[sdiy] Yet more questions: best VC drive approach

Wed Feb 16 15:37:58 CET 2022

Thanks! With this new info I did some more experiments on my Nord Modular
this morning and indeed came closer to an analog sounding type of
saturation and associated harmonic locking! What I did was subtracting a
single stage LPF output from its input signal (basically deriving a hpf),
full wave rectifying and inverting it and using it to modulate (expo)
cutoff frequency of the same filter stage. While certainly not mathatically
equivalent, the sound character is in the right ballpark! The limitations
of the Nord platform unfortunately mean that it does not work near higher
frequencies, due to aliasing, but it is encouraging enough to try it in
analog circuitry.

Rutger

Op wo 16 feb. 2022 13:07 schreef Andrew Simper <andy at cytomic.com>:

> 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
>>> >>>>> _______________________________________________
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>>> >>>>> http://synth-diy.org/mailman/listinfo/synth-diy
>>> >>>
>>> >>>
>>> >>>
>>> >>> Links:
>>> >>> ------
>>> >>> [1] http://jacobwatters.com/
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>>>
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