost the low frequencies it actually attenuates the high frequencies
as you may have guessed.
I propose to use the following circuit:

|----Inductor----|
+ -----| |-------|
|----Resistor----| |
Driver
|
- ------------------------------|

This means that at high frequencies the inductor will be of high
impedance and therefore the network will have high impedance (up to
the value of the resistor) and therefore suppress the high
frequencies. At low frequencies the inductor will have little
resistance (to a minimum of theoretically 0 ohms) and therefore no
resistance.

The impedance of the driver is nominally 4 ohms.
Therefore, to assume the worst case:

50 Watts @ 4 ohms
P = I^2 R
therefore I = 3.5 amps

If R = 5 and L = 2 mH
then the power for the resistor is 12.5*5 = 62.5 Watts

As for the inductor, at high frequencies it will have a high impedance
therefore all the current will flow through the resistor and therefore
very little power will be disipated by the inductor. At low
frequencies all the current will flow through the inductor but its DC
resistance is small so again it will disipate little power. When the
inductive impedance = resitance = 5 ohms then both components will
have 1.75 amps flowing through them, therefore (P=I^2 R) power
disipated by the inductor is 15.3 watts.

Does this seem correct?

Thanks for the help
Ivan

And your question is further complicated by the fact that you also
have an inductor, and you give us no idea of the value of the
inductor nor the topology of the circuit, both crucial to determining
the power dissipation of the resistor (the power dissipation of the
inductor is, rest assured, almost totally irrelevant. It's DC
resistance is very low compared to the driver, and so it will be
dissipating very little power. For example, if its resistance is
0.25 ohms, it will be dissipating less than 2 watts).

Okey Dokey, I'll make things a little clearer. My circuit does not
boost the low frequencies it actually attenuates the high frequencies
as you may have guessed.
I propose to use the following circuit:

|----Inductor----|
+ -----| |-------|
|----Resistor----| |
Driver
|
- ------------------------------|

This means that at high frequencies the inductor will be of high
impedance and therefore the network will have high impedance (up to
the value of the resistor) and therefore suppress the high
frequencies. At low frequencies the inductor will have little
resistance (to a minimum of theoretically 0 ohms) and therefore no
resistance.

The impedance of the driver is nominally 4 ohms.
Therefore, to assume the worst case:

50 Watts @ 4 ohms
P = I^2 R
therefore I = 3.5 amps

If R = 5 and L = 2 mH
then the power for the resistor is 12.5*5 = 62.5 Watts

As for the inductor, at high frequencies it will have a high impedance
therefore all the current will flow through the resistor and therefore
very little power will be disipated by the inductor. At low
frequencies all the current will flow through the inductor but its DC
resistance is small so again it will disipate little power. When the
inductive impedance = resitance = 5 ohms then both components will
have 1.75 amps flowing through them, therefore (P=I^2 R) power
disipated by the inductor is 15.3 watts.

Does this seem correct?

Thanks for the help
Ivan

And your question is further complicated by the fact that you also
have an inductor, and you give us no idea of the value of the
inductor nor the topology of the circuit, both crucial to determining
the power dissipation of the resistor (the power dissipation of the
inductor is, rest assured, almost totally irrelevant. It's DC
resistance is very low compared to the driver, and so it will be
dissipating very little power. For example, if its resistance is
0.25 ohms, it will be dissipating less than 2 watts).

Okey Dokey, I'll make things a little clearer. My circuit does not
boost the low frequencies it actually attenuates the high frequencies
as you may have guessed.
I propose to use the following circuit:

|----Inductor----|
+ -----| |-------|
|----Resistor----| |
Driver
|
- ------------------------------|

This means that at high frequencies the inductor will be of high
impedance and therefore the network will have high impedance (up to
the value of the resistor) and therefore suppress the high
frequencies. At low frequencies the inductor will have little
resistance (to a minimum of theoretically 0 ohms) and therefore no
resistance.

