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From: Jim Thompson on 12 Jan 2010 11:41
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
--
| 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.
From: Tim Wescott on 12 Jan 2010 14:02
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.

--
www.wescottdesign.com
From: Ian Bell on 12 Jan 2010 14:33
Tim Wescott wrote:
> On Tue, 12 Jan 2010 10:22:17 +0000, Ian Bell 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
>
>
> .--||---UUU--. ____
> Vin o--o ____ o----|____|----.
> '---|____|---' |
> ===
> gnd
>
> Like this? How about simple circuit analysis?
>

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

Cheers

Ian
From: Jim Thompson on 12 Jan 2010 14:40
On Tue, 12 Jan 2010 13:02:01 -0600, Tim Wescott <tim(a)seemywebsite.com>
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.

Baxandall, as in http://www.duncanamps.com/technical/tonestack.html

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.
From: Ian Bell on 12 Jan 2010 14:43
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
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