crossover attenuation

[mr. jamie]

Hello, forum goers,

I have a pair of home built speakers, they are lovely, other than that, to my ears, they sound too bright.

I am looking into using an L-pad resistor system to cut a bit of the treble out.

The additive effect of the high range driver having an spl greater than that of the bass driver, inductor resistance and baffle step-up results in a need to get rid of a few decibels above the treble crossover point.

I am quite apt with mathematics, and have found equations with which to choose the resistors I need.

However, these equations only seem to take into consideration the attenuation caused by the resistor in series. The resistor in parrellel only being there to retain the original impedance.

This seems a little flawed. Surely current is "leaking" though the parallel resistor, and hence the driver is recieving less power than calculated from the series resistor alone.

Or is this current "leak" insignificant.

Any comments or advice would be very much appreciated.
I would also like to hear about any crossover mods of the same variety that you have done.

Thank you,
Jamie

[tinitus]

Originally posted by mr. jamie
Hello, forum goers,

However, these equations only seem to take into consideration the attenuation caused by the resistor in series.

The resistor in parrellel only being there to retain the original impedance.

Jamie

So it is

But you can "fineadjust" by ear, and choose whatever sounds best

[mr. jamie]

http://sound.westhost.com/lr-passive.htm#s6.0

This webpage has the general eqautions I used.

[Dr. Ram]

Originally posted by mr. jamie

This seems a little flawed. Surely current is "leaking" though the parallel resistor, and hence the driver is recieving less power than calculated from the series resistor alone.

Or is this current "leak" insignificant.
Thank you,
Jamie

The parallel resistor effect need not be insignificant (I can have a large L-pad attenuation), and it is the overall combination that is designed in a way so as to present the same original impedance looking into the L-Pad. I see that the page you referred to starts with the dB attenuation required and then works out the series and parallel resistor values - which is the standard approach of course. Not sure where you found a paradox.

Can you narrow down the apparent flaw to a specific equation or sentence?

-Ram

[mr. jamie]

After the db attenuation needed is calculated, the value of a resistor in series (to cause the needed attenuation) is calculated.
Although this gives the attenuation required it changes the impedance (and thus the crossover frequency), to retain the impedance a resistor of a specific value must be placed in parellel to the driver.

This is all well and good. Nice and simple.

However, the resistor in parellel will alter the attenuation as some current will move though it rather than the driver. This "short circuit" effect seems to be ignored by the equations.

Is the extra attenuation caused by the parellel resistor small enough to be ignored?

[tinitus]

In that sense you are right though

A proper L-pad is supposed to attenuate better than a simple series resistor
If only a series resistor is used, it would be bigger than the series resistor of a L-pad

Or at least, so I am told

[Dr. Ram]

Originally posted by mr. jamie

However, the resistor in parellel will alter the attenuation as some current will move though it rather than the driver. This "short circuit" effect seems to be ignored by the equations.

Is the extra attenuation caused by the parellel resistor small enough to be ignored?

No it is not ignored. The series resistor has been calculated using the known net attenuation across the parallel combination of Rp and Z. So the effect of Rp is not ignored at all. The calculations are correct. Once you have figured the series resistor (or, equivalently, the parallel combination impedance), you calculate Rp since you know the parallel combination and know Z.

[Dr. Ram]

Originally posted by tinitus
In that sense you are right though

A proper L-pad is supposed to attenuate better than a simple series resistor
If only a series resistor is used, it would be bigger than the series resistor of a L-pad

Or at least, so I am told

Well - the idea is that if you vary a simple series resistor you also change the overall input impedance. If you vary an L-pad you keep the overall input impedance constant and equal to the original driver impedance (assumed constant with a conjugate network etc but that is a different story).

[mr. jamie]

Very well. I'm still not quite sure I get it, but thank you.

Another thing that has been intrigueging me, if anyone has the time to indulge,

Ignoring the change in impedance, how would placing the series resistor (value calculated for traditional L-pad) between the parallel resistor (value calculated for traditional L-pad) and the driver effect attentuation?
(i.e the parallel resistor would effectively be parallel to both the driver and the series resistor)

[infinia]

If you rearranged Rp to the input side of Rs it would not affect atten. , remembering the goal (importantly) is not to change the impedance seen by the crossover network.

The atten would be 20LOG * Z/(Rs+Z) where Z is the imp of the tweeter in ohms

Note> This is only true when using audio amps that are constant voltage (ie voltage feedback). Which is the case for the majority of audio products.