The impedance of the driver is nominally 4 ohms.
Therefore, to assume the worst case:

50 Watts @ 4 ohms
P = I^2 R
therefore I = 3.5 amps

If R = 5 and L = 2 mH
then the power for the resistor is 12.5*5 = 62.5 Watts

As for the inductor, at high frequencies it will have a high impedance
therefore all the current will flow through the resistor and therefore
very little power will be disipated by the inductor. At low
frequencies all the current will flow through the inductor but its DC
resistance is small so again it will disipate little power. When the
inductive impedance = resitance = 5 ohms then both components will
have 1.75 amps flowing through them, therefore (P=I^2 R) power
disipated by the inductor is 15.3 watts.

Does this seem correct?

Thanks for the help
Ivan

(Ivan) writes:

And your question is further complicated by the fact that you also
have an inductor, and you give us no idea of the value of the
inductor nor the topology of the circuit, both crucial to determining
the power dissipation of the resistor (the power dissipation of the
inductor is, rest assured, almost totally irrelevant. It's DC
resistance is very low compared to the driver, and so it will be
dissipating very little power. For example, if its resistance is
0.25 ohms, it will be dissipating less than 2 watts).

Okey Dokey, I'll make things a little clearer. My circuit does not
boost the low frequencies it actually attenuates the high frequencies
as you may have guessed.
I propose to use the following circuit:

|----Inductor----|
+ -----| |-------|
|----Resistor----| |
Driver
|
- ------------------------------|

This means that at high frequencies the inductor will be of high
impedance and therefore the network will have high impedance (up to
the value of the resistor) and therefore suppress the high
frequencies. At low frequencies the inductor will have little
resistance (to a minimum of theoretically 0 ohms) and therefore no
resistance.

The impedance of the driver is nominally 4 ohms.
Therefore, to assume the worst case:

50 Watts @ 4 ohms
P = I^2 R
therefore I = 3.5 amps

If R = 5 and L = 2 mH
then the power for the resistor is 12.5*5 = 62.5 Watts

As for the inductor, at high frequencies it will have a high impedance
therefore all the current will flow through the resistor and therefore
very little power will be disipated by the inductor. At low
frequencies all the current will flow through the inductor but its DC
resistance is small so again it will disipate little power. When the
inductive impedance = resitance = 5 ohms then both components will
have 1.75 amps flowing through them, therefore (P=I^2 R) power
disipated by the inductor is 15.3 watts.

Does this seem correct?

Yes, your calculations look good. In the worst case with a functional
driver, you'll have about 63W. Actually, in reality the max will
potentially be a bit more than because a 4ohm nominal speaker has some
frequencies at which its impedance is less than 4ohms.

Still, what you're proposing to do is rather inelegant. You'd be
better served by doing tone shaping at the preamp level using an
equalizer or active crossover. You'd be even better served with a
speaker design that doesn't need such EQ compensation.

Full range speakers have their limitations and typically sound quite
marginal...and as the saying goes, "You can't polish a turd."

Best of luck in your design, though!

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

(Ivan) writes:

And your question is further complicated by the fact that you also
have an inductor, and you give us no idea of the value of the
inductor nor the topology of the circuit, both crucial to determining
the power dissipation of the resistor (the power dissipation of the
inductor is, rest assured, almost totally irrelevant. It's DC
resistance is very low compared to the driver, and so it will be
dissipating very little power. For example, if its resistance is
0.25 ohms, it will be dissipating less than 2 watts).

Okey Dokey, I'll make things a little clearer. My circuit does not
boost the low frequencies it actually attenuates the high frequencies
as you may have guessed.
I propose to use the following circuit:

|----Inductor----|
+ -----| |-------|
|----Resistor----| |
Driver
|
- ------------------------------|

This means that at high frequencies the inductor will be of high
impedance and therefore the network will have high impedance (up to
the value of the resistor) and therefore suppress the high
frequencies. At low frequencies the inductor will have little
resistance (to a minimum of theoretically 0 ohms) and therefore no
resistance.

