Need help building a R-C fixed LPF for crossover

[Jim Saurman]

I need to wire up a fixed point crossover for some KF300 racks I have.
Its been so long since I have done this that my memory is rusty... so
someone please help me out and check my figures:

I'm coming out of a EAW MX250 processor, and normally the low will just
go to a power amp for the low end in the KF300, but I want to wire up a
switch that when flipped will split the signal, one will go thru a HPF
and go on to the power amp, the other end of the split will go thru a LPF
and go to XLR Panel Mounts for feeding another amp to drive subwoofers.

Input impedence on the Sub amp will be 12K (QSC PL236). I don't know
what the output impedence on the MX250 is, but I don't think that
matters.

Here is what I've got as a rough draft:

Crossover point 100hz
Slope 12db
R1=1000 ohms (does it matter what wattage these are? will 1/4 watt
work?)
C1=1.59 microfarads (maybe a 1 ufd and a .47 ufd wired in parallel?)
R2=10000 ohms
C2=.159 microfarads (I think they sell .150 ufd)

The diagram is here:
http://www.t-linespeakers.org/tech/filters/passiveHLxo.html

I'd like to keep the components as small as possible as I'm going to try
to fit them in one of those male-female XLR barrels that are about 4
inches long.

Anyone see any flaws so far?

Thanks in advance!

JIM

[Scott Dorsey]

Jim Saurman > wrote:
>I'm coming out of a EAW MX250 processor, and normally the low will just
>go to a power amp for the low end in the KF300, but I want to wire up a
>switch that when flipped will split the signal, one will go thru a HPF
>and go on to the power amp, the other end of the split will go thru a LPF
>and go to XLR Panel Mounts for feeding another amp to drive subwoofers.
>
>Input impedence on the Sub amp will be 12K (QSC PL236). I don't know
>what the output impedence on the MX250 is, but I don't think that
>matters.

The numbers you have are pretty much in the ballpark, but what you need
to watch out for is that there will be substantial loss through that
network and you may not have enough gain to make it up. Also, it will
be a very shallow filter slope.

Active filters will buy you a LOT in this application, and the Jung
Audio Op-Amp Cookbook has a bunch of designs you can just lift completely.
The sharper the slope, the less you have to worry about the speakers
interfering with one another over the crossover region because the narrower
the crossover region.

But build it and see. You may find that the amp input Z isn't what the
manufacturer says it is either, but that's easy to change. Toss it
together on perfboard, then actually measure it and see if the corner
frequency is really where you want it to be. Then tweak a shunt resistor
on the output until it is.
--scott

--
"C'est un Nagra. C'est suisse, et tres, tres precis."

[Chris Hornbeck]

On 24 Aug 2006 03:41:24 GMT, Jim Saurman >
wrote:

>Here is what I've got as a rough draft:
>
>Crossover point 100hz
>Slope 12db
>R1=1000 ohms (does it matter what wattage these are? will 1/4 watt
>work?)
>C1=1.59 microfarads (maybe a 1 ufd and a .47 ufd wired in parallel?)
>R2=10000 ohms
>C2=.159 microfarads (I think they sell .150 ufd)

One thing to keep in mind is that the amp's 12K ohm input
is in parallel with the crossover's 10K resistor, moving
the second pole up almost an octave.

>The diagram is here:
>http://www.t-linespeakers.org/tech/filters/passiveHLxo.html

The article discusses this issue very well; just a
matter of choosing a resistor for R2 that totals to
10K when paralled with 12K. Piece of cake.


FWIW, a second order Linkwitz-Riley high pass may or
may not be a good choice for your particular use. The
system's total transfer function includes the response
of the drivers, who are doing all kinds of whacky stuff
in this region.

Also note that an idealized LR2 crossover gives flat
magnitude (frequency response on axis) with the low and
high pass drivers in opposite "DC" polarity; that is
a battery across the terminals gives opposite cone travel,
(in vs. out).

The idealized case gives a perfect null (*no* summed output
on axis) with the same DC polarity. Most real world
loudspeaker systems are so amazingly far from ideal that
it's somewhat difficult to believe this based on an
actual test.

That's just part of what makes speakers so much fun.

All good fortune,

Chris Hornbeck
"History consists of truths which in the end turn into lies,
while myth consists of lies which finally turn into truths."
- Jean Cocteau

[Richard Nixon]

Chris Hornbeck > wrote in
:

> On 24 Aug 2006 03:41:24 GMT, Jim Saurman >
> wrote:
>
>>Here is what I've got as a rough draft:
>>
>>Crossover point 100hz
>>Slope 12db
>>R1=1000 ohms (does it matter what wattage these are? will 1/4 watt
>>work?)
>>C1=1.59 microfarads (maybe a 1 ufd and a .47 ufd wired in parallel?)
>>R2=10000 ohms
>>C2=.159 microfarads (I think they sell .150 ufd)
>
> One thing to keep in mind is that the amp's 12K ohm input
> is in parallel with the crossover's 10K resistor, moving
> the second pole up almost an octave.
>
>
> The article discusses this issue very well; just a
> matter of choosing a resistor for R2 that totals to
> 10K when paralled with 12K. Piece of cake.

So I should change it to an 8K resistor and work backwards to update C1,
R1, and C2 to the appropriate values, or should I leave C1, R1, and C2 as
is?



> FWIW, a second order Linkwitz-Riley high pass may or
> may not be a good choice for your particular use. The
> system's total transfer function includes the response
> of the drivers, who are doing all kinds of whacky stuff
> in this region.

Well this is for hotel A/V, and as much as I'd like to set up a better
system, I'm constrained by a budget of zero dollars. Anything would be
better than what we've got now. I'm not as worried as I would be if this
was for a nightclub or band.

>
> Also note that an idealized LR2 crossover gives flat
> magnitude (frequency response on axis) with the low and
> high pass drivers in opposite "DC" polarity; that is
> a battery across the terminals gives opposite cone travel,
> (in vs. out).

So when these are feeding the sub amp, the phase needs to be reversed?
I'd totally forgotten that. Thanks for that info.

[Chris Hornbeck]

On 25 Aug 2006 04:02:07 GMT, Richard Nixon >
wrote:

>> The article discusses this issue very well; just a
>> matter of choosing a resistor for R2 that totals to
>> 10K when paralled with 12K. Piece of cake.
>
>So I should change it to an 8K resistor and work backwards to update C1,
>R1, and C2 to the appropriate values, or should I leave C1, R1, and C2 as
>is?

The 8K number comes from a math error. We want something
in the small five-figure area. Shoot, just use 100K; plenty
close enough for loudspeaker work.

For that matter, you may be just as happy or even happier
with just a single series'd capacitor. 0.12uF should do
the trick, or something in that ballpark.

All good fortune,

Chris Hornbeck
"History consists of truths which in the end turn into lies,
while myth consists of lies which finally turn into truths."
- Jean Cocteau