Hi,
I'm looking for some help regarding shelving high pass filter design.
I've got an open baffle speakers, with two large (identical) woofers and a small full range driver. Two LF woofers combined are about 5 dB more sensitive, compared to the full range.
Simulating in VituixCad, high pass shelf with -6dB (at around 100ish Hz) output lowers the SPL of LF woofers to FR's level, while not altering the frequency below 100Hz.
The idea is to avoid a simple driver attenuation (that would require then significant bass boost - since it's an open baffle roll-off), but to simply cut the LF driver excess SPL.
Is is possible to design such an active filter, using op amps?
I'm looking for some help regarding shelving high pass filter design.
I've got an open baffle speakers, with two large (identical) woofers and a small full range driver. Two LF woofers combined are about 5 dB more sensitive, compared to the full range.
Simulating in VituixCad, high pass shelf with -6dB (at around 100ish Hz) output lowers the SPL of LF woofers to FR's level, while not altering the frequency below 100Hz.
The idea is to avoid a simple driver attenuation (that would require then significant bass boost - since it's an open baffle roll-off), but to simply cut the LF driver excess SPL.
Is is possible to design such an active filter, using op amps?
Indeed, but if you take a look at math on one of attachments below HPS filter - I don't see how to make a negative version.
Edit - is the inverting variant used for that?
What is 'a negative version' ? Do you mean a circuit which applies only attenuation to LF and has unity gain at HF?
I don't need an increase in SPL above the set frequency, but a decrease - a gain of - 6 dB.
Edit:
No, I need a circuit that attenuates the high pass section.
The only way to get attenuation with an opamp stage is to use the inverting configuration. So you would need an inductor in the feedback path to get lower gain at LF. The inductor value might not be a practical one though, given this is at LF.
This gives -3dB at 100Hz, -6dB around 10Hz.
This gives -3dB at 100Hz, -6dB around 10Hz.
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@Dennnic , could you sketch the desired frequency response, just to make sure there won't be any miscommunication about what you want?
It isn't clear to me what you mean by this sentence. Do you want to cascade the filter discussed in this thread with a bass boosting filter? If so, maybe there is an elegant to combine them. If applicable, could you also sketch the combined response?
Sorry for the delay... Here's a screenshot from Vituixcad:
I tweeked the filter in the simulator a bit, turns out, it requires quite an attenuation. FR looks quite messy, its taken outside, with no gating.
Thank you for drawing the schematics.
I tweeked the filter in the simulator a bit, turns out, it requires quite an attenuation. FR looks quite messy, its taken outside, with no gating.
Is that 1.4 Henrys? I'd need to get a bigger case for that 😆 I think I saw Linkwitz's op amp replacement for an inductor, however, it really gets complex then.
Thank you for drawing the schematics.
It is indeed - but its still relatively a practical value in this circuit given its very low Q. I would guess (though haven't done the math) that it would be achievable in a P18 gapped ferrite core (18mm diameter).
Here's a potential alternative:
Increasing the FR's gain by 5 dB. However, if you take a look at the downward LF slope, you can see that 45 Hz (anticipated speaker extension) response needs up to 9 dB of boost in order to achieve smooth bass response. Compared to previous post - double the gain needed.
Since its a digital source (Volumio), final FR (boosting and adjustment for current room) is planned to be achieved digitaly, prior to preamp. Still, I've got around 80w for LF woofers.
So you are looking for a first-order filter with 0 dB gain at 0 Hz, -10 dB at high frequencies and a -5 dB point at 150 Hz. The simplest way to make that is a resistive voltage divider with 10 dB of attenuation and with a capacitor in series with the bottom resistor. Depending on the output impedance of the source and the input impedance of the amplifier, you may or may not need buffers before and after it. If the buffers are needed, you end up with something like this:
Due to rounding to E12 values, E6 for the capacitor, your 10 dB has changed to 9.834415266 dB and the frequency where the attenuation (in dB) is half of that has shifted to 154.45568 Hz (theoretical values, obviously).
If the amplifier is DC coupled with no DC servo, you may also need a DC blocking capacitor at the output.
What I just described does not invert the polarity of the signal. Is there a requirement that it does? If so, why not swap the loudspeaker terminals?
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Or do you need the opposite, -10 dB at 0 Hz, 0 dB at high frequencies? You can make that with a resistive voltage divider with the upper resistor shunted by a capacitor. With buffers and everything, it becomes something like this. Attenuation 10.08974878 dB, halfway at 158.0488521 Hz.
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The first option - here's a more simple way of describing it:
That exact filter would make wonders with my setup. Or with any system, that includes multiple woofers in an open baffle configuration.
Resistive voltage divider seems like a rather simple solution, is there any ''catch'' to it? I haven't seen it in use in active filters. First order might be too shallow to counter the rising LF response, and it might require to move the crossover point somewhat lower. No need of signal inversion, at least that I know of.
So now it's 0 dB at 0 Hz, -8 dB at high frequencies, -4 dB at 100 Hz.
Chances are that you don't need the input buffer around U5B. A typical active Linkwitz-Riley filter can drive the 15 kohm - 10 kohm - 100 nF correction circuit directly. The theoretical value for the difference between the gains for low and high audio frequencies is 7.958800173 dB and it is theoretically halfway (in dB) at 100.6584242 Hz.
Chances are that you don't need the input buffer around U5B. A typical active Linkwitz-Riley filter can drive the 15 kohm - 10 kohm - 100 nF correction circuit directly. The theoretical value for the difference between the gains for low and high audio frequencies is 7.958800173 dB and it is theoretically halfway (in dB) at 100.6584242 Hz.
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Does it always provide a 6 dB / oct slope - similar to one with a first order crossover? Does it affect phase response?
Please note that I updated the values, the round-off errors are much smaller with 15 kohm, 10 kohm, 100 nF.
The theoretical asymptotic slope is -6 dB/octave (or -6.020599... dB/octave to be precise), but it never gets anywhere near the asymptotic slope when the difference between the low- and high-frequency gain is only 8 to 10 dB. The phase response is affected, but I don't know if the overall phase response gets better or worse.
In any case, according to post #10, the IIR filter you use for your simulation is first order. It should behave almost exactly the same as the circuit. Or is the purple line of post #15 not coming from a first-order IIR filter?
The theoretical asymptotic slope is -6 dB/octave (or -6.020599... dB/octave to be precise), but it never gets anywhere near the asymptotic slope when the difference between the low- and high-frequency gain is only 8 to 10 dB. The phase response is affected, but I don't know if the overall phase response gets better or worse.
In any case, according to post #10, the IIR filter you use for your simulation is first order. It should behave almost exactly the same as the circuit. Or is the purple line of post #15 not coming from a first-order IIR filter?
If you need something steeper, you could try a biquad of some kind. The Linkwitz transform circuit is a popular one, but there are many ways to make biquads.