The impedance of the driver is nominally 4 ohms.
Therefore, to assume the worst case:

50 Watts @ 4 ohms
P = I^2 R
therefore I = 3.5 amps

If R = 5 and L = 2 mH
then the power for the resistor is 12.5*5 = 62.5 Watts

As for the inductor, at high frequencies it will have a high impedance
therefore all the current will flow through the resistor and therefore
very little power will be disipated by the inductor. At low
frequencies all the current will flow through the inductor but its DC
resistance is small so again it will disipate little power. When the
inductive impedance = resitance = 5 ohms then both components will
have 1.75 amps flowing through them, therefore (P=I^2 R) power
disipated by the inductor is 15.3 watts.

Does this seem correct?

Yes, your calculations look good. In the worst case with a functional
driver, you'll have about 63W. Actually, in reality the max will
potentially be a bit more than because a 4ohm nominal speaker has some
frequencies at which its impedance is less than 4ohms.

Still, what you're proposing to do is rather inelegant. You'd be
better served by doing tone shaping at the preamp level using an
equalizer or active crossover. You'd be even better served with a
speaker design that doesn't need such EQ compensation.

Full range speakers have their limitations and typically sound quite
marginal...and as the saying goes, "You can't polish a turd."

Best of luck in your design, though!

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

(Ivan) writes:

And your question is further complicated by the fact that you also
have an inductor, and you give us no idea of the value of the
inductor nor the topology of the circuit, both crucial to determining
the power dissipation of the resistor (the power dissipation of the
inductor is, rest assured, almost totally irrelevant. It's DC
resistance is very low compared to the driver, and so it will be
dissipating very little power. For example, if its resistance is
0.25 ohms, it will be dissipating less than 2 watts).

Okey Dokey, I'll make things a little clearer. My circuit does not
boost the low frequencies it actually attenuates the high frequencies
as you may have guessed.
I propose to use the following circuit:

|----Inductor----|
+ -----| |-------|
|----Resistor----| |
Driver
|
- ------------------------------|

This means that at high frequencies the inductor will be of high
impedance and therefore the network will have high impedance (up to
the value of the resistor) and therefore suppress the high
frequencies. At low frequencies the inductor will have little
resistance (to a minimum of theoretically 0 ohms) and therefore no
resistance.

The impedance of the driver is nominally 4 ohms.
Therefore, to assume the worst case:

50 Watts @ 4 ohms
P = I^2 R
therefore I = 3.5 amps

If R = 5 and L = 2 mH
then the power for the resistor is 12.5*5 = 62.5 Watts

As for the inductor, at high frequencies it will have a high impedance
therefore all the current will flow through the resistor and therefore
very little power will be disipated by the inductor. At low
frequencies all the current will flow through the inductor but its DC
resistance is small so again it will disipate little power. When the
inductive impedance = resitance = 5 ohms then both components will
have 1.75 amps flowing through them, therefore (P=I^2 R) power
disipated by the inductor is 15.3 watts.

Does this seem correct?

Yes, your calculations look good. In the worst case with a functional
driver, you'll have about 63W. Actually, in reality the max will
potentially be a bit more than because a 4ohm nominal speaker has some
frequencies at which its impedance is less than 4ohms.

Still, what you're proposing to do is rather inelegant. You'd be
better served by doing tone shaping at the preamp level using an
equalizer or active crossover. You'd be even better served with a
speaker design that doesn't need such EQ compensation.

Full range speakers have their limitations and typically sound quite
marginal...and as the saying goes, "You can't polish a turd."

Best of luck in your design, though!

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

(Todd H.) wrote in message ...
(Svante) writes:

(Todd H.) wrote in message ...
(Ivan) writes:
Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors.
Hmm... Strictly, no. Consider this network:
U1 U0
----------Inductor----------------
| |
Capacitor Speaker
| |
----------------------------------
If you can provide me a value of L, C, R and omega for which that
transfer function is 1 and I'll concede that you've managed to
_boost_ anything without the benefit of an active component.

The network you've described is a 2nd order passive network that will
only attenuate high frequencies. It's not going to boost the lows
over the value they'd have in the absence of the network, unless you
rely on something really odd the amplifier is doing... but that'd be
cheating as the amplifier is a big box of active electronics. :-)

Well, with a slight change to the topology and given a real speaker
load, you will get significant boost.

