Help to determine formulae for inverse/reverse log tapered potentiometers

dougr

New member
Joined
Aug 20, 2015
Messages
3
I am trying to solve for a reverse/inverse log taper for a potentiometer. I have a formula for a "log" output (yes, technically it is an "exponential" output but it is called a "log" in the terminology) from the following link:
http://www.geofex.com/article_folders/potsecrets/potscret.htm ... "The Secret Life of Pots" ... Copyright 1999 R.G. Keen

attachment.php



Hopefully the picture is large enough... otherwise, please refer to the web page above or the following text version of the formula:
1 / (((1 - a)/a) + ((1 - a)/b)) + 1

The variables:

a => 0 to 100 ... the range 0 to 100% rotation of the Potentiometer (yes, I know pots can't be rotated 100%, but theoretically)
b = 6 ... the "taper" value ... this value ranges from 0 to 6 (actually near 0 i.e. .0001 for example) ... with 0 producing a linear output to any value but with 6 as a practical limit.

This calculates perfectly and generates a graph output ... using a "b" = 0 to 6:
attachment.php



the ""inverse" or "reverse" log output is the mirror image of this ... simply (pictorially, at least) swapping the X and Y axis:

attachment.php


However, I have been unsuccessful in "reverse engineering" the "log" formula to create the "inverse log" X - Y output for each (any) of the "taper" values (1 to 6).. obviously, taper 0 is linear in each formulation.

Please ... whatever assistance anyone could provide would be enormously helpful!
 

Attachments

  • log_taper.jpg
    log_taper.jpg
    16.6 KB · Views: 13
  • LOG_TAPER_OUT_0_6.jpg
    LOG_TAPER_OUT_0_6.jpg
    21.1 KB · Views: 13
  • inverse_LOG_TAPER_OUT_0_6.jpg
    inverse_LOG_TAPER_OUT_0_6.jpg
    21.3 KB · Views: 14
Top