next topic »

[on:]

I am interested in performing the typical which battery lasts longest experiment with some AA and AAA alkaline batteries. The idea was to simply put maybe a 20ohm resistor in circuit with it then measure the voltage. 1.5v / 20ohms = 75mA so my current in mA = analogRead / (1024 / 5 * 1.5) * 75... Or something like that, just a general idea.

My question is, when the voltage of the battery drops, is this equation still valid or will it not account for a dying battery? Because obviously that's the point.

Secondary question, I might not even worry about it, but can I drop the vRef to 3.3v by simply wiring it on the breadboard or PCB and connecting vRef to a 3.3v source? Then obviously the equation would change.

Thanks.

[Reply #1 on:]

I would hook  a resistor across the battery read the volts drop across it till it dropped below the 75mA level and yes you can change the  vRef to 3.3 but i would leave it.

[Reply #2 on:]

http://mad-science.wonderhowto.com/inspiration/diy-arduino-battery-tester-reveals-secret-capacity-disposable-batteries-0134638/

Enjoy

[Reply #3 on:]

While this would definitely suit my needs, I was hoping to get the actual current. That way every few seconds I could see what the current was then at the end I could get an actual mAh reading rather than just a voltage/time number. Could I still do that?

[Reply #4 on:]

While this would definitely suit my needs, I was hoping to get the actual current. That way every few seconds I could see what the current was then at the end I could get an actual mAh reading rather than just a voltage/time number. Could I still do that?

If you discharge the battery through a known resistance and measure the voltage across the resistor then you can calculate the current through the resistor; by integrating the current over time you can calculate the charge capacity. By integrating the voltage * current over time you can calculate the energy capacity.