Consider the following network:

o------ 400 uF ----+--------o
|
15 mH
|
o------------------+--------o

This constitutes about a 64 Hz high-pass filter. Into even a
resistive load, look at the transfer function (calculated
every 1/12th octave from 30 to 100 Hz):

30 -12.2 dB
31.7 -11
33.6 -9.8
35.6 -8.7
37.8 -7.5
40 -6.3
42.4 -5.
44.9 -3.8
47.6 -2.6
50.4 -1.5
53.4 -0.4
56.6 0.6
59.9 1.4
63.5 2.1
67.3 2.6
71.3 2.9
75.5 3
80 3
84.8 2.9
89.8 2.7
95.1 2.53
101 2.33

Yup, that's right, from about 55 Hz till above 100 Hz, this PASSIVE
circuit has voltage gain.

Now, let's replace the resistor with a real speaker whose resonant
frequency is 65 Hz. Now, electrically, that speaker looks like
a parallel RLC tank circuit in series with a resistor. Doing THAT
experiment reveals the following transfer function:

30 -13.1 dB
31.7 -12
33.6 -10.8
35.6 -9.6
37.8 -8.31
40 -6.94
42.4 -5.47
44.9 -3.85
47.6 -2.05
50.4 0.013
53.4 2.44
56.6 5.4
59.9 9.13
63.5 13.3
67.3 14
71.3 10.9
75.5 8.04
80 6.01
84.8 4.54
89.8 3.45
95.1 2.62
101 2

Goodness gracious! 14 dB of voltage gain out of a passive circuit!

Check it out for yourself: here's thge complete spice net list
for the simulation:

* Passive "equalizer"

..SUBCKT H3043 1 10
Re 1 2 7.55
Lces 2 3 19.66MH
Cmes 2 3 291.94UF
Res 2 3 24.54
Lvc1 3 4 0.38MH
Rs1 3 4 2
Lvc2 4 5 0.26MH
Rs2 4 5 21
Lvc3 5 10 0.09MH
..ENDS

VIN 1 0 AC SIN 1.0 0.0

C11 1 2 400UF
L12 2 0 15MH
XWoof1 2 0 H3043

..AC OCT 12 20 20K
..PRINT AC VDB(2,0)
..END

(Todd H.) wrote in message ...
(Svante) writes:

(Todd H.) wrote in message ...
(Ivan) writes:
Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors.
Hmm... Strictly, no. Consider this network:
U1 U0
----------Inductor----------------
| |
Capacitor Speaker
| |
----------------------------------
If you can provide me a value of L, C, R and omega for which that
transfer function is 1 and I'll concede that you've managed to
_boost_ anything without the benefit of an active component.

The network you've described is a 2nd order passive network that will
only attenuate high frequencies. It's not going to boost the lows
over the value they'd have in the absence of the network, unless you
rely on something really odd the amplifier is doing... but that'd be
cheating as the amplifier is a big box of active electronics. :-)

Well, with a slight change to the topology and given a real speaker
load, you will get significant boost.

Consider the following network:

o------ 400 uF ----+--------o
|
15 mH
|
o------------------+--------o

This constitutes about a 64 Hz high-pass filter. Into even a
resistive load, look at the transfer function (calculated
every 1/12th octave from 30 to 100 Hz):

30 -12.2 dB
31.7 -11
33.6 -9.8
35.6 -8.7
37.8 -7.5
40 -6.3
42.4 -5.
44.9 -3.8
47.6 -2.6
50.4 -1.5
53.4 -0.4
56.6 0.6
59.9 1.4
63.5 2.1
67.3 2.6
71.3 2.9
75.5 3
80 3
84.8 2.9
89.8 2.7
95.1 2.53
101 2.33

Yup, that's right, from about 55 Hz till above 100 Hz, this PASSIVE
circuit has voltage gain.

