0
\$\begingroup\$

I'm using rf-tools.com LC filter designer to design a passive 4 pole filter for an active subwoofer.

The response is not what I expect, it's not showing any attenuation which seems to be good to be true with the input and output resistors. Maybe the resistors are not included in the simulation for some reason?

I would like to keep the input impedance of filter as high possible to minimise load on the source, as long as the resistor values are kept equal it doesnt seem to add any attenuation, only inductor value increases which will be the limiting factor. Inductors are something I dont understand very well and I know they can store energy so would like to confirm what's going and If i'm interpreting the sims incorrectly.

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ I've had previous encounters with rf-tools where they didn't include attenuations, and also sometimes, they presumed the corner frequency to be not what I wanted. So their graphs may be correctly plotted, but off in numbers. \$\endgroup\$ Commented May 28, 2021 at 15:25
  • \$\begingroup\$ I'm guessing the program is normalizing the filter output response to make it easier to evaluate the response. If you simulate this in Spice, you'll see that the low-freq gain is -6dB. Blindly accepting what comes out of a tool can get you in trouble if you don't understand how the tool operates. Fortunately, you questioned the tool. \$\endgroup\$
    – qrk
    Commented May 28, 2021 at 17:08
  • \$\begingroup\$ Ok, so not accounting for attenuation would explain the behaviour if the values are kept at the default equally weighted values, but the effects from changing these values are making less sense. If I treat the resistors as a voltage divider and increase the output resistor value to lower attenuation then the graph starts showing increasing attenuation??? it is the opposite of the intended effect. Luckily getting a high input impedance is not really a problem, just requiring larger inductors, but I have no idea what the output level of the filter will be with these confusing results. \$\endgroup\$
    – Jay
    Commented May 28, 2021 at 17:47
  • \$\begingroup\$ It's common in passive filter design texts to compare the attenuation to the minimum attenuation caused by Rs. An active filter - even a DSP filter - in front of the bass speaker amp would give to you much more flexibility to take into the account how the speaker + its cabinet actually work in different frequencies. Their frequency response is anything but flat. \$\endgroup\$
    – user136077
    Commented May 28, 2021 at 19:41

1 Answer 1

1
\$\begingroup\$

LC filters are a pain in this regard: you need to account for all the possible loads, in and out, and once calculated they stay that way. Therefore a change in either the input or the output needs recalculating the values.

Then you have an even order Chebyshev. As you probably know, these filters have the peculiarity of equiripples in the passband (or stopband for inverse kind). For active filters, this is not a problem, but passive filters are tricky: even orders don't have unity gain at DC (see this answer for more details), and that means there is a maximum value over which the response is no longer a Chebyshev.

But then, this line in your OP seems strange:

design a passive 4 pole filter for an active subwoofer.

If it's active, you don't need to use LC filters, use active filters which avoid the use of inductors. This is how active speakers are built: with a filter at the input, where the signal doesn't need power. Because, honestly, I don't know how you'll build those values in practice: 800 mH and 1.3 H are huge, unless you consider adding a magnetic core, but then there are all sorts of nonlinearities since this is audio, so you'll need an "air" core (i.e. no core). And having a 1 H inductance will take lots of turns, which means lots of parasitic capacitance, and space/weight.

So, rf-tools is tricky, therefore always verify the design with some SPICE tool (plenty free available, one right here on this site). And if you choose to go with passive LC filters, know that an LC filter is supposed to be lossless, therefore the attenuation is given by the resistive divider formed by the input and output resistances. Ideally, because in practice the inductors will have a resistance and capacitors, too, will contribute with their ESR.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.