Here were my connections : PA, PB and EA to three of arduino pins
to 5V arduino
to its ground
B+ and B- to my external alimentation (12V,8A)
Motor to M1 and M2
when i gave the code to the arduino ( it was simply based on a digital digitalwrite HIGH on EA and two analogwrite and EA and EB.) the D2 had a red color and like 2 second after i turned the external alimentation on, my computer suddently shut down and now i can't get any display from it anymore, wtf ? Isn't the USB port supposed just to turn off itself and then that's all ?
I looked at the data sheet briefly but see no mention of a voltage regulator that puts the +/- pins at 5 volts dc. What voltage do you measure at those pins? In fact is specifies that the + pin is for output to an EXTERNAL regulator so it looks like you fed your 12 volts to an input pin 'post regulator' and simply overwhelmed the whole board.
Ok so i don't really understand what are these two pins for ? if i use something else than an arduino?
So what should i have done ? No arduino alimentation, just ground on the - ?
You think i burned it and it wont work anymore ? And arent the usb ports on my computer supposed to be protected against this ?
USB ports don't have much protection - certainly not against over-voltage. They usually tolerate too much
current being drawn from the computer's 5V, but that about it.
yann10:
Ok so i don't really understand what are these two pins for ? if i use something else than an arduino?
So what should i have done ? No arduino alimentation, just ground on the - ?
You think i burned it and it wont work anymore ? And arent the usb ports on my computer supposed to be protected against this ?
I'll measure this tonight, thanks for your answer
Just the ground would be fine. If you want to run the Arduino from it you would have to put it back into the barrel jack so that it goes through the regulator. Is t ruined? Don't know. Plug it in and see but it doesn't sound good. I don't know about USB protection but suspect the previous poster is correct. They protect against what they expect an errant device would do with a short circuit at 5 volts being the worst. You supplied 12 volts capable of 8 amps.