Now, let's replace the resistor with a real speaker whose resonant
frequency is 65 Hz. Now, electrically, that speaker looks like
a parallel RLC tank circuit in series with a resistor. Doing THAT
experiment reveals the following transfer function:

30 -13.1 dB
31.7 -12
33.6 -10.8
35.6 -9.6
37.8 -8.31
40 -6.94
42.4 -5.47
44.9 -3.85
47.6 -2.05
50.4 0.013
53.4 2.44
56.6 5.4
59.9 9.13
63.5 13.3
67.3 14
71.3 10.9
75.5 8.04
80 6.01
84.8 4.54
89.8 3.45
95.1 2.62
101 2

Goodness gracious! 14 dB of voltage gain out of a passive circuit!

Check it out for yourself: here's thge complete spice net list
for the simulation:

* Passive "equalizer"

..SUBCKT H3043 1 10
Re 1 2 7.55
Lces 2 3 19.66MH
Cmes 2 3 291.94UF
Res 2 3 24.54
Lvc1 3 4 0.38MH
Rs1 3 4 2
Lvc2 4 5 0.26MH
Rs2 4 5 21
Lvc3 5 10 0.09MH
..ENDS

VIN 1 0 AC SIN 1.0 0.0

C11 1 2 400UF
L12 2 0 15MH
XWoof1 2 0 H3043

..AC OCT 12 20 20K
..PRINT AC VDB(2,0)
..END

(Todd H.) wrote in message ...
(Svante) writes:

(Todd H.) wrote in message ...
(Ivan) writes:
Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors.
Hmm... Strictly, no. Consider this network:
U1 U0
----------Inductor----------------
| |
Capacitor Speaker
| |
----------------------------------
If you can provide me a value of L, C, R and omega for which that
transfer function is 1 and I'll concede that you've managed to
_boost_ anything without the benefit of an active component.

The network you've described is a 2nd order passive network that will
only attenuate high frequencies. It's not going to boost the lows
over the value they'd have in the absence of the network, unless you
rely on something really odd the amplifier is doing... but that'd be
cheating as the amplifier is a big box of active electronics. :-)

Well, with a slight change to the topology and given a real speaker
load, you will get significant boost.

Consider the following network:

o------ 400 uF ----+--------o
|
15 mH
|
o------------------+--------o

This constitutes about a 64 Hz high-pass filter. Into even a
resistive load, look at the transfer function (calculated
every 1/12th octave from 30 to 100 Hz):

30 -12.2 dB
31.7 -11
33.6 -9.8
35.6 -8.7
37.8 -7.5
40 -6.3
42.4 -5.
44.9 -3.8
47.6 -2.6
50.4 -1.5
53.4 -0.4
56.6 0.6
59.9 1.4
63.5 2.1
67.3 2.6
71.3 2.9
75.5 3
80 3
84.8 2.9
89.8 2.7
95.1 2.53
101 2.33

Yup, that's right, from about 55 Hz till above 100 Hz, this PASSIVE
circuit has voltage gain.

Now, let's replace the resistor with a real speaker whose resonant
frequency is 65 Hz. Now, electrically, that speaker looks like
a parallel RLC tank circuit in series with a resistor. Doing THAT
experiment reveals the following transfer function:

30 -13.1 dB
31.7 -12
33.6 -10.8
35.6 -9.6
37.8 -8.31
40 -6.94
42.4 -5.47
44.9 -3.85
47.6 -2.05
50.4 0.013
53.4 2.44
56.6 5.4
59.9 9.13
63.5 13.3
67.3 14
71.3 10.9
75.5 8.04
80 6.01
84.8 4.54
89.8 3.45
95.1 2.62
101 2

Goodness gracious! 14 dB of voltage gain out of a passive circuit!

Check it out for yourself: here's thge complete spice net list
for the simulation:

* Passive "equalizer"

..SUBCKT H3043 1 10
Re 1 2 7.55
Lces 2 3 19.66MH
Cmes 2 3 291.94UF
Res 2 3 24.54
Lvc1 3 4 0.38MH
Rs1 3 4 2
Lvc2 4 5 0.26MH
Rs2 4 5 21
Lvc3 5 10 0.09MH
..ENDS

VIN 1 0 AC SIN 1.0 0.0

C11 1 2 400UF
L12 2 0 15MH
XWoof1 2 0 H3043

..AC OCT 12 20 20K
..PRINT AC VDB(2,0)
..END

(Dick Pierce) writes:

Yup, that's right, from about 55 Hz till above 100 Hz, this PASSIVE
circuit has voltage gain.

