[sdiy] Moog filter self-FM questions

[cheater cheater]

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.
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