[sdiy] Yet more questions: best VC drive approach

Wed Feb 16 13:07:05 CET 2022

for an ota single stage it's roughly:
ic = ictrl * tanh(in - vc)

in a mood ladder single stage it's roughly:
ic = ictrl * (tanh(in) - tanh(vc))

Andy

On Wed, 16 Feb 2022 at 03:56, Rutger Vlek via Synth-diy <
synth-diy at synth-diy.org> wrote:

> Amazing! Thanks for taking the time to put this into words. Once again, I
> want to say I really love this mailinglist!
>
> The only question in my mind that remains is how similar or different the
> situation is in a ladder filter. But I can probably figure that out with
> the help of Spice now I better understand the OTA case.
>
> Thanks!
>
> Rutger
>
> Op di 15 feb. 2022 17:04 schreef Richie Burnett <
> rburnett at richieburnett.co.uk>:
>
>> The OTA in a cascaded OTA low-pass filter (like the Roland Juno 106
>> filter)
>> works like this...  In each stage the OTA inputs measure the
>> instantaneous
>> difference between the input voltage and the capacitor voltage.  This
>> measurement is then scaled by the control current and finally converter
>> to a
>> current output that is fed to the capacitor.  In this way the behaviour
>> of
>> the OTA approximates that of a resistor between the input and the
>> capacitor.
>> i.e. The current out of the OTA is a scaled version of the voltage
>> difference between the input voltage and the capacitor voltage.  And
>> that's
>> exactly what a resistor would do if it was also connected between the
>> input
>> voltage and the capacitor.  The difference here is that the scaling
>> factor
>> (or "gain") of the OTA is variable and is determined by the control
>> current
>> into the OTA.
>>
>> Now all of that assumes that the OTA's input stage behaves completely
>> linearly.  But it doesn't!  The front end of the OTA is a long-tailed
>> pair
>> differential amplifier and exhibits the tanh() distortion we mentioned
>> earlier in this thread.  For small signals everything said in the first
>> paragraph holds true.  But as the signals get larger we start to see a
>> reduction in the incremental gain of the OTA as we start to use more of
>> the
>> tanh() curve.  The effect is that even for a fixed control current
>> (cutoff
>> frequency setting,) the scaling factor for the output current ("gain" of
>> the
>> OTA) decreases when there is a large difference between the input voltage
>> and the capacitor voltage.  If the scaling factor decreases it is as if
>> the
>> resistor value somehow got larger for big signals.  And an increase in
>> this
>> "virtual resistance" moves the instantaneous cutoff frequency of the
>> resulting RC filter downwards to a lower frequency.
>>
>> The scaling resistors around the inputs of OTA circuits are an attempt to
>> make the differential input voltage relatively small compared to the
>> thermal
>> voltage so that the OTA mostly operates in the reasonably linear region
>> around the origin of the tanh() curve.  But if you drive the filter hard
>> enough or mess about with the resistor values, you can easily drive the
>> OTA
>> input stage with signal large enough to drive into the saturation regions
>> of
>> the tanh() curve.
>>
>> I hope this explanation helps.  And I hope others also think it is valid
>> and
>> technically sound.  The key takeaway from this analysis for me is that it
>> is
>> not the signal amplitude itself that saturates, but rather the rate of
>> change of the signal that is limited when the OTA input is driven hard.
>> This results in a more subtle distortion than basic saturation.
>>
>> -Richie,
>>
>>
>>
>> -----Original Message-----
>> From: Rutger Vlek
>> Sent: Tuesday, February 15, 2022 2:43 PM
>> To: Richie Burnett
>> Cc: SDIY
>> Subject: Re: [sdiy] Yet more questions: best VC drive approach
>>
>>
>> Thanks for the helpful replies since I brought this thread back to life!
>>
>> @Richie Burnett  Thanks for the wonderful summary of knowledge on this! I
>> was already aware of the tanh characteristic involved, but mainly
>> struggle
>> to understand the interaction of it with the capacitor (capacitive load)
>> in
>> each stage of the ladder filter. Is the math behind it described
>> somewhere?
>> The audio-rate modulation of the filter's cutoff frequency that occurs as
>> a
>> consequence of this interaction (if I'm right?) as what fascinates me
>> particularly. I already did some experiments on a Nord Modular to see if
>> I
>> could somewhat replicate this type of saturation with a feedback (from
>> output) or feedforward (from input) signal at each filter stage to the
>> control input for frequency (thereby modulating cutoff at audio rate with
>> the input or output of each stage). It sounds interesting, when applied
>> in
>> modest amounts, but does not quite get to Moog smoothness territory. I'd
>> like to understand if there are ways to get closer to the actual
>> behaviour
>> inside a ladder filter, and if I could extrapolate that to other (analog)
>> topologies, such as a 2164-based filter.
>>
>> The case for an OTA-C filter seems a bit different, if I understand
>> correctly, as it's saturation in the output stage that interacts with the
>> capacitor to cause the cutoff modulation described above. Additionally
>> there's more tanh saturation happening in the input stage of the OTA,
>> without consequences for the cutoff frequency? I presume that this last
>> effect is happening in almost all OTA filters, since inputs are so
>> sensitive, but the first effect (cutoff modulation) is only happening in
>> some OTA filters, depending on OTA output loading?
>>
>> Rutger
>>
>>
>>
>> Op ma 14 feb. 2022 om 12:14 schreef <rburnett at richieburnett.co.uk>:
>> Hi Rutger,
>>