You have convinced me of the time varying wonders of reactive
elements. Your point is well made, and you're quite correct, in the
context of a speaker crossover network where an active amplifier is
pushing the circuit, you can boost the voltage of certain frequencies
at the expense of others with a reactive element. Power, of course is
still conserved.

VIN 1 0 AC SIN 1.0 0.0
L12 2 0 15MH

These are the components that make it happen--a perfect AC source with
infinite current sourcing capability, and and inductor that can
generate as much voltage as the AC source can generate a time varying
current (di/dt).

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

(Dick Pierce) writes:

Yup, that's right, from about 55 Hz till above 100 Hz, this PASSIVE
circuit has voltage gain.

You have convinced me of the time varying wonders of reactive
elements. Your point is well made, and you're quite correct, in the
context of a speaker crossover network where an active amplifier is
pushing the circuit, you can boost the voltage of certain frequencies
at the expense of others with a reactive element. Power, of course is
still conserved.

VIN 1 0 AC SIN 1.0 0.0
L12 2 0 15MH

These are the components that make it happen--a perfect AC source with
infinite current sourcing capability, and and inductor that can
generate as much voltage as the AC source can generate a time varying
current (di/dt).

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

(Dick Pierce) writes:

Yup, that's right, from about 55 Hz till above 100 Hz, this PASSIVE
circuit has voltage gain.

You have convinced me of the time varying wonders of reactive
elements. Your point is well made, and you're quite correct, in the
context of a speaker crossover network where an active amplifier is
pushing the circuit, you can boost the voltage of certain frequencies
at the expense of others with a reactive element. Power, of course is
still conserved.

VIN 1 0 AC SIN 1.0 0.0
L12 2 0 15MH

These are the components that make it happen--a perfect AC source with
infinite current sourcing capability, and and inductor that can
generate as much voltage as the AC source can generate a time varying
current (di/dt).

Best Regards,
--
/"\ ASCII Ribbon Campaign | Todd H
\ / | http://www.toddh.net/
X Promoting good netiquette | http://triplethreatband.com/
/ \ http://www.toddh.net/netiquette/ | "4 lines suffice."

(Todd H.) wrote in message ...
(Svante) writes:

(Todd H.) wrote in message ...
(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

Hmm... Strictly, no. Consider this network:

U1 U0
----------Inductor----------------
| |
Capacitor Speaker
| |
----------------------------------

Oops! This should be:

U1 U0
---------Capacitor---------------
| |
Inductor Speaker
| |
---------------------------------

Let us for the sake of argument assume that the speaker is resistive.
The transfer function will be:


U0 s^2*LC
-- = -------------------
U1 1+s*L/R+s^2*LC

set s=jw and take its absolute value:


|U0| w^2*LC
|--| = -----------------------------
|U1| sqrt((w*L/R)^2+(1-s^2*LC)^2)

(Hope I got it right)

Obviously I got it wrong, should be:

|U0| w^2*LC
|--| = -----------------------------
|U1| sqrt((w*L/R)^2+(1-w^2*LC)^2)
^
|

If you can provide me a value of L, C, R and omega for which that
transfer function is 1 and I'll concede that you've managed to
_boost_ anything without the benefit of an active component.

Given my corections above, the output level at w=w0=sqrt(1/LC) will be
Q.
If Q1 so will the output voltage.

Example: R=8 ohm, L=6.4mH, C=400uF would give Q=2 and f0=100Hz
At 100 Hz the speaker voltage would be Q=2 times the the input
voltage, but the impedance seen by the amplifier would be only
R/(Q^2)=8/(2^2)=2 ohms, which of course would be near a disaster.

The network you've described is a 2nd order passive network that will
only attenuate high frequencies. It's not going to boost the lows
over the value they'd have in the absence of the network, unless you
rely on something really odd the amplifier is doing... but that'd be
cheating as the amplifier is a big box of active electronics. :-)

Yeah, my diagram described a LP filter, and my equations something
that I don't know what it was. I hope it is a HP filter now, and that
the equations are right. I don't rely on some odd amplifer phenomena,
it is just an effect of the reactive components (L&C).

