I am going to try to implement the circuit that Admin has in the Battery Monitor Circuit diagram on my Axon and I want to make sure my math is correct and my hook up procedure is also correct. (I would hate to fry something in my Axon).
First I will be using a Axon and it looks for a upper limit of 5v on a analog input so Vlog-1 would be 4. then I am using a 7.2 volt supply and if it is fully charged it is actualy about 8.6 volts so 8.6 divided by 4 equals 2.15 and 50,000 divided by 2.15 means I need a 23,200 ohm resistor in the R2 location. If this is all right I plan on doing the following.
I connect P1 to my positive lead between the battery and my Axon Positive Bat input pin (P2). then I connect P3 to a analog input ADC1(inside pin) on my axon and P4 to the ground (outside pin) on the axon?
My image of my schematic is not inserting properly (my bad) please look at the PDF attached.