Hi.
I have been working with 5v pro micro Arduboy clones. The way I wired them up they were powered by 4xAAA batteries via the RAW connection. So the internal power regulator dropped the extra voltage. However, when connecting the board to the PC at the same time via the USB I am assuming that there will be no danger of damaging the PC or the pro micro.
Is this true? Or should I be expecting smoke?
Thanks
Steve.
Looks like D10 should have you covered. You should be fine. (I don’t have the specifications for D10 but I’m assuming it can handle 6 volts)
Assuming the schematic that @bateske posted above:
No danger to the PC or Pro Micro, but possibly to the battery. (I use the term battery to describe your 4 AAA cells).
As long as the battery voltage remains above the USB voltage minus the D10 diode drop, you will still be running off the battery when plugged into USB. If the diode is a regular type, the drop will be about 0.7V. If the diode is a Schottky type, the drop will be about 0.3V. Assuming a USB voltage of 5V, this means a battery voltage above 4.3V for a regular diode and 4.7V for Schottky.
If the battery voltage falls below the above 4.7V or 4.3V the USB voltage will feed current into the battery. For a non-rechargable battery, this isn’t good (since you should never try to recharge a non-rechargable battery). For a rechargeable battery, this also isn’t good since there’s nothing to limit the charge current (other than the F2 fuse or a fuse in the PC), so you could end up overheating or otherwise damaging the battery.
I would put a Schottky diode in series, between the battery + and RAW input, to protect the battery.
If you want to actually power the circuit from USB, I’d consider also adding a switch to allow you to disconnect the battery.
In my previous post, I wrongly was thinking about a 3.3V Pro Micro. With a 5V Pro Micro, things are a bit worse. If the output of the U5 5V regulator ends up higher than the USB voltage, current will flow through the closed solder jumper, at the top of the schematic, into the USB port of the PC. This is unlikely to damage the PC but it will discharge the battery faster and, depending on the amount of current, could cause the regulator to overheat. The regulator has thermal protection, so it might not be damaged, but it certainly isn’t very good for it. Fuse F2 could also end up “tripping”, which isn’t very good either.
There isn’t a good solution to this other than making sure the battery is disconnected when the Pro Micro is attached to USB.
I would go for the schottky diode in series with the battery as @MLXXXp suggested and make sure the solder jumper near the micro USB connector is open (They are not soldered on most Pro Micro clones but you can’t blindly trust they’re always open)
The diode will also protect against reverse polarity of wrongfully inserted batteries.
If the jumper is left open then the circuit will always be powered through the regulator even when on USB. And, assuming D10 is a Schottky diode, you’ll loose another 0.3V on the input of the regulator. Therefore, with a 5V USB voltage, you’ll be feeding 4.7V to a 5V regulator. Assuming about a 0.1V dropout voltage across the regulator, the circuit will be powered at about 4.6V. This will probably work fine, but the jumper is there to assure a full 5V when powering from USB (and also from RAW, as long as the input voltage is above the regulator dropout voltage).
Of course, if you use 4 NiMH cells, you have the same situation since their nominal voltage will be about 5V.
