[sdiy] DCO with OTA fiddlings (semi-beginner question)

[Richie Burnett]

As Steve Taylor said, an integrator will ultimately integrate up any tiny 
error and head off to infinity given long enough, if there isn't some 
negative feedback around it to bring it's DC gain down to something finite.

You can get away with using an integrator for the ramp (sawtooth) waveform 
because you are periodically resetting it back to a known hard value. 
(Although at very low frequencies you might still see some variation in 
amplitude of the sawtooth because the long period of integration might be 
sufficient for the error component to integrate up to something significant 
again.)

I would recommend generating your triangle wave from the sawtooth output by 
waveshaping.  Ultimately if you have gone to the trouble of making a 
constant amplitude DCO that makes a highly linear sawtooth ramp, then you 
can get a decent triangle waveform from that just be full-wave rectifying 
the sawtooth waveform.  That's how may of the old analogue synths did it. 
Making a triangle waveform up from an integrator is asking for trouble if 
you base the switching instants on time, but if you make it switch direction 
at accurately controlled voltage thresholds then this will keep the 
amplitude bounded, as others have done in their tri-core VCOs.

-Richie,


-----Original Message----- 
From: Steve Taylor
Sent: Wednesday, March 23, 2016 10:49 AM
To: synth-diy at dropmix.xs4all.nl
Subject: Re: [sdiy] DCO with OTA fiddlings (semi-beginner question)

Hi Steve,

the integrator gradually drifts off without the reset to zero used by the
saw. Think about the maths of integration. Any tiny error will eventually
integrate to infinity. I wonder whether a DC servo loop might be able to
control that but I haven't tried it yet.

Steve.


On Wed, 23 Mar 2016 10:09:59 -0000, Steve <sleepy_dog at gmx.de> wrote:

> Hey there.
>
> I, a software guy dabbling in electronics, recently read a bit of how 
> OTAs work
> and started to experiment.
>
> From an inspiration of some years ago, on Tom Wiltshire's website 
> describing how
> the Juno DCOs worked, I thought myself: Hey, why not do that with an  OTA, 
> which
> might even enable to do some more funny stuff...
>
> Feeding a pulse wave from my microcontroller of choice...
> So I first built the simplest of things, which works fine: a ramp DCO 
> using an
> OTA to constant-current charge a capacitor and abruptly discharging it 
> with an
> NPN for the last percent or so of the waveform period.
> The amplitude compensation that's necessary to keep the wave at the same 
> peak is
> done via the current input of the OTA with a signal from the MCU.
>
>
> I then proceeded to do a triangle wave with pretty much the same setup, 
> and a
> minor change:
> I fed a 50:50 pulse to the OTA, swinging positive and negative. My 
> thinking was,
> I first charge the capacitor with a certain current, then discharge it 
> with the
> same current inverted, for the same time, and should thus land at zero 
> voltage
> again. (no dedicated discharge transistor involved)
>
>
> But this quickly wanders off towards one of the rails - I thought, ok, 
> probably
> imperfections in parts, a slight bias towards one direction, so I just 
> put a 10Meg
> resistor across the capacitor, large enough to not deform the wave.
> This seemed to do it at first, but it's very frequency dependent, when  it 
> looks
> good at 200 Hz, it will have a large offset at 800 Hz or so. (this was 
> actually
> breadboarded, not just in spice)
>
>
> This is a bummer, as I thought I could even build something that morphs 
> between
> saw and tri seamlessly by feeding a different duty cycle pulse,  switching 
> not
> only the polarity but also magnitude of (dis-)charging current for the 
> rising (a)
> vs. falling (b = (-1) * (1 - a) = a - 1 : using the OTA also as 
> subtractor) ramp
> parts of the period to get the same overall time. But given that the 
> simpler
> scenario of a perfect triangle doesn't even work, I'm not so sure  anymore 
> :-)
>
>
> Can someone give me some pointers here?
>
>
> Regards,
> Steve
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