[sdiy] Moog filter self-FM questions

[Tom Wiltshire]

There's interesting analysis of the moog filter in some of the papers  
dealing with implementing it digitally. They maybe provide a slightly  
different angle.
You could try:

http://www.qvoaudio.net/pdf/moog_filter.pdf
(Antti H's original Moog filter paper, and a great piece of work)

http://www.rs-met.com/documents/notes/ 
ResonanceTuningForTheDigitalMoogFilter.pdf

HTH,
Tom

On 28 Mar 2010, at 14:16, cheater cheater wrote:

> Simon,
> Thanks for the writeup. Interesting thoughts. Would you be able to
> point to a resource that explains how the principle you're mentioning
> works exactly (hopefully with a derivation of the formula)? My
> googling for formulas didn't bring up a lot.
>
> In your analysis, I think what you're looking at (tail current) is the
> current in a long-tailed pair after common-mode rejection, right?
>
> But notice that this isn't exactly a long-tailed pair. In an LTP the
> emitters are tied together (DC coupled) while in the moog ladder
> filter the emitters are only AC coupled. This is, as I understand, to
> prevent common-mode rejection of the control DC(*). In that case, I
> think your analysis could be a little bit off - what do you think? I'm
> no good with electronics, just worried if we're looking at this
> through the right lens.
>
> So I guess my question should be: in the collector current of one side
> of the differential pair on a stage of the moog filter what is the
> max/min control DC and max/min audio AC in amperes? (phew that's long)
>
> I might be thick, but I'm having a bit of a problem understanding your
> analysis; on the one hand because you don't seem to be relating it to
> the control current (but it's quite probable I don't understand
> electronics well enough to see it indirectly), and on the other hand
> because it just doesn't *seem* like the audio modulates the filter so
> much.. in my imagination it would sound different - especially at low
> cutoff it would be very metallic instead of producing an almost
> sinewave-like quality at higher resonance settings.
>
> ----
>
> * this brings up the question of: if AC control current gets rejected
> after the first stage, wouldn't it be better to instead DC-couple the
> tail, and then ac-couple it to each collector of the input stage
> separately, injecting the control current after the cap? This way
> you're re-injecting the control on every stage, not just once.
>
>
> BTW, I totally don't understand why the base in the differential pairs
> gets the control current too...
>
> Thanks a lot
> D.
>
> On Sun, Mar 28, 2010 at 09:31, Simon Brouwer <simon.oo.o at xs4all.nl>  
> wrote:
>> Hi,
>>
>> If  you take Rick's schematic
>> (http://dropmix.xs4all.nl/rick/Emusic/Moog/moogvcf_schematic.gif)  
>> as a
>> reference, you
>> can see that a 2.5V p/p input voltage is attenuated to 62.5 mV p/p  
>> at the
>> input of differential pair Q14-Q15. So at a peak in the signal, the
>> differential voltage
>> will be 31 mV.
>>
>> As the differential output current of such a differential pair is  
>> given as
>> Ic = Ie*tanh(Vd/(2*Vt)), with Vt=26 mV and Vd=31 mV, Ic=0.6 * Ie.
>> That means the current in one transistor goes from 0.5*Ie to  
>> 0.8*Ie, and in
>> the other from 0.5*Ie to 0.2*Ie, where Ie is the tail current.
>>
>> That's 60% modulation of the corner frequency (up in one leg, and  
>> down in
>> the other, which however does not simply cancel out). Obviously  
>> the Moog
>> filter can not be regarded as a straightforward lowpass filter as  
>> it has
>> some interesting nonlinearities going on near the corner frequency.
>>
>> Best regards
>> Simon
>>
>>
>>
>>
>> cheater cheater schreef:
>>>
>>> Hi guys,
>>> I have a question about the control current on the transistors in  
>>> the
>>> moog ladder filter.
>>>
>>> First I'll explain how I see the filter, this is how Richard  
>>> Atkinson
>>> explained this to me (thanks Richard!) - I'm not that great with
>>> electronics so I might be screwing up something he told me.
>>>
>>> I understand the moog ladder is (basically) 2x butterworth filter,
>>> connected in parallel in differential mode, with the resistors
>>> replaced with transistors; the transistors change resistance  
>>> based on
>>> the DC current across them; that same current contains an AC signal
>>> that is the audio. Because the whole thing is differential, once the
>>> signals are mixed together, the DC current which is common mode gets
>>> rejected whereas the audio which is flipped on the left side gets
>>> amplified (via normal differential operation).
>>>
>>> The key thing is that on each side of the ladder the signal and the
>>> control current are the same signal. In this case the audio  
>>> modulates
>>> the cutoff to some extent too; on one side of the ladder it  
>>> modulates
>>> it positively, on the other negatively. The question is to what  
>>> extent
>>> this happens.
>>>
>>> I understand that this is different depending on the implementation
>>> (again pointed out by Richard); so when talking about this let's
>>> mention what synth we mean and if it's modular or custom-made then
>>> also what specific transistor model and manufacturer.
>>>
>>> Questions: 1. what is the swing of the (common-mode) DC control
>>> current? What is the maximum and minimum?
>>> 2. what is the swing of the (differential-mode) current that
>>> represents audio input?
>>> 3. if audio is patchable into the cutoff control, what's the  
>>> swing of
>>> the current control that can be created this way?
>>> 4. are the ladder sides biased by some DC that's always present?
>>> 5. what's the DC injected at the minimum and maximum cutoff setting?
>>>
>>>
>>> BTW, when you mix down the filter sides they will actually be
>>> (somewhat) different, as I understand it; to the extent to which the
>>> audio is actually FMing the filter.
>>>
>>> Thanks a lot
>>> D.
>>> _______________________________________________
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>>>
>>>
>>>
>>
>>
>> --
>> Vriendelijke groet, Simon Brouwer.
>> | http://nl.openoffice.org | http://www.opentaal.org |
>>
>>
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