Im powering a nano and servo from a 7.4V battery pack. So Ive used an LM7805 to trim the voltage down to 5 for the nano and servo.
The thing is that part of the project also uses an LCD to display the voltage (as a proxy to SOC for my battery). So originally the LCD will display the 5V but Im wondering how that will vary as energy in the pack is consumed and voltage drops. Is it linear?
So will a drop from 5->4.7V (6% drop) = 6% * 7.4 = a 0.44 drop, resulting in 6.96V?