# Analog potentiometer input current

When I hook up an analog potentiometer as an input, such as here: http://www.arduino.cc/en/Tutorial/Potentiometer

Does anybody have an idea how much current that draws from the power source? i.e. how much power it consumes (so I can know what the power requirements would be).

You don't mean power you mean current. Ohm's law states I = E/R, so if you have a 10K pot across 5V it draws 5 / 10000 = 0.5 mA

Grumpy_Mike: You don't mean power you mean current. Ohm's law states I = E/R, so if you have a 10K pot across 5V it draws 5 / 10000 = 0.5 mA

Is that no matter what position the 10k pot is in it would always draw 0.5mA? Does that mean a 100k pot would draw 0.05mA? Is there any "proper" value to use for a pot as an analog input to Arduino? If not it would make sense to use a larger value to draw less current.

Is that no matter what position the 10k pot is in it would always draw 0.5mA?

Yes, providing the wiper is connected to a high impedance as the analogue input would be.

Is there any "proper" value to use for a pot as an analog input to Arduino?

Yes it is 10K. It optimists the input impedance for the sample and hold on the A/D converter.

Thanks much for the help!

You can use a higher value pot if you need to reduce current consumption. Using a 10K pot, the worst case source resistance seen by the analog input pin is 2.5K. The pot could be up to 40K and you would still stay within the 10K recommended max source resistance. You can even go to 100K or higher still, since you don't need a high absolute accuracy in this application, although you might want to connect a 1000pF capacitor between the pin and ground as a precaution against noise.

However, the normal current consumption of the mcu is much higher than the 0.5mA drain by a 10K pot. You'd need to have the mcu spend a substantial part of its time in sleep mode for a high value pot to make a significant different to the current draw.