I am working on a project with a pro mini to be powered by batteries and I am trying to reduce power consumption as much as possible. I am using a power down library available and it works great. For my application I need to take a measurement every few seconds or even minutes, so the MC could be in power down most of the time and consume little. After removing the power LED and while in power down mode, my multimeter measures 0.6 mA. The thing is, part of the circuit is a 10k potentiometer which I have connected to the Vcc pin, GND and A0, as I have read everywhere. I have noticed that if I remove the potentiometer the current drops to 0.1 mA. My question is if something speaks against connecting the pot not to the Vcc pin but to a digital pin set as output. That way, changing the code accordingly I could feed the pot the 5V from the digital pin for a short time while the measurement is made and then bring the pin LOW again. Would there be something wrong with that? Maybe the pin could be damaged if the pot is turned too much in one direction?
Another idea was leave the circuit as it is and use a 100k pot to reduce current through it.
Thanks for answering. I am making a humidistat to switch a fan on/off depending on air humidity. The pot is to set the humidity at which the fan state changes. So for example, setting the pot at the middle would set the humidity threshold at 50 %.
That is my concern, that too much current will flow through the digital pin if the wiper is turned too close to the 5V lead of the pot. In the page https://www.arduino.cc/en/Tutorial/AnalogInput it says: "The circuit based on a photoresistor uses a resistor divider to allow the high impedence Analog input to measure the voltage. These inputs do not draw almost any current, therefore by Ohm's law the voltage measured on the other end of a resistor connected to 5V is always 5V, regardless the resistor's value." Does that mean that not much current will flow between the digital pin and the analog pin regardsless of the wiper position?
The total resistance of the pot (outer leg to outer leg) from Digital out to Gnd will determine the current flow from the digital pin.
Current into the analog pin from the wiper will be 1uA max. Next to nothing.
I build a home humidification system--similar to yours. I used a DHT22 to sense humidity and when below a certain amount a humidifier would start (the reverse of what you are doing, but same in principal). the DHT22 is connected to analog inputs where a library interprets the data, such as TheHumidity = dht22.humidity. Then i set a digital output to trigger a solenoid to turn on the humidifier.
franzdrs:
That is my concern, that too much current will flow through the digital pin if the wiper is turned too close to the 5V lead of the pot. In the page https://www.arduino.cc/en/Tutorial/AnalogInput it says: "The circuit based on a photoresistor uses a resistor divider to allow the high impedence Analog input to measure the voltage. These inputs do not draw almost any current, therefore by Ohm's law the voltage measured on the other end of a resistor connected to 5V is always 5V, regardless the resistor's value." Does that mean that not much current will flow between the digital pin and the analog pin regardsless of the wiper position?
Thanks.
You're kind of mixing up a few different concepts there. The analog input consumes so little current that you could leave it out of your circuit for the purpose of analyzing current and voltage. So, look at your schematic for your pot. Normally one end is connected to 5V and the other to ground. The analog input is on the wiper in the middle, but we are ignoring that for now.
Where does the current flow? How much? It's pretty simple as there's only one resistor.
If the ground end of the pot accidentally became disconnected, then the analog input will get 5V regardless of the position of the wiper.