Logarithmic potentiometers


Thread Starter


Hi, I hope there is somebody outhere who can set me on the right trail.

I would like to have a formula, to calculate the resistance ratio of a log pot, depending upon its wiper posistion.

Example: the wiper of a 10K log pot is at it's midway position (50%), what is then the calculated value at that point. I know a linear pot is at 50% position 50/100xRt, but how do I do this with a logarithmic pot?

Your help will be very much appreciated.
tnx Flyer.

Robert Scott

If p is the wiper position, then the resistance is:

b * log(a * p)

where a and b depend on the particular pot.

Robert Scott
Real-Time Specialties
Embedded Systems Consulting

If it is a pot used for audio volume, it is actually exponential (logarithmic is a misnomer)

First, log pots are/were used for audio volume control in audio appliances - because the human ear senses sound level logarithmically.

Second, they were never extremely accurate.

Third, they are actually exponential.

Fourth, I am not aware of any spec that defines how many dB per angle of rotation are the factor of the pot and one cannot calculate the pot resitance at a particular angle without this figure.

I would use a safe bet of 40 to 60 dB total range (volume at the ear). Also remember that a potentiometer controls voltage, which has a square effect on volume (power to the ear) - so a 10 dB step in power only requires 5 dB (1/2) step in voltage.

A log pot does not need to reach zero Ohm just a very low value.

Example: So if you assume, say, 60 dB total power range (100%) and 1 Mohm total resitance, you get a 30 dB voltage (resistance) range i.e., a 1000:1 resitance ratio.

100% 30 dB 1 Mohm
80% 24 dB 500 Kohm
60% 18 dB 250 Kohm
40 % 12 dB 125 Kohm
20 % 6 dB 62.5 Kohm
0% 0dB 31.25 Kohm
Have you ever measured the resistance of such pots?

They are not especially consistent from unit to unit.
That was a serious answer. If you grasp the importance of log functions in electronics, all becomes clear. Even the wide tolerance of the pot curves. Plus or minus 5 or 10 percent is almost meaningless. You have to think in decibels (which see). And it is far better understood from
study than a brief explanation. Without that understanding, it's just a funny curve.


Thanks guys for all your contributions.

If these pots vary from type to type then I'm lost. This made me realise that I better look for a different approach.

anyway thanks for your support.
Responding to Flyer's Jan 17 and Joseph's Feb 26 queries. I can help if you are still interested in a solution. For example:

1) Do you need greater accuracy than can be accomplished with an audio-taper pot found in Radio-Shack?

2) Do you need greater precision using a multi-turn pot?

3) Do you want a formula between resistance available, in Ohms, between the wiper and end-terminal, and shaft position, in degrees, similar to that found on a slide rule's A-scale and D-scale?

4) Do you want a logarithmic relationship between the input and output voltages, Ein and Eout, respectively?

Note, it is also possible to provide the above relationships, to a reasonable degree of accuracy, using linear pots.

Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])
Responding to earlier responses to Flyer's original Jan 17th post:

Meir and Curt... I'm surprised at your "TIC" semantic attacks... log vs
exp vs nvat. So here is a little more light on the subject:

While it is true that resistance vs wiper position is an exponential
one, the plot of such a function is virtually useless if the range of
the dependent variable, Y, is much much greater than the independent
variable, X.

Let Y = A(B^X), where,
Y = resistance.
X = wiper position.
A and B are constants to be explained later.

The plot of the above on a rectangular coordinate graph does not yield
interpretable Y values for small X values. However, when plotted on a
semilogarithmic coordinate graph the plot yields a straight line
relationship that is easier to analyze. The X-axis has a linear scale,
and the Y-axis scale is proportional to logarithms of the resistance
values, Taking the logarithms of both sides of the above 'exponential'
equation yields a straight line of the form,

U = C + DX, where,
U = log(Y).
C = log(A),
D = log(B).

When constructing a table of the two variables it can be seen that while
the X values increase arithmetically, the Y values increase
geometrically. Obviously, A = the value of Y when X = 0. But, what is
B? Not so obvious, is that it is the multiplier coefficient for
successive Y terms!

And now for the "little light" I alluded to in my opening sentence...
lighten up fellas. Life is short enough!

Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])
Hi Phil The real problem is that these devices are not intended for precision and so are notoriously inaccurate to any curve or function. They are for volume control which is pretty non-critical. If you did calibrate one you might not find another to match. In fact many tandem controls used for stereo are audibly mismatched, sometimes so much as to be annoying. If one needs accuracy to a non-linear function, I would take it into the digital domain and do the math. Or use a log Op Amp to do this with reasonable accuracy. Which approach would be better depends a lot on the application. Once upon a time there may have been instrumentation quality pots for analog computers etc. But these days the old slidewire is becoming a rarity except for special cases. Even cheap boomboxes often have encoder knobs. My point was not regarding the math although it seems logs are a class of exponential to my poor recollection.