does linear POT have better tracking that a log one?

Jun 28, 2003 at 12:26 PM Post #2 of 4
Absolutely. Log pots are usually made by - get this! - splicing together two different linear taper resistance elements!
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So what do you do if you want a volume control?? Simple. Put a resistor across the wiper and grounded leg of the pot that is 15% of the value of the pot (e.g. - a 100k pot would need a 15k resistor, and the combination would act as a 13k pot). This will approximate a log function better than a log pot will 99 times out of 100 and you get the bonus of far better tracking than even the dual-gang linear pot could deliver because the shunt resistor swamps resistance variations in the pot elements.

Given the superior characteristics of this simple modification, it's surprising that so few commercial products use it....
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Jun 28, 2003 at 12:40 PM Post #3 of 4
Quote:

Originally posted by jeffreyj

Given the superior characteristics of this simple modification, it's surprising that so few commercial products use it....
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thanks for the reply.

As regarding the commercial usage of such, some products require 100K or 250K POT. based on your computation saying a 100k/15k resistor = 13K log, a 100k or 250k log POT will require a huge value of a linear POT which I don't think is available.
 
Jun 28, 2003 at 1:13 PM Post #4 of 4
That's definitely no problem. 1M linear pots are available with excellent linearity (large resistance values are available, but it rapidly becomes more difficult to maintain a linear resistance across the element). A 1M pot + 150k resistor gives a 130k Thevenin equivalent resitance, which should certainly be high enough for any audio application (probably too high, w/r/t Johnson noise and EMC/RFI susceptibility).
 

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