Could I read USB data from the mini USB port on an Arduino board, or is that only for installing Sketches?
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2\$\begingroup\$ What do you mean by "read USB data"? \$\endgroup\$– endolithCommented Jun 20, 2011 at 18:33
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4 Answers
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What do you mean by "read USB data"? If you are referring to devices like keyboards and barcode readers, then no. You'd have to have a board with "USB host" capabilities. That said, apparently there are shields that can provide this functionality.
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1\$\begingroup\$ If it's an arduino Leonardo, you can have it emulate a USB keyboard/mouse/whatever. \$\endgroup\$ Commented Nov 15, 2012 at 23:11
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You can read data from your Arduino, using a program like hyper-terminal on windows. To do that set up hyper-terminal to listen to which ever com port that the Arduino is on with the correct baud rate (Same as in your sketch)
The Arduino is a simple USB to Serial device as Aoi Karasu said.
answered Jun 21, 2011 at 11:07
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If you are on Linux or OSX, you can use the command:
screen /dev/ttyACM0 9600
ttyACM0 is the name of the device on my system, you may change it according to your system.
9600 is the baud rate, again you can change it.
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Arduino is not a genuine USB device. It acts as a Serial to USB adapter. You have to scan all COM ports in the system to find which one is it and then talk to Arduino (perform read/write operations) using standard serial port communications that come with your development environment.
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2\$\begingroup\$ A USB serial adapter is very much a genuine USB device, unless it has been mis-implemented. One normally uses an operating system level driver which bridges USB to the operating system's serial port API, but that's not mandated by the hardware - one could treat it as a raw USB device with something like libusb and exchange the USB data at application level. \$\endgroup\$ Commented Nov 15, 2012 at 20:50
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1\$\begingroup\$ This is not true for the Arduino Leonardo. The leonardo uses a USB-bootloader, and has no intermediate device. It can emulate any USB device the ATmega USB PHY can. \$\endgroup\$ Commented Nov 15, 2012 at 23:12