I have an Arduino Uno driving a scrolling display on a MAX7219 4x8x8 LED display. All powered via USB. Works great, and the Uno can handle 500 mA, which is enough for everything.
But the Nano can, I understand, only handle less (I believe I read 100 mA?) from its +5V pin. This would not be enough for that LED display.
So how can I do this? If I had access to the raw USB power, I would be good, but I understand that I not the case. Even if I use a separate 9V Supply, I think I still can't access that directly - correct?
Any advice welcome, as always.
Michael
PS yes I did read up and search first, but the replies leave me more confused than I was
The datasheet says otherwise. 100 mA max current output. If I could attach a screen print here, I would.
So are you saying that this does not apply to the +5V pin? Ie the actual CHIP output is 100 mA max, but the 5V output is only limited to the USB max output?
If you power on Vin with e.g. 7V, the 5V comes from the regulator which only can deliver a small amount of current due to the lack of heat sink. If you use the USB, the power goes through the diode. According to the datasheet, that diode can handle 1A.
Result of the diode is that 5V will be reduced a little to around 4.5V; the MAX7219 spec states that Vcc should be between 4 and 5.5V so I don't expect problems there.
Question is which datasheet you're refering to; you can always post the link (url).
It prevents the 5V from the regulator flowing into the USB port which could (would) damage the USB power source (usually your PC).
All signals marked +5V are connected on the board. So there is only one +5V; it's connected to e.g. the 328P, the TTL-to-USB converter and the 5V pin on the header.
