Ivan wrote:


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

Close enough for usenet.


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


What does L = 2 mH have to do with it? You seem to be
confused about inductance and inductive impedance.

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.

So far, so good ...

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?

Sorry, no. For one thing, a pure inductor doesn't dissipate
power. Secondly, when you are on the path of AC
enlightenment, you'll be using complex math; that is, numbers
with real and imaginary parts.

Ivan wrote:


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

Close enough for usenet.


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


What does L = 2 mH have to do with it? You seem to be
confused about inductance and inductive impedance.

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.

So far, so good ...

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?

Sorry, no. For one thing, a pure inductor doesn't dissipate
power. Secondly, when you are on the path of AC
enlightenment, you'll be using complex math; that is, numbers
with real and imaginary parts.

Ivan wrote:


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

Close enough for usenet.


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


What does L = 2 mH have to do with it? You seem to be
confused about inductance and inductive impedance.

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.

So far, so good ...

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?

Sorry, no. For one thing, a pure inductor doesn't dissipate
power. Secondly, when you are on the path of AC
enlightenment, you'll be using complex math; that is, numbers
with real and imaginary parts.

(Ivan) wrote in message . com...
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?

No. First, ideal inductors do not dissipate power, they just store it
for part of the oscillatory cycle and then deliver it back to the
network it is connected to. However, real inductors have a series
resistance and THAT can consume power and make the temperature raise.
So, you cannot calculate the active power dissipated in a coil from
its inductance, you must know its resistance. Assuming that the
resistance of the coil is low, the maximum current that will float
through the coil is 14/4=3.5 ampere. This is the CURRENT the coil
should be able to handle.

Second, your speaker system would be rated as a 4 ohm system, since
the impedance at low frequencies (400Hz) will be 4 ohms (assuming a
purely resistive driver, which may actually be far from the case).
This means it can be connected to an amplifier that can deliver 50 W
into 4 ohms. This corresponds to a voltage from the amplifier of
U=sqrt(P*R)=14 volts. Toward higher frequencies (900 Hz) 14 volts
would produce 14/9=1.6 ampere in 4+5=9 ohms. 1.6 ampere would
dissipate 1.6*1.6*5=12.8 watts in 5 ohms.

Third, at high frequencies, music does not contain as much energy as
at low frequencies, which means that the power estimate above is
probably too high.
Another example of this is tweeters, that usually only can take about
1/10 of the rated system power. That is why they mostly specify a
crossover frequency and a maximum "system power", rater than the
actual power that will burn the tweeter.

(Ivan) wrote in message . com...
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?

No. First, ideal inductors do not dissipate power, they just store it
for part of the oscillatory cycle and then deliver it back to the
network it is connected to. However, real inductors have a series
resistance and THAT can consume power and make the temperature raise.
So, you cannot calculate the active power dissipated in a coil from
its inductance, you must know its resistance. Assuming that the
resistance of the coil is low, the maximum current that will float
through the coil is 14/4=3.5 ampere. This is the CURRENT the coil
should be able to handle.

Second, your speaker system would be rated as a 4 ohm system, since
the impedance at low frequencies (400Hz) will be 4 ohms (assuming a
purely resistive driver, which may actually be far from the case).
This means it can be connected to an amplifier that can deliver 50 W
into 4 ohms. This corresponds to a voltage from the amplifier of
U=sqrt(P*R)=14 volts. Toward higher frequencies (900 Hz) 14 volts
would produce 14/9=1.6 ampere in 4+5=9 ohms. 1.6 ampere would
dissipate 1.6*1.6*5=12.8 watts in 5 ohms.

Third, at high frequencies, music does not contain as much energy as
at low frequencies, which means that the power estimate above is
probably too high.
Another example of this is tweeters, that usually only can take about
1/10 of the rated system power. That is why they mostly specify a
crossover frequency and a maximum "system power", rater than the
actual power that will burn the tweeter.

(Ivan) wrote in message . com...
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?

No. First, ideal inductors do not dissipate power, they just store it
for part of the oscillatory cycle and then deliver it back to the
network it is connected to. However, real inductors have a series
resistance and THAT can consume power and make the temperature raise.
So, you cannot calculate the active power dissipated in a coil from
its inductance, you must know its resistance. Assuming that the
resistance of the coil is low, the maximum current that will float
through the coil is 14/4=3.5 ampere. This is the CURRENT the coil
should be able to handle.

Second, your speaker system would be rated as a 4 ohm system, since
the impedance at low frequencies (400Hz) will be 4 ohms (assuming a
purely resistive driver, which may actually be far from the case).
This means it can be connected to an amplifier that can deliver 50 W
into 4 ohms. This corresponds to a voltage from the amplifier of
U=sqrt(P*R)=14 volts. Toward higher frequencies (900 Hz) 14 volts
would produce 14/9=1.6 ampere in 4+5=9 ohms. 1.6 ampere would
dissipate 1.6*1.6*5=12.8 watts in 5 ohms.

Third, at high frequencies, music does not contain as much energy as
at low frequencies, which means that the power estimate above is
probably too high.
Another example of this is tweeters, that usually only can take about
1/10 of the rated system power. That is why they mostly specify a
crossover frequency and a maximum "system power", rater than the
actual power that will burn the tweeter.

(Todd H.) wrote in message ...
(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.

Well, it might seem obvious, but if the signal ain't time varying,
then the components ain't reactive, eh?

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.

