passive EQ with boost/cut and equal Q

[rafafredd]

I am working on a mastering passive EQ circuit for my friends studio.

So I am studying and doing some simulations with passive eqs, and all the circuits I tried had this same issue...

I am trying to have a EQ that all bands can be switched to boost or cut, like Manley's Massive Passive, Knif Soma and so many others.

The problem is that all circuits I use, based on Pultec variations, NYDave, RuffRecords, etc, have the same Q issue. It works good on boost, but when switched to cut, Q turns very steep, high Q curves.

So I found some curves of the commercial EQs and none of them has this problem. The Qs of the boost and cut seems exactly the same in these.

Do any of you know what circuit does they use for maintaining the Q on switchable boost/cut EQ? Maybe just a direction?

I came up with a solution by isolating boost and cut, and working with different impedance (and reacende) with an amplifier in between. But it is an expensive solution, and still not a perfect one, as the Qs still don't match perfectly, but are close.

What bothers me is that Manley and others seems to get perfectly mirrored Qs for boost and cut with a much simpler approach.

Thanks a lot for this.

 

[ruffrecords]

This effect occurs in many passive EQs.  In general, a passive EQ us based arounf an attenuator. Imagine a 20dB attenuator consisting of a 9K in series with a  1K resistor to ground.  The input is fed to the yop of the 9K and the output is from the junction of the 1K and the 9K. The attenuator is followed by a 20dB amplifier so the circuit has overall unity gain.

Boost is achieved by a frequency selective network connected across the 9K. If the network is a series LC the you get a peaking response and the Q you get depends on L and the 9K. In many passive EQ, the same LC circuit is switched across the 1K resistor to achieve cut. In this case the Q of the dip is determined by the vle of L and the 1K. For this reason, the Q of the dip is typically 6 to 9 times that of the boost.

To  achieve the same Q for cut as for boost, you need to use different values of LC for boost and cut and each frequency.  If you use the same ones for boost and cut the Q will change. My Pultec 3 band EQ and my Helios EQ both use the same inductor for boost and cut. My REDD EQ uses different ones for boost and cut so the Q is the same for boost and cut.

Cheers

Ian

 

[rafafredd]

Thanks for this. I was with the impression that could be some "magic circuit" that would address this.

So I am on the right track...

Regards.

 

[rafafredd]

By looking at this again, a constant Q EQ could be built with single filters and only 6dB boost and cut...

 

[abbey road d enfer]

You may switch the LC between a series configuration and a shunt configuration. But the impedances around them will be seriously diferent. See attached.

 

[ruffrecords]

By looking at this again, a constant Q EQ could be built with single filters and only 6dB boost and cut...

Exactly, because that is the only case where the resistors are equal..

Cheers

Ian

 

[joaquins]

If 6dB is good enough for your application (which may be, in mastering 6dB may be enough, depending on the application, but certainly useful)

The easiest way I found to build that in a circuit is the one I posted on the universal EQ thread, I made a proto and measure but it was about 14dB max boost/cut and the plots will show there the difference between one side and the other. For 6dB you just need to use two equal resistors as the attenuator and linear pots, for the plots shown I used log pots and matched the attenuation at mid point. Those are measurements of the actual circuit, so losses in the inductor are seen if you look at the difference in max boost vs max cut and a little shift in frequency. Also, due to difference in impedance HF shelf freq is changed quite a bit. Making it for 6dB max boost/cut should minimize those effects. If using rotary switches instead of pots, leaving out the attenuator and using the mid point at each switch instead would give perfect null at 0 position without need of further trim.

http://groupdiy.com/index.php?topic=50484.msg640488#msg640488

I don't know the behavior of the circuits like API553 but it may be more closer to what you are looking for... Try a sim on that to see what happens, is not passive but it uses LC filters.

JS

 

[Monte McGuire]

There's another way to look at that asymmetric behavior as 'more correct' than a symmetric boost cut curve. What the passive EQ designs do is filter the original signal with a second order resonant filter, with the Q of your choosing (Q being defined in the proper electrical engineering sense, how damped that second order resonator is, and not the 3dB points of the completed filter).

If you then take the output of the second order resonator and add or subtract it from the flat signal, you get the same curves as the Pultec, and as you note, the composite response curve is not symmetric with boost and cut.

However, let's say you're trying to tame a resonance in a cheap acoustic guitar. If you can get the second order filter to have the same center frequency and Q as the resonance in the cheap guitar, and you subtract that from your guitar recording, you've actually corrected the resonance perfectly, in amplitude _and_ phase. An EQ with symmetric curves will make the resonance smaller, but it cannot perfectly compensate for it.

So, in many ways, symmetric boost and cut curves are not really symmetric, they're just pretty to look at, on a log-log scale, they seem to be complementary, and it's easy for some circuits to produce that response. IMHO the additive/subtractive 2nd order resonator designs, like the Pultec, are the most ideal, especially for taming LF problems that result from resonances.

 

[joaquins]

Symmetric is a strong word, this usually mean same Q for boost and cut, constant Q Symmetric EQs is not what we usually look for.

I think he is looking for mirrored filters, symmetric filters look thiner on cut than boost, for example API550 has mirrored filters and proportional Q, which is what we are most used to see in graph on plugins, which tends to make the EQ more usable, the more you cut/boost the higher the Q gets, and it's useful since in extreme use you are probably trying to correct some problem, while soft use is more used for shaping of the sound, where lower Q is usually better.

JS

 

[abbey road d enfer]

However, let's say you're trying to tame a resonance in a cheap acoustic guitar. If you can get the second order filter to have the same center frequency and Q as the resonance in the cheap guitar, and you subtract that from your guitar recording, you've actually corrected the resonance perfectly, in amplitude _and_ phase.

If I understand correctly, you're saying that in order to compensate for the typical resonance of a "cheap guitar", one needs a rather narrow cut, with which I tend to agree.

An EQ with symmetric curves will make the resonance smaller, but it cannot perfectly compensate for it.

I believe you presume that a symmetrical EQ will not be capable of providing the narrow cut.

So, in many ways, symmetric boost and cut curves are not really symmetric,

? Having tested dozens of them, I know that most symmetrical B/C EQ's ARE really symmetric.

they're just pretty to look at, on a log-log scale, they seem to be complementary, and it's easy for some circuits to produce that response.

Not only pretty to look at; they are indeed complementary; identical boost and cut sum flat.

IMHO the additive/subtractive 2nd order resonator designs, like the Pultec, are the most ideal, especially for taming LF problems that result from resonances.

They would be ideal if and only if the actual BW of the cut was perfectly complementary with the BW of the resonance. What are the odds for a guitar resonance to be exactly the complementary of a Pultec cut response?
IMNSHO, the only ideal substractive EQ is parametric, where one can precisely tune the center frequency and width, in addition with depth, of course.