[sdiy] 50k Tapped Pots

[JH.]

>Doesn't the SEM use a center tap, without a detent, to sweep the response
of
>the filter?

If I had the choice, I'd go _without_ center detent, too. (Or with a detent
that
could be removed, even better).
It's easy to adjust a knob dead center to 12 o'clock position, especially
when
the center tap will cause a little dead zone where the resistance doesn't
change
(does it?). But a detent gets in the way when you want to make a smooth
sweep across that spot.

JH.


Foggy and can't find me SEM schemos,
Scott

----- Original Message -----
From: "Michael Bacich" <weareas1 at earthlink.net>
To: "Tim Parkhurst" <tim.parkhurst at gmail.com>
Cc: "synth diy" <synth-diy at dropmix.xs4all.nl>
Sent: Friday, August 04, 2006 7:42 PM
Subject: Re: [sdiy] 50k Tapped Pots


>
> On Aug 4, 2006, at 5:25 PM, Tim Parkhurst wrote:
>
> > Why not 100K? Could be handy (especially if you want to do the
> > SEM-style 'turn to the left to get mod source A, turn to the right to
> > get mod source B). Or am I getting my input summer math wrong?
>
> 50K would probably work just as well.  If you're thinking about input
> summing resistors on a CV control input stage, it's probably best to
> do those with matched, fixed resistors, anyway.  Just use the pot as
> a voltage divider stage to feed voltage into the fixed resistors at
> the CV summing node -- don't try to use the resistance of the pot as
> your summing resistor.
>
> I will admit that I don't remember how it's exactly done on the SEM
> -- I don't have that schematic handy.
>
> 50K is probably a good value for a lot of purposes.  I think 100K
> starts getting kind of high for audio purposes, at least from the
> standpoint of resistor self-noise.
>
> Mike B.