Well, specifically, energy is conserved, but since energy is power
integrated over time ...

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).

It need not be a "perfect AC source," merely a reasonable
approximation thereof. Insert a small resistance in series with
the voltage source, oh, 0.5 ohms, and similarly with the inductor.
You won't get as high a boost, but you will get one still.

(Todd H.) wrote in message ...
(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.

Well, it might seem obvious, but if the signal ain't time varying,
then the components ain't reactive, eh?

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.

Well, specifically, energy is conserved, but since energy is power
integrated over time ...

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).

It need not be a "perfect AC source," merely a reasonable
approximation thereof. Insert a small resistance in series with
the voltage source, oh, 0.5 ohms, and similarly with the inductor.
You won't get as high a boost, but you will get one still.

(Todd H.) wrote in message ...
(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.

Well, it might seem obvious, but if the signal ain't time varying,
then the components ain't reactive, eh?

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.

Well, specifically, energy is conserved, but since energy is power
integrated over time ...

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).

It need not be a "perfect AC source," merely a reasonable
approximation thereof. Insert a small resistance in series with
the voltage source, oh, 0.5 ohms, and similarly with the inductor.
You won't get as high a boost, but you will get one still.

(Ivan) wrote in message . com...
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, it's all reasonably correct but, to summarize the comments
of others, so what?

The circuit you propose provides about an 8 dB "boost" starting
at about 127 Hz. In fact, it does so by providing a "lack" of
an 8 dB drop in efficiency below that frequency.

Unfortunately, your scheme has lots of deleterious side effects,
not the least of which is a broadband 8 dB loss in efficiency.
I doubt that, on these units, you're going to find such a gross
loss in broadband efficency as tolerable.

Secondly, you are ignoring the fact that with such a large series
resistance, the frequency response of the majority is now more
more dependent on the frequency-dependent impedance of the speaker
than before. For example, you may find your scheme backfires
because the impedance of the driver rises at higher frequencies
thus negating your circuit.

And, again to echo the comments of others: a silk purse made from
a sow's ear makes for both a poor ear and a lousy purse. The end
result is useless and serves to really **** off the sow.

(Ivan) wrote in message . com...
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, it's all reasonably correct but, to summarize the comments
of others, so what?

The circuit you propose provides about an 8 dB "boost" starting
at about 127 Hz. In fact, it does so by providing a "lack" of
an 8 dB drop in efficiency below that frequency.

Unfortunately, your scheme has lots of deleterious side effects,
not the least of which is a broadband 8 dB loss in efficiency.
I doubt that, on these units, you're going to find such a gross
loss in broadband efficency as tolerable.

Secondly, you are ignoring the fact that with such a large series
resistance, the frequency response of the majority is now more
more dependent on the frequency-dependent impedance of the speaker
than before. For example, you may find your scheme backfires
because the impedance of the driver rises at higher frequencies
thus negating your circuit.

And, again to echo the comments of others: a silk purse made from
a sow's ear makes for both a poor ear and a lousy purse. The end
result is useless and serves to really **** off the sow.

(Ivan) wrote in message . com...
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, it's all reasonably correct but, to summarize the comments
of others, so what?

The circuit you propose provides about an 8 dB "boost" starting
at about 127 Hz. In fact, it does so by providing a "lack" of
an 8 dB drop in efficiency below that frequency.

Unfortunately, your scheme has lots of deleterious side effects,
not the least of which is a broadband 8 dB loss in efficiency.
I doubt that, on these units, you're going to find such a gross
loss in broadband efficency as tolerable.

Secondly, you are ignoring the fact that with such a large series
resistance, the frequency response of the majority is now more
more dependent on the frequency-dependent impedance of the speaker
than before. For example, you may find your scheme backfires
because the impedance of the driver rises at higher frequencies
thus negating your circuit.

And, again to echo the comments of others: a silk purse made from
a sow's ear makes for both a poor ear and a lousy purse. The end
result is useless and serves to really **** off the sow.

(Dick Pierce) wrote in message

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):

--------Snip

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!

Could you also, for the sake of the poster's amplifier, also post the
electrical input impedance of the cirquit?

(Dick Pierce) wrote in message

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):

--------Snip

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!

Could you also, for the sake of the poster's amplifier, also post the
electrical input impedance of the cirquit?

(Dick Pierce) wrote in message

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):

--------Snip

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!

Could you also, for the sake of the poster's amplifier, also post the
electrical input impedance of the cirquit?

(Ivan) wrote in message


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





The circuit you drew will make a nice 7 dB shelf response. The
disadvantage is you will lose 55% of the power from the amplifier. The
2 mH inductor seems a little low. That would make the 3 dB "boost"
frequency 300 Hz, and the 6 dB "boost" frequency 100 Hz.


Bob Stanton

(Ivan) wrote in message


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





The circuit you drew will make a nice 7 dB shelf response. The
disadvantage is you will lose 55% of the power from the amplifier. The
2 mH inductor seems a little low. That would make the 3 dB "boost"
frequency 300 Hz, and the 6 dB "boost" frequency 100 Hz.


Bob Stanton

(Ivan) wrote in message


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





The circuit you drew will make a nice 7 dB shelf response. The
disadvantage is you will lose 55% of the power from the amplifier. The
2 mH inductor seems a little low. That would make the 3 dB "boost"
frequency 300 Hz, and the 6 dB "boost" frequency 100 Hz.


Bob Stanton