[sdiy] Dual oscillator cores?

[Jerry Gray-Eskue]

-----Original Message-----

Trying to match the caps will help, and as Ian has mentioned using two in
parallel to make matched cap pairs improves your chances of getting a good
match.

What you are trying to do is match the time constant of both integrators.
Using a multi turn pot in the current feed path of one will allow you to
trim it to match the other side. Using this setup you can build it with less
precision in the capacitor matching and still achieve a matched pair of
integrator time constants.

- Jerry


From: synth-diy-bounces at dropmix.xs4all.nl
[mailto:synth-diy-bounces at dropmix.xs4all.nl]On Behalf Of cheater cheater
Sent: Sunday, October 18, 2009 12:04 PM
To: synth-diy
Subject: [sdiy] Dual oscillator cores?


Hi guys,
I was wondering if a design like this was tried somewhere already.. it
seems basic enough that someone should have thought of it before :-)
I'm not an expert at any rate, but I think it could work well to
remove HF problems existent in some popular approaches at designing
oscillator cores..

Saw core. Two accumulators instead of one - call them A and B. Both
are empty at the beginning. They both go out through one JFET each,
that JFET works as a voltage controlled switch. They also both have
their 'inputs' connected to one voltage controlled switch made out of
a JFET each. (so two parallel chains of JFET -> Acc -> JFET). When you
turn on the oscillator, accumulator A's input and output JFET are on
and the accumulator starts being loaded. Each accumulator has one
comparator each. When Acc A reaches the trigger level for comparator
A, two things happen: 1. the input switch A is turned off 2. The input
switch B is turned on 3. The output switch A is turned off. 4. The
output switch B is turned on. At that point accumulator B is at level
0 and starts moving up towards comparator B's trip level. 5.
Accumulator A gets drained. There is no delay while it's being
drained, because it's B that is outputting at that point.

Of course accumulator A becomes drained long before B reaches its trip
point. Once B reaches its trip point, it gets swapped with A again.

What are the main problems with such an approach that you guys can see?

Thanks a lot
Damian
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