[sdiy] backwards pots. Whats the deal?

[Donald Tillman]

> On Apr 1, 2015, at 10:23 AM, mark verbos <mverbos at earthlink.net> wrote:
> 
> I have noticed on some mixer circuits that the pots seem to be wired backwards from a conventional voltage divider. i.e. signal into the wiper and out of the CW end. What is the point of this? Should the pot still be an audio taper?

Good question!

Let's assume we're talking about the classic mixer circuit, with resistors summing signals into an inverting opamp stage.  Okay?

A backwards pot operates with the same curve as a standard pot.  You can prove it if you do the math.

The difference between the two is that the backwards pot presents a much lower variation in input resistance with pot rotation.

"Huh?"

Let's try an example: 100K pot, 100K input resistor.

For a standard pot configuration the input resistance goes:
  100K at full CCW
   83K at 50% resistance
   50K at full CW
For a 2:1 ratio.

For a backwards pot configuration, the input resistance goes:
  100K at full CCW
  125K at 50% resistance
  100K at full CW
For a 1.25 : 1 ratio.

That's a significant difference.

   -- Don

--
Don Tillman
Palo Alto, California
don at till.com
http://www.till.com