[sdiy] 1-quadrant multiplier with 2164

Neil Johnson neil.johnson71 at gmail.com
Tue Jan 18 13:54:47 CET 2022


> This is certainly true but note also the importance of zero when multiplying. The zero signal stays zero no matter what you multiply by. In Rutger's case that zero is in fact -5V, so the origin of Neil's graph should be at -5V signal and zero control voltage. That is why the level-shifting solution is so effective and it is also why I believe Rutger is correct to call this a one quadrant multiplier.

Yes, this is just a bit of algebraic juggling.

If we take Dave's approach:
- convert the bipolar +/- 5V input to a unipolar 0 to -10V input
- add a -5V offset to the output _after_ the VCA (so no bearing on the
quadrantiness of the VCA itself)

With a unipolar CV and a unipolar signal ... a 1-quadrant VCA.
And don't forget that as-drawn the linearised VCA is inverting.

Cheers,
Neil




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