I do have a problem with my arduino uno ..
Just bought it last week, having a nice project going..
I had trouble reading my SD card because i just used the USB power, so i decided to connect a
transformer wich i had layng around, it was a 12v 500mA dc supply wich gave me an output of 16v
nothing connected to it.
Before connecting i measured all the pins and and the + was on the center pin, negative on the outer ring.
In the specs it says max/min voltages are 6 to 20v , so my question is WHY TF thit it frie the
328 as well the mega 8u2 ????
has anybody else experienced this ?
Hello
I have a similar experience using a tranformer to power it.Im my case I dont damage because I notice it before any damage.
When I connect it to 12 V transformer I notice that the onboard regular heats to much as normal.
On the web I found the arduino can be powered at this voltage but I think the regulator cant handle it.
I decide to power it with just 6 -9 Volts and now the heat is less and works fine now.
Hope you can repair it
thanks ,
but i don think its reparable becouse of the mega 8u2.
( I dont have the equipment for that .. ).
Have to order a new one .. or maybe 2 lol ..
Or buy the parts as 2 components - an FTDI Basic for programming, and a RBBB, Ardweeny, or similar for the processor. Then if you fry the processor you can easily replace it, and you only have the USB interface connected while programming and powered from the USB interface.
I had trouble reading my SD card because i just used the USB power
This doesn't sound right. If the only hardware you have connected is a SD card, USB provides plenty of current.
In the specs it says max/min voltages are 6 to 20v
The regulator has a spec that allows up to 20V. However, the regulator has almost no heat sinking. So the practical limit is 12V, but entirely depends on how much current you are drawing.
How are you connecting an SD card? You have a shield for it?
Everything I've seen says they run on 3.3V, and can require a lot of current at times.
I had a design that used a pretty low current capable regulator, SD cards wouldn't work. Had to step to a higher current capable part like the adafruit shield design uses. http://www.ladyada.net/make/logshield/design.html
I have an SD card with its own 3,3v regulator (from dfrobot.com),
and that's wy i think it draws more current .. (and it wasn't the only thing connected)..
I also had 3 temp sensors connected and 3 led's so its just a bit more than USB can handle ...
A block diagram and/or schematic of your total circuit would probably be helpful. Just asking about a single component can lead you in the wrong direction.
so its just a bit more than USB can handle ...
"A bit more" isn't a number value. The amount of calculated or measured current is a better number to use.
In either case, your statement doesn't make sense. USB provides 500mA. Your power supply is rated as a 12V, 500mA. So in either case, your design can't be using "a bit more than USB can handle" because both options supply the same amount of current. Neither supply would result in a functional design.
In terms of the on board regulator, I suspect you were using a supply that wasn't well regulated. Look at the power dissipation at 12V and 16V:
12V: (12V-5V) * 0.5A = 3.5Watts.