I was using "Arduino Uno" and "Mini L298N" (MX1508) to control a motor. They both had different power supplies (USB to arduino and 12V battery to the H-Bridge) however I forgot to ground, the circuit is attached.
Reading this post I understood the importance of sharing ground. However, my circuit was working just fine, how? As far as I know it would just not work at all. But on that post he also said:
"Without that link the voltage difference between ground on the 5V side and 12V on the 12V side could potentially be many hundreds of volts"
Why is that?
And my last question(s) (that I dont really know how to ask), if they are not sharing the same ground, the current will flow from the "signal" to the battery ground (instead arduino ground) correct? Or no? I know the signal is 5V compared to the arduino ground, but it is just impossible to know how many volts compared to the ground from the battery? (another image attached)
According to your Fritzy thing there are 4 wires excluding the ground between the Uno and MX1508. If it's working then the signal will be finding it's way back through protection diodes and whichever wires it can.
Your schematic is not accurate because you only show one wire not the 4 that are actually there.
The suggestion that the current can return to a different ground is nonsense.
Often different power supplies do have common grounds via mains earth, but if they are
isolated from one another a single wire is not going to carry current (other than at RF where
it allows each circuit to act like an antenna for the other!)
PerryBebbington:
According to your Fritzy thing there are 4 wires excluding the ground between the Uno and MX1508. If it's working then the signal will be finding it's way back through protection diodes and whichever wires it can.
Your schematic is not accurate because you only show one wire not the 4 that are actually there.
The suggestion that the current can return to a different ground is nonsense.
