Mbira said:
I understand how that works in relationship to our ears...
What I'm wondering is how when looking at a circuit that does not indicate which taper to use, you know which to use...
You generally want a logarithmic taper for channel faders in a standard circuit like this but you need to look at a few other things to determine what pot is needed.
As John said, a linear pot will attenuate by 6dB @ the 50% rotation point, so most of the control will be in the top 75% (12dB) of the pot's travel, with the last area rapidly attenuating the signal as it approaches the "zero" point. This is assuming there's negligible load on the wiper.
Here, we don't have a significant load but we do have
"some" load on the wiper in the form of the 10K build-out resistors and, to a lesser extent, the pots and build-out resistors from the other channels in parallel
It's difficult to analyze exactly because of interaction but, assuming a few channels are still connected (let's say they're turned off), for a simple thumb calculation: ...
10K build-outs (plus a few other parallel channels worth of 10K build-out 's @ ground potential on the pot side) results in an attenuation of closer to 8dB @ 50% rotation of a 5K linear pot rather than the ideal 6dB.
So the pot's linear taper has been changed (faked), albeit a little, and is leaning more towards a log taper.
To get closer still we'd need to lower the value of the wiper load, and, a wiper load value of about half of the pot's value would give a pretty good approximation to a log taper. However, the load on the source is now also lower so some number crunching of assumed levels and current would be in order.
In this circuit with these values: linear or log? I don't know. Depends how it's being used or where
you want most of the control.
Just a wee small caveat that, because of the wiper load, although it's not a drastic deviation, neither a linear nor a logarithmic pot. will quite behave as stated on the can.
Edited for a mistake in thumb mathematics.