Reverse Log Pots

Hi, at the old hangout there was a discussion about being able to emulate a reverse log pot with a resistor and linear pot? . Is this the case for use in the calrec eq?? Can't find the info, can't find the pots, can't finish my project :sad:

Ghengis

http://www.geofex.com/Article_Folders/potsecrets/potscret.htm

Has some info....

-Jay

Some more info....

http://www.wavefront.mcmail.com/pot.htm

And here is some info on growing, err... I mean rebuilding pots to rlog.

http://www.gyraf.dk/gy_pd/calreq/magnus_j/rebuilding_pot.htm

elco

hi !

that parallel resistor method looks just fine ! I did the simulation on excel with and the curve looks like some kind of neg log curve ! that seems like it's working so perfect that there MUST be drawbacks... i admit i'm sometimes a kind of pessimistic bastard  :

i tried to do the calculation for the Calrec EQ freq pot C100k. it's used as a variable resistor so i think the problem is easier to solve.

refer to this page for the notations (R3 etc). please scroll down about the 2/3 of the page.
http://www.geofex.com/Article_Folders/potsecrets/potscret.htm
let Rout be R2|R3 and N be the fraction of total rotation of the pot.

we want Rout = 100 000 for N=1. so N*R*R3/(N*R+R3) = 100 000 with R3 = R/10. that leads to R = 100 000*11 = 1.1MOhm.

so we'll use a dual 1.1M lin pot with a parallel resistor of 110k for each of the two "inner pot". much easier (and cheaper...) to find than a dual lin pot than a dual rev log pot !!!

the excel file with the curve is here :
http://insidecircus.free.fr/neglogfun.xls

PLEASE CORRECT ME IF I'M WRONG !!!
if i'm not, i hope this will help !

does anyone know the mathematic formula f(x) of neg log ? i'd like to trace it in order to find the value of R3/R that sticks to it the best...

thanx in advance !

You can also just order the pots from Omeg and or collin's shop.

I think I went direct and it was a little under $100 USD for my pots for 2 Eq's.

> that parallel resistor method looks just fine !

No.
No.
No.
No.
No.

> calculation for the Calrec EQ freq pot C100k.

I don't know this specific box. But nearly ALL frequency dials attached to resistors obey a law like F = 1/RC. Bigger R, lower frequency.

And we usually want frequency on "Octave spacing". We need to cover 1,000:1 of frequency, and while we might want 1Hz resolution at 20Hz, we sure don't want 1Hz resolution at 20KHz.

So take a straight 100K pot. Assume you pick C so that it works at 100Hz. Then the resistor value versus frequency looks like:

100K  = 100Hz
90K = 111Hz
...
50K = 200Hz
...
20K= 500Hz
10K = 1KHz
zero = infinity

The octave from 100hz to 200 Hz fills half the dial. The four+ octaves from 200Hz to 1KHz are crammed into "5" to "1" on a 0-10 dial. The pretty important octave 500Hz-1KHz is crammed into "2" to 1", whereas the "10" to "9" range covers only two semitones.

Your proposed pot is worse. At "1" it isn't 10K like straight linear, it is 55K. The octave from 500Hz to 1Khz is, I think, crammed into "0" to "0.24", on a 1-10 dial.

If you want to cover a 10:1 range, nicely, you want "5" to be about 0.316 instead of 0.5; to cover a 25:1 range you want "5" to be about 0.2.

And you can't get there by slugging a reostat.

No....  at the halfpoint (50% of the track), the resistance needs to be about 10-15% of the total.

So, a 100k pot needs to be at 1k to 1k5 to give a reasonable spread of frequencies around the pot.

For example - based on the standard f= 1/(2*pi*R*C) formula:

R fixed = 159R
C = 155 nF

Freq         R total (Pot + Rfixed)         Pot
100 Hz 10159 R                         10000 R 100,0% maximum
200 Hz   5079 R                           4921 R 49,2%
400 Hz   2540 R                           2381 R 23,8%
800 Hz   1270 R                           1111 R 11,1% mid point
1600 Hz   635 R                           476 R 4,8%
3200 Hz   317 R                           159 R 1,6%
6400 Hz   159 R                               0 R 0,0%         minimum

See - at midpoint the value for the pot is just 11% of its nominal value.


C.