Q of passive audio equaliser

Hello Ian,

The LC series circuit "sees" the upper and lower resistor in parallel. This means when Rupper or Rlower (or both) are zero, the Q factor will be infinite (assuming lossless L and C).

So the Q factor of the loaded LCcircuit will be: Q = 2*pi*f*L/(Rp) Rp = Rupp parallel to Rlow (that is Rp = 0.5*Rlow, when Rlow = Rupper).

However this Q factor does not directly relate to the frequency response. So the -3dB BW of your circuit is not equal to BW = (center frequency)/Q. This can be seen when Rupper = 0 and Rupper > 0 (all- pass filter).

Given your circuit, when you double the value of L (and halve the value of C to enable same resonant frequency), the width of the peak halves.

Best regards,

Wim PA3DJS

the email address is OK, but without abc

Yes, that was my first thought but that still gives the wrong Q value.

I think you meant the second to be Rlower, but I understand what you are saying.

Yes, I already realised that. What I don't understand is how the Q of the filter relates to the circuit values.

I have just simulated the first example again and I suspect the problem arises because the 3dB points turn out not to be symmetrical about the resonant frequency. The simulation yields a resonant frequency close to

3.1KHz and -3db points at about 1.1KHz and 8.5KHz which gives a Q of about 0.42 (assuming Q = deltaf/f). Feeding this back in to find the effective resistance of the RLC gives R = 6974 ohms which is rather higher than the expected 5000 ohms.

I am still puzzled.

Cheers

Ian

Wim, I just changed Rlower to 1K which gives simulation results much closer to what I expected i.e Q = 3, (resonance near 3KHz, -3dB points at 2.5K and 3.5K respectively) and if you calculate Q from w*L/R where R ie 10K//1K you get Q = 3.

Cheers

Ian

.--||---UUU--. ____ Vin o--o ____ o----|____|----. '---|____|---' | === gnd

Like this? How about simple circuit analysis?

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Try this (for a starter)...

...Jim Thompson

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| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
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I love to cook with wine.     Sometimes I even put it in the food.

Jim touches on something in that PDF that I thought but didn't post -- what are you _really_ trying to do? Are you truly trying for an all- passive equalizer with lots of sliders, or are you just looking for a tone control?

'cause if it's a one- or two-knob tone control you want, do a web search on "tone control" -- this is a long solved problem, and there are various solutions with varying compromises between sound/cost/difficulty/etc.

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The analysis is simple enough. Deriving the effective Q is not. That's what I need help with really.

Cheers

Ian

Baxandall, as in

Better results (and less interaction) can be had by going "active".

Once you go active, "octave" equalizers are easy to do. ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.

What I am really trying to do is to understand how to determine the circuit parameters of that network given R, centre frequency and the Q of the final response.

I am not looking to make a passive equaliser with lots of sliders and I do not want a 'tone control'.

That topology is one element at the heart of several passive audio equalisers that are much revered in the pro audio world (with various ratios of the pot divider). I want to understand them well enough to be able to design my own.

Cheers

Ian

Thanks Jim, a very elegant analysis. As you rightly point out, defining Q is not straightforward but in this context I mean it to be the centre frequency divided by the half power bandwidth of the network. So how do I determine Q from the circuit parameters?

Cheers

ian

Well, asymptotic gain is 1/2, as can be seen from the circuit or its transfer function. That's not the textbook case of Q, where the band edges drop off towards 0 asymptotically. It's also not quite a parallel or series resonant circuit, though textbooks can still define Q for that case.

If you want to brute force it, of course, you can just solve for H = sqrt(2)/2 and see where that goes. Downside is solving the polynomial roots, but it's only quadratic.

Solving for s, I got s = -1.103*R / L +/- sqrt(1.218*R^2 / L^2 - 1/LC)

Using values of R = 10k, L = 1.5H and C = 1.8nF (mind the inductor may have as much parallel capacitance itself), I got s = -7353 +/- sqrt(54.13M - 370.4M) = -7353 +/- j17783 Hmm, that looks more like ordinary poles than a frequency you're supposed to have. Maybe it's sideways? A center frequency of 18kHz with bandwidth

+/-7.3kHz wouldn't be too unbelievable.

Tim

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Website: http://webpages.charter.net/dawill/tmoranwms

rk

t d

=A0 =A0 ...Jim Thompson

There is the damped resonant frequency and the natural frequency.

I don't think there is a way to keep a passive network centered.

I see no reason for pro-audio to like this passive circuit. Physical inductors are kind of crappy components. They are no where near as ideal as capacitors. Inductors pick up hum and the signals from nearby inductors.

I'm pretty sure RANE publishes their schematics. See what they use.

When you make one for sale, make sure you use gold plated wire for the (obviously air-core for max linearity) inductor to achieve that desirable gold-plated sound (and price).

LOL. I have no intention of doing either.

Cheers

Ian

I'd see it as a Q 'definition' oddity. If the pot is set at say the 80%, ie 4k/16k position then there's an output baseline of 0.8V which simply can't be measured using the normal method of 71% of peak voltage. The skirts are just too high and the resulting curve shape 'looks' decidedly low Q. But, subtract the 0.8V offset and remeasure Q as normal, this gives the 'real' working Q value of XL/R. ('R' in this case appears as 3200ohms hence the real Q is simply proportional to to the L or C reactance value at resonance. ). Set the pot at say the 10% position and the standard Q definition becomes more and more valid.