Q of passive audio equaliser

Discussion in 'Electronic Design' started by Ian Bell, Jan 12, 2010.

1. Ian BellGuest

Suppose you have a simple pot divider consisting of two equal value
resistors, say 10K each. Across the upper one you connect a series LC
circuit that resonates at 3KHz or thereabouts. If you drive this network
from a low impedance source and plot the response across the bottom
resistor, the Q of the resulting peak is not the Q of the series LC but
rather is determined by the pair of LC values. For example, choosing
l=150mH and C=18nF gives a Q of just over 2. Choosing L= 1.5H and
C=1.8nF gives a much higher Q.

What I need is a simple means of calculating L and C given the pot
divider resistor value and desired Q and f (assuming the Q of the LC
itself is much higher).

Cheers

Ian

2. Ian BellGuest

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

3. Ian BellGuest

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

4. Ian BellGuest

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

Cheers

Ian

5. Ian BellGuest

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

6. Ian BellGuest

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

7. Tim WilliamsGuest

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

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

8. Ian BellGuest

LOL. I have no intention of doing either.

Cheers

Ian