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From: Ian Bell on 12 Jan 2010 14:47
Jim Thompson wrote:
> On Tue, 12 Jan 2010 10:22:17 +0000, Ian Bell <ruffrecords(a)yahoo.com>
> wrote:
>
>> 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
>
> Try this (for a starter)...
>
> www.analog-innovations.com/SED/For_Ian_Bell_2010_01_12_A.pdf
>
> ...Jim Thompson


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
From: Tim Williams on 12 Jan 2010 20:29
"Ian Bell" <ruffrecords(a)yahoo.com> wrote in message
news:hiijk3$ql1$2(a)localhost.localdomain...
> 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?

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

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms


From: miso on 13 Jan 2010 01:03
On Jan 12, 11:47 am, Ian Bell <ruffreco...(a)yahoo.com> wrote:
> Jim Thompson wrote:
> > On Tue, 12 Jan 2010 10:22:17 +0000, Ian Bell <ruffreco...(a)yahoo.com>
> > wrote:
>
> >> 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
>
> > Try this (for a starter)...
>
> >www.analog-innovations.com/SED/For_Ian_Bell_2010_01_12_A.pdf
>
> >                                         ...Jim Thompson
>
> 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

There is the damped resonant frequency and the natural frequency.
http://www.websters-online-dictionary.org/da/damped_natural_frequency.html
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.
From: Robert Baer on 13 Jan 2010 04:42
Ian Bell wrote:
> Tim Wescott wrote:
>> On Tue, 12 Jan 2010 09:41:55 -0700, Jim Thompson wrote:
>>
>>> On Tue, 12 Jan 2010 10:22:17 +0000, Ian Bell <ruffrecords(a)yahoo.com>
>>> wrote:
>>>
>>>> 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
>>> Try this (for a starter)...
>>>
>>> www.analog-innovations.com/SED/For_Ian_Bell_2010_01_12_A.pdf
>>>
>>> ...Jim Thompson
>>
>> 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.
>>
>
> 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
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).
From: Ian Bell on 13 Jan 2010 05:08
Robert Baer wrote:
> Ian Bell wrote:
>> Tim Wescott wrote:
>>> On Tue, 12 Jan 2010 09:41:55 -0700, Jim Thompson wrote:
>>>
>>>> On Tue, 12 Jan 2010 10:22:17 +0000, Ian Bell <ruffrecords(a)yahoo.com>
>>>> wrote:
>>>>
>>>>> 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
>>>> Try this (for a starter)...
>>>>
>>>> www.analog-innovations.com/SED/For_Ian_Bell_2010_01_12_A.pdf
>>>> ...Jim Thompson
>>>
>>> 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.
>>>
>>
>> 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
> 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
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