Muff's Modules & More

marcn · Posted: Fri Mar 30, 2012 9:27 am Post subject: Modding a log pot to linear pot ?

Well I did a classic mistake and ordered a bunch of 100k log pots stead of linear pots d'oh!

After a bit of research I just put a 100k between lugs 1-3 so I got it down to 50k (which is still fine for my project) and the log curve is also a tad more linear now but well, not exactly perfect.
Was just wondering if anyone has done a similar thing and how it worked out in their project or if there's a better procedure to do this ?
(searched the almighty net and nothing found about log to linear)

ps. since I'm new around here gotta say great stuff in the diy section, much respect to what you people build and the fact that it's shared for everyone to see.

sduck · Posted: Fri Mar 30, 2012 9:29 am Post subject:

I think you can mod linear to log, but not vice versa (edit - see Daverj's post - it seems you CAN get pretty close!). I also have a ton of unloved log pots here...

CLee · Posted: Fri Mar 30, 2012 10:33 am Post subject:

I think you can bend a linear pot to a somewhat exponential response, so if you did that to a log pot you should get close to something linear.

Google "the secret life of pots". You'll find a whole section on bending response curves.

We all have some pile of unloved pots don't we... Mine are the 1K that were supposed to be 100K

keninverse · Posted: Fri Mar 30, 2012 1:48 pm Post subject:

daverj · Posted: Fri Mar 30, 2012 2:53 pm Post subject:

I just did a quick spreadsheet calculation.

The red line is a linear pot. The blue line is an audio taper (log) pot.

The yellow line is what you get with a 100k linear pot that has a 10K resistor from the center wiper to ground (counterclockwise pin).

The green line is what you get with a 100K audio taper pot with a 12K from the center wiper to the top pin (clockwise pin).

This is a crude calculation, but certainly worth trying with the real parts. Not a perfect conversion to linear, but a lot closer than using the audio taper pot in a place that calls for a linear one.

If you use this as a voltage source (outer pins to power rails and/or ground) the extra resistor just draws a little extra power. But if you try to use it as an attenuator the input is going to see a load that varies from about 10.7K to 100K (in parallel with whatever load the pot wiper is connected to). So not so ideal as a linear attenuator, but not too bad as a linear voltage source.

marcn · Posted: Fri Mar 30, 2012 3:37 pm Post subject:

Thanks for your replies guys.

To Lee and Ken, excellent site, I've been over it earlier but read rather quickly as it explains quite a lot about pots but as far as I've seen it doesn't mention changing log to linear (might have missed it though).

To Dave, yes, that's what I've sortta been trying, found another thread around where there were tons of more spreadsheet calculations and that got me thinking I might get it somehow close to linear.
Now I'm not really good at the mathematical part of this so I didn't figure out what values get you to where but I've tried a few things and it seems the closest to linear works with the 100k between the 2 outer lugs (1 and 3).

I think another addition is that i use Alpha pots and I guess it's the 'cheaper' log version with the 'fake' log curve, so the more you bring down the total resistance as with the 12k, the more pronounced the 'rectified' curve is going to be, at least that's what my DMM shows comparing the 100k with the 12k version.

Anyways, I'll try a few more things and will report how it worked out in the build if I consider it relevant for others who want to get rid of their unused log pot stash. Rasta-nana

pugix · Posted: Fri Mar 30, 2012 5:42 pm Post subject:

Why not sell them and buy linear pots?

daverj · Posted: Fri Mar 30, 2012 7:00 pm Post subject:

Here's how the 12k would be connected to do the green line in my chart above:

The "clockwise" end is the pin that the center wiper connects to when you turn the pot shaft clockwise.

And you can't measure the linearity of this type method with just an ohm meter. You need to feed a voltage into the first one or a signal into the second one and see how the voltage changes or the amplitude changes.

daverj · Posted: Fri Mar 30, 2012 7:08 pm Post subject:

marcn · Posted: Sat Mar 31, 2012 5:07 am Post subject:

Thanks Dave, didn't know there was a difference to the curve when reading ohms or testing with voltage.
I'll give that a try.

daverj · Posted: Sat Mar 31, 2012 11:27 am Post subject:

When reading with ohms you're only reading between two terminals. But all 3 terminals are involved when adjusting a voltage or signal level. If you are trying to make a variable resistor rather than a variable voltage source or attenuator, then you could measure it with an ohm meter.

JJ · Posted: Mon Apr 02, 2012 2:16 pm Post subject:

Umm...

How does one calculate this? seriously, i just don't get it

I have a 1MΩ lin pot that I would prefer turning into a log...

(Buchlidian Voltage Control Processor Glide function comes in a bit too steep with a lin pot, only thing I had in stock (9mm pot) )

Do I just multiply with 10 and go with a 100k resistor?

daverj · Posted: Mon Apr 02, 2012 5:09 pm Post subject:

For turning linear into log, turn on Java in your browser and head over to:

http://www.mindspring.com/~clist/PotGraph.html

JJ · Posted: Mon Apr 02, 2012 5:16 pm Post subject: