Help me understand the Analog Alchemy Taper Calculator

[lars-musik]

Maybe a trivial question, but still.
I'd like to change the taper of a potentiometer from linear to (ultra-)logarithmic. The Analog Alchemy calculator obviously will help me determine the values. BUT: How do I interpret the percentage on the page. Does 10% mean that 10% of the rotation give me 50% of the pot value? Or the other way round? Or am I completely off?

Here is the corresponding link:

https://www.diystompboxes.com/analogalchemy/emh/emh.html

THANKS! 

[PRR]

Potentiometer connection or Rheostat connection? Makes a real difference.

First work it out on a matchbook. Simple parallel resistors should be a snap. If you can't rough it out by hand, those "smart" calculators will LIE to you.

50% is linear. Half-turn gives half voltage. See what values it gives.

"Audio taper" is nominally 10% voltage at half-turn.

"Ultra Audio taper" might be 1% at half-turn? There are few signal-processing applications where that makes sense. On that calculator, selecting 1,000K pot and 100K target, what you really get is a pot with input resistance which varies from 110K to 1K. And it does not tell you.

Lastly(?), you do not just want "10" "5" and "0" to be specified ratios, you want the in-between ranges to be smooth and smoothly tapered. Heavy loading tends to give little action until you get in the crack between "1" and "0".

[Rob Strand]

Maybe a trivial question, but still.

There's some important things about that page:
- You start with a pot with value "Actual Pot Value".   I believe it is linear.
- You end up with a pot with value "Desired Value".   
-  and when the pot is set to half-way the resistance between the counter clock wise pot connection (ie. the lower one on the figure) and the wiper  is  "Taper %" of the "Desired Value"

If you set Actual = 250k, Desired = 200k and Taper = 10%  your R2 will be negative.  That means your Desired is too high.   

If you have a 250k pot, when it is at the centre the top pot resistance (pot wiper to pot top) will be 125k and the bottom pot resistance  (pot wiper to pot bottom)  will be 125k; I'm taking about the raw pot here without R1 and R2.   For Taper = 10%, and with no R2 present, the top pot resistance corresponds to 90% of the target resistance.  So 90% = 125k  and  that means the 10% must be 125k * (10 / 90) = 13.89k.
When you add those two together you get 138.89k.  So that means the smallest "Desired" pot value that makes sense is 138.89k  and this results in no R2.   The R1 value can be calculated   1/R1 + 1/125k = 1/13.89k  which will give you R1 = 15.63k.

So now go to the page and plug in Actual = 250, Desired = 138.88 (a bit smaller than the exact value), Taper = 10% and will see it produces a large R2 which is means it's essentially open.   Now try  Desired =138.89 and you will see R2 is again large but this time negative indicating your Desired value is too large.

The problem with the calculator is it lets you set Actual = 250 and Desired = 10.  This will cause R1 and R2 to be small values.   The reason it's a problem is with such small R1 and R2 values the pot behaves more like a three position switch than a pot, the 25% and 75% rotation points are cramped very close to the 50% rotation point.

If you calculate the way I've done it,  ie. remove R2,  you will end-up with a smoother behaviour.  And the "Desired" pot value is as high as possible.

Here's a comparison for the 138.88k Desired (R2=open) case and the 10k Desired case:

Click to Enlarge, then click again:


Both design pass through the same values at pot settings of 0 0.5 and 1.0 but you can see the 10k Desired case doesn't do much in the middle positions.