Logarithmic pots, formula for resistance

[Fred]

This all started when I needed a 500k log pot and Craplin doesn't sell miniature versions of them.  The upshot is that I've quickly knocked up an Excel spreadsheet giving the response of a linear pot with a couple of resistors from the centre lug to each of the outer lugs.  I was going to use this to graph out and tailor the response of my pot.

I was planning on taking the difference between that and the response for a perfect log pot and then minimising the square of the differences using Excel to calculate the optimum values for the resistors.

This was a cunning plan that all relied somewhat on my being able to write a formula for a log pot, something which I have failed to do.  Maybe I'm just being thick, most probably.  So, can anyone help?  Let us call 'x' the fraction of rotation of the pot.  So I'm guessing the taper of the resistance goes something like a^x, where 'a' is, er, 10?  This is where I am coming rapidly unstuck!

Thanks for any help.

[axg20202]

I've been looking at this kind of problem myself recently. Google "Secret life of pots" and read RGs article.

[Fred]

OK, I've read through, but it isn't really helping that much.  There's a very nice 'logarithmic or audio taper' graph, but no real indication of where it comes from.  I guess I could just fit an inverse logarithm roughly in the range required, but I'm still none the wiser as to whether it's base 10, e, whatever for a 'logarithmic' pot, or at least what the manufacturers are trying to approximate.

Thanks again.

[spudulike]

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

[dschwartz]

i made this excel spreadsheet, and found the same problem.. its made to see how accurate was the lin to revlog taper mod was..

the "real log" pot has a base of 1.1, just cause i seemed to have a nice exponential curve at the graph..

i hope this helps....
the file is here:

http://diynoise.googlepages.com/pot_calc1.xls

[dschwartz]

did someone take a look to my spreadsheet???

[gaussmarkov]

fwiw:  http://gaussmarkov.net/wordpress/parts/resistors/resistors-5-potentiometers/#moreinfo

i made up a formula, but it's not official.     it's based on a rule of thumb R.G. mentions on geofex.com:  to get half the perceived loudness use one-tenth the voltage.

[pcarew]

@gaussmarkov,
On your web page (  http://gaussmarkov.net/wordpress/parts/resistors/resistors-5-potentiometers/#moreinfo ),
you suggest (reasonably so) for audio/logarithmic Pots, that "...to get half the perceived loudness use one-tenth the voltage".

Would this not suggest an exponential progression that is a function of x^10 ?
Your example however uses an exponential that is a little over x^3. What was your basis for choosing this? Was it a 'trial and error' process to get a curve that sort of 'looked right'? Nothing wrong in that, it's just that I'm curious as to your reasoning.

Thanks
Paul.

[Rob Strand]

In acoustics it's pretty common to treat double the perceived loudness as 10dB  (which is 10 times the *power* and 3.2 times the voltage).

http://www.sengpielaudio.com/calculator-levelchange.htm

As for pots they vary a lot.  Many are almost two straight lines with a bit of a curve between the two.  You can see this on Gaussmarkov's site.  You can over do it getting caught-up fitting curves (we have all been there!).   Many digital controls have steps in dB's, like mp3 players, and this sounds pretty even to me - even though the steps could be finer on some units!  You can use this as a model but you might want to play around with the part of the curve near zero as a step of infinite dB doesn't fit the pattern.