>> Wow, that is an old thread.  I've slept since then! ;-)
>>
>> They are both based around a "long tailed pair," which is a differential
>> amplifier made up of two transistors.  If you read up about this
>> arrangement you will find that there is a tanh() function in its
>> transfer function, that leads to a soft saturation behaviour.  Both the
>> Moog ladder filter and the input stage of a bare OTA exhibit similar
>> tanh() soft distortion.  Although the exact effect on the cutoff
>> frequency in each type of filter may be subtly different.
>>
>> There are some good papers out there discussing the non-linearities in
>> the Moog ladder filter arrangement.  Ones by Antti Huovilainen, Tim
>> Stinchcombe and to a lesser extent Tim Stilson are the ones that
>> immediately come to mind.  Some of the stuff in those papers is about
>> making a digital "DSP" model of the filter, but the first bits about how
>> the analogue filter works discusses the non-linear behaviour for large
>> signals.
>>
>> I don't have a reference for OTA operation immediately at hand but I
>> would have thought the non-linearities for large input signal amplitudes
>> would be discussed in the datasheet, or documented somewhere.  As I
>> said, the first stage is just a long-tailed pair diff amp, so assuming
>> the OTA doesn't have any fancy linearising diodes, it will have a tanh()
>> shape to its transfer function that starts to kick in once the
>> differential input signal amplitude goes over a few tens of millivolts.
>>
>> Hope this helps...
>>
>> -Richie,
>>
>>
>>
>> On 2022-02-12 20:15, Rutger Vlek wrote:
>> > Hi Richie,
>> >
>> > I hope you don't mind me bumping up an old thread. I was reading back
>> > what you wrote in 2018 and wondered if you could refer me to more
>> > background information on filter saturation. I'd like to understand
>> > what happens in a ladder filter, and weather something musically
>> > similar could also be recreated in other ways (e.g. in other
>> > topologies than a ladder). If you have an opinion on the latter,
>> > please share!
>> >
>> > Regards,
>> >
>> > Rutger
>> >
>> > Op vr 9 nov. 2018 10:51 schreef <rburnett at richieburnett.co.uk>:
>> >
>> >> When you over-drive OTA based 1-pole "leaky integrator" stages, you
>> >> actually get a signal dependent shift in the cutoff frequency as the
>> >> OTA
>> >> saturates, rather than what you would typically describe as
>> >> "clipping".
>> >> This behaviour is down to the way in which the OTA and filter
>> >> capacitor
>> >> are wrapped up inside a negative feedback loop.  The behaviour is
>> >> quite
>> >> like how the cutoff frequency of the Moog ladder filter changes
>> >> dynamically with drive signal level.  It is much more musical that
>> >> simple signal clipping.
>> >>
>> >> -Richie,
>> >>
>> >> On 2018-11-09 08:48, Rutger Vlek wrote:
>> >>> Hi Jacob,
>> >>>
>> >>>> This also ensures that the clipping happens in the chip used for
>> >> the
>> >>>> integrators, and not in the OTA's, which sound bad when
>> >> overdriven.
>> >>>
>> >>> I presume you refer to the OTA in the VCA that controls the drive
>> >>> level? Or do you mean OTAs inside your integrators? In the latter
>> >> case
>> >>> I don't understand what you're saying (sorry)...
>> >>>
>> >>> Rutger
>> >>>
>> >>>> JACOB WATTERS
>> >>>> Web & Multimedia Specialist
>> >>>>
>> >>>> JacobWatters.com [1]
>> >>>> Tel: 226-886-3526
>> >>>>
>> >>>> On Thu, Nov 8, 2018 at 3:32 PM Rutger Vlek <rutgervlek at gmail.com>
>> >>>> wrote:
>> >>>>
>> >>>>> Hi guys,
>> >>>>>
>> >>>>> I've been wondering about many things lately, hence the flood of
>> >>>>> emails to the list :). I also have to admit feeling a bit stupid
>> >>>>> about having to ask this.. but here goes:
>> >>>>>
>> >>>>> What's the best approach to designing a voltage controlled
>> >>>>> overdrive? The obvious solution I can think of is having a
>> >>>>> saturation element preceded by a VCA. While I haven't fully done
>> >>>>> my homework on it yet, my guts tell me that this isn't the best
>> >>>>> circuit in terms of noise behaviour, as it would require the
>> >>>>> saturation element to be at high gain constantly while the VCA
>> >>>>> various input level. Meaning that any VCA noise would be
>> >> amplified
>> >>>>> by the full gain of the saturation element. In guitar stomp
>> >> boxes,
>> >>>>> some design place a pot in the feedback loop of an opamp to
>> >> change
>> >>>>> gain. Could a similar approach work well for a VC-drive unit and
>> >>>>> would it perform better/worse than the first solution I
>> >> proposed?
>> >>>>>
>> >>>>> Finally, I've been thinking about making drive level voltage
>> >>>>> controlled via power supply to the saturation element
>> >> (transistor
>> >>>>> in this case). Feeding the control voltage into a buffer that
>> >> puts
>> >>>>> out the supply to a transistor would also allow to change drive
>> >>>>> level.
>> >>>>>
>> >>>>> What do you think? How it this typically done? I just bought a
>> >>>>> Novation Peak, and am impressed with it's three stages of
>> >>>>> overdrive although it suffers from noise issues due to the
>> >> amounts
>> >>>>> of again at hand. It makes me wonder how I would design such a
>> >>>>> stage myself.
>> >>>>>
>> >>>>> Rutger
>> >>>>> _______________________________________________
>> >>>>> Synth-diy mailing list
>> >>>>> Synth-diy at synth-diy.org
>> >>>>> http://synth-diy.org/mailman/listinfo/synth-diy
>> >>>
>> >>>
>> >>>
>> >>> Links:
>> >>> ------
>> >>> [1] http://jacobwatters.com/
>> >>> _______________________________________________
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>>
>>
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