[sdiy] CV inputs with bipolar "amount" knobs

[Mattias Rickardsson]

On 25 July 2010 07:51, David G. Dixon <dixon at interchange.ubc.ca> wrote:
>
> If you use a standard one-opamp, inverting/non-inverting, differential-amp
> configuration, then it is very easy to do this.  You can do it two ways:
>
> 1.  With resistors:
>
> Connect two equal resistors between the centre tap and each end tap of the
> pot.  With no resistors, the gain response is linear.  This response gets
> flatter in the centre and steeper at the ends (like the tan function between
> pi/2 and 3*pi/2) as the resistors get smaller with respect to the pot
> resistance.  For example, with a 100-k pot, two 47k resistors will give a
> nicely flattened response.  For an even flatter response, use smaller
> resistors.

Yes, that's what I was sketching on the other day. :-)
I haven't got the calculation in front of me now, but I wouldn't
expect equal pot/resistor values to give much of a "log"-like
flattening. The shunting resistor needs to be much smaller than the
pot... an example can be seen here:

  http://sound.westhost.com/project01.htm

With "log"-like flattening I mean a range of 20 decibels or more
(i.e., a factor of 10). Shunting a 100k pot with a 100k resistor
probably just gives something like a factor of 2. On the other hand,
"log" response over a large range is more of a hi-fi amplifier volume
knob feature; in a synthesizer the signal level variations are more
moderate and we can live with a smaller "log"-like range on our audio
level adjustments. CV level adjustments are probably even less
critical. :-)

But an important point of flattening curves is still there - finding
zero on bipolar knobs!

/mr  -  with flattering curves