[sdiy] Re: vactrol end resistance

[harrybissell]

OK I agree with that.

In this case it means the Vactrol is a voltage divider... ie the output desired

is a voltage that could come from an opamp buffer.

If the resistance is actually a circuit element, it does not work the same way.

My "morphlag" circuit is a good example. The 'shape' pot really need to be a
good
potentiometer... which almost ALL mechanical units are.  You might scale the
impedances WAY up and try it, but the result is likely to be poor.

The two 'time' pots could very easily be single vactrols... they don't need to
go
really low in resistance... the 'end resistance' would actually be a feature
because
you don't wnat it to go to zero anyway.

Vactrols make really good feedback resistors in state-variable filters.

moral...  you pays your money and you takes your chances   :^P

H^) harry

"JH." wrote:

> > Another problem...  Potentiometers have a very low 'end' resistance...
> they
> > go essentially to zero ohms.  Vactrols will never make it under maybe 100
> ohms...
> > more like 500 ohms to 1K ohms for common units.
>
> Yes, but when you build a potentiometer from two vactrols, the _other_ side
> can get very high impedance, so you get excellent attenuation even with an
> on-resistance of several hundred ohms!
>
> That's what I meant with the "vactrol pot" changing its end-to-end
> resistance.
> If you design that behaviour into your circuit (basically, that means no
> heavy loading of the vactrol-pot), you can turn it into a feature.
>
> JH.