I have a 10k rotary potentiometer which turns from 0° to 270°. I wire it to an Arduino with the middle pin (2) connected to an analog input (A0) and the other two pins (1 and 3) connected to 5V and GND.


simulate this circuit – Schematic created using CircuitLab

I know the potentiometer is a voltage divider between pin 1, 2 and 3 as above so as expected I get 5 volts on pin 2 when the potentiometer is turned left (0°), 0 volts when the potentiometer is turned right (270°) and something in between in the middle. The problem comes when I try to calculate the angle - voltage function. I expected that it would be a linear function that can be calculated from the two given points (0°, 5V) and (270°, 0V) but when I turn the potentiometer to 135° which is supposed to be the middle point I don't get 2.5 volts instead I get something around 1.5 volts. I also noticed that when close to 270° a single ° in rotation corresponds to a bigger voltage change than when closer to 0°. So is this normal or is my potentiometer busted? And if it is normal why is it so?


I did some measurements and came up with the table below as Umar suggested

Degrees   |  Ohms  |   Delta
   0°     |    3   |     -
   45°    |   480  |    480
   90°    |   972  |    492
  135°    |  1704  |    732
  180°    |  3423  |   1719
  225°    |  7280  |   3857
  270°    | 10300  |   3020

So Kevin White is correct it seems that I have a tapered audio potentiometer and the function is logarithmic (the blue curve in Kevin's answer). Thanks, everybody for your time and effort.

  • \$\begingroup\$ Any current draw at pin 2 will cause a voltage drop through your pot. A voltage follower at pin 2 may reduce or eliminate this effect. \$\endgroup\$ Nov 22, 2019 at 1:03
  • \$\begingroup\$ @Optionparty So what would be the correct way to relate the angle of the pot with the voltage on pin 2? \$\endgroup\$ Nov 22, 2019 at 1:09
  • \$\begingroup\$ Add the datasheet link of the potentiometer. \$\endgroup\$
    – User323693
    Nov 22, 2019 at 1:22
  • 1
    \$\begingroup\$ It might be a log taper pot \$\endgroup\$ Nov 22, 2019 at 1:24
  • \$\begingroup\$ @Umar, unfortunately, I don't know model or maker so I can not find the datasheet the only thing on it says A10K \$\endgroup\$ Nov 22, 2019 at 1:27

1 Answer 1


It is almost certainly an audio taper version. They have a pseudo-logarithmic law that works better for audio applications because of the way that we perceive loudness.

The A10K means audio taper potentiometer 10 kilo-ohms. One marked B10K would have a linear law that does what you expect.

Pot laws

The picture was taken from this site Resistor Guide


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