It may seem strange that a passive network can boost voltage. The key
to understanding this is that it does not boost POWER. Since the input
impedance of the network is lowered the power drawn from the amplifier
is increased by the same amount as the power to the speaker in
increased. Reactive components cannot consume or produce (active)
power.

(Todd H.) wrote in message ...
(Svante) writes:

(Todd H.) wrote in message ...
(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

Hmm... Strictly, no. Consider this network:

U1 U0
----------Inductor----------------
| |
Capacitor Speaker
| |
----------------------------------

Oops! This should be:

U1 U0
---------Capacitor---------------
| |
Inductor Speaker
| |
---------------------------------

Let us for the sake of argument assume that the speaker is resistive.
The transfer function will be:


U0 s^2*LC
-- = -------------------
U1 1+s*L/R+s^2*LC

set s=jw and take its absolute value:


|U0| w^2*LC
|--| = -----------------------------
|U1| sqrt((w*L/R)^2+(1-s^2*LC)^2)

(Hope I got it right)

Obviously I got it wrong, should be:

|U0| w^2*LC
|--| = -----------------------------
|U1| sqrt((w*L/R)^2+(1-w^2*LC)^2)
^
|

If you can provide me a value of L, C, R and omega for which that
transfer function is 1 and I'll concede that you've managed to
_boost_ anything without the benefit of an active component.

Given my corections above, the output level at w=w0=sqrt(1/LC) will be
Q.
If Q1 so will the output voltage.

Example: R=8 ohm, L=6.4mH, C=400uF would give Q=2 and f0=100Hz
At 100 Hz the speaker voltage would be Q=2 times the the input
voltage, but the impedance seen by the amplifier would be only
R/(Q^2)=8/(2^2)=2 ohms, which of course would be near a disaster.

The network you've described is a 2nd order passive network that will
only attenuate high frequencies. It's not going to boost the lows
over the value they'd have in the absence of the network, unless you
rely on something really odd the amplifier is doing... but that'd be
cheating as the amplifier is a big box of active electronics. :-)

Yeah, my diagram described a LP filter, and my equations something
that I don't know what it was. I hope it is a HP filter now, and that
the equations are right. I don't rely on some odd amplifer phenomena,
it is just an effect of the reactive components (L&C).

It may seem strange that a passive network can boost voltage. The key
to understanding this is that it does not boost POWER. Since the input
impedance of the network is lowered the power drawn from the amplifier
is increased by the same amount as the power to the speaker in
increased. Reactive components cannot consume or produce (active)
power.

(Todd H.) wrote in message ...
(Svante) writes:

(Todd H.) wrote in message ...
(Ivan) writes:
I've built some speakers using Jordan JX92s full range drivers. To
boost the low end I plan to place a resistor/inductor network in
series with the driver (resistor and inductor in parallel).

Thought #1 (feel free to consider as nitpicky): You will not
successfully boost the low end using only passive components like
inductors and resistors. At best, you can attenuate the highs and
thereby make the low end appear more pronounced in comparison. To
truly boost frequency ranges, you need active components
(e.g. transistors).

Hmm... Strictly, no. Consider this network:

U1 U0
----------Inductor----------------
| |
Capacitor Speaker
| |
----------------------------------

Oops! This should be:

U1 U0
---------Capacitor---------------
| |
Inductor Speaker
| |
---------------------------------

Let us for the sake of argument assume that the speaker is resistive.
The transfer function will be:


U0 s^2*LC
-- = -------------------
U1 1+s*L/R+s^2*LC

set s=jw and take its absolute value:


|U0| w^2*LC
|--| = -----------------------------
|U1| sqrt((w*L/R)^2+(1-s^2*LC)^2)

(Hope I got it right)

Obviously I got it wrong, should be:

|U0| w^2*LC
|--| = -----------------------------
|U1| sqrt((w*L/R)^2+(1-w^2*LC)^2)
^
|

If you can provide me a value of L, C, R and omega for which that
transfer function is 1 and I'll concede that you've managed to
_boost_ anything without the benefit of an active component.

