Hi all,
can anyone explain the purpose of diode D3 on the Cerbuino NET and the 12V use?
The power connector states 6 - 9V, so there never will be 12V on the 12V point if powered from a wall adapter. If 12V were connected somehow on the 12V then the power adapter won’t be necessary? Am I correct?
It confuses me :-S
It is needed to project the board in case of reverse power error.
What are you trying to do? Powering the board from 12v is okay temporary for testing but it will overheat the regulator in extended use.
Ah the joys of auto suggest on phones 
Gus meant to say “protect”. When the diode is forward biased current flows into the board else not.
I’m not trying to do anything. I’m just trying to understand the board.
What is the purpose of the 12V. Is it output or input?
From Gus’ reply I think it’s not meant as input, so it must be output? But how can 12V be ‘made’ out of 9V max on the power jack? The 12 confuses me, it won’t be 12 if powered from 9V?
I see. Please ignore that it says 12v. This signal is Vin. What you put it is what you get out.
The only place “12v” appears is the schematic, which is just a name given to a particular line. As Gus says, that’s essentially the “protected” VIN line from the barrel plug. It’s certainly not an output line, and the silkscreen is more correct in that you should only supply 6v - 9v input.
Does this use the same circuit as the USB Power EDP module? I haven’t looked… But since they both have the same 6-9V limits then my guess would be that they do. If that’s the case then this is not good advice. I know from experience that the EDP module cannot handle more than 2-3 seconds of 12V power before one of the diodes will turn into [blue] flames.
Yes and actually the regulator can accept more than 12v but heat will be a problem.
Ok, bit confusing though ??? VIn as on the arduino boards would have been clearer.
Thanks for the help!