Given my corections above, the output level at w=w0=sqrt(1/LC) will be
Q.
If Q1 so will the output voltage.

Example: R=8 ohm, L=6.4mH, C=400uF would give Q=2 and f0=100Hz
At 100 Hz the speaker voltage would be Q=2 times the the input
voltage, but the impedance seen by the amplifier would be only
R/(Q^2)=8/(2^2)=2 ohms, which of course would be near a disaster.

The network you've described is a 2nd order passive network that will
only attenuate high frequencies. It's not going to boost the lows
over the value they'd have in the absence of the network, unless you
rely on something really odd the amplifier is doing... but that'd be
cheating as the amplifier is a big box of active electronics. :-)

Yeah, my diagram described a LP filter, and my equations something
that I don't know what it was. I hope it is a HP filter now, and that
the equations are right. I don't rely on some odd amplifer phenomena,
it is just an effect of the reactive components (L&C).

It may seem strange that a passive network can boost voltage. The key
to understanding this is that it does not boost POWER. Since the input
impedance of the network is lowered the power drawn from the amplifier
is increased by the same amount as the power to the speaker in
increased. Reactive components cannot consume or produce (active)
power.

(Todd H.) wrote in message ...


If you can provide me a value of L, C, R and omega for which that
transfer function is 1 and I'll concede that you've managed to
_boost_ anything without the benefit of an active component.

The network you've described is a 2nd order passive network that will
only attenuate high frequencies. It's not going to boost the lows
over the value they'd have in the absence of the network, unless you
rely on something really odd the amplifier is doing... but that'd be
cheating as the amplifier is a big box of active electronics. :-)



Here is a network that will boost the low end by 6 dB, (at 32 Hz)


------------ C = 5000 uF----------------------------
| |
L = 5 mH |
| 8 Ohms
R = 0.33 Ohms (coil) |
| |
----------------------------------------------------


The 0.33 Ohms is the *wire resistance* of the inductor. If you can get
the wire resistance lower, the boost will be even greater. (For
example, at 0.1 Ohms the low end boost would be 12 dB.)

Bob Stanton

(Todd H.) wrote in message ...


If you can provide me a value of L, C, R and omega for which that
transfer function is 1 and I'll concede that you've managed to
_boost_ anything without the benefit of an active component.

The network you've described is a 2nd order passive network that will
only attenuate high frequencies. It's not going to boost the lows
over the value they'd have in the absence of the network, unless you
rely on something really odd the amplifier is doing... but that'd be
cheating as the amplifier is a big box of active electronics. :-)



Here is a network that will boost the low end by 6 dB, (at 32 Hz)


------------ C = 5000 uF----------------------------
| |
L = 5 mH |
| 8 Ohms
R = 0.33 Ohms (coil) |
| |
----------------------------------------------------


The 0.33 Ohms is the *wire resistance* of the inductor. If you can get
the wire resistance lower, the boost will be even greater. (For
example, at 0.1 Ohms the low end boost would be 12 dB.)

Bob Stanton

(Todd H.) wrote in message ...


If you can provide me a value of L, C, R and omega for which that
transfer function is 1 and I'll concede that you've managed to
_boost_ anything without the benefit of an active component.

The network you've described is a 2nd order passive network that will
only attenuate high frequencies. It's not going to boost the lows
over the value they'd have in the absence of the network, unless you
rely on something really odd the amplifier is doing... but that'd be
cheating as the amplifier is a big box of active electronics. :-)



Here is a network that will boost the low end by 6 dB, (at 32 Hz)


------------ C = 5000 uF----------------------------
| |
L = 5 mH |
| 8 Ohms
R = 0.33 Ohms (coil) |
| |
----------------------------------------------------


The 0.33 Ohms is the *wire resistance* of the inductor. If you can get
the wire resistance lower, the boost will be even greater. (For
example, at 0.1 Ohms the low end boost would be 12 dB.)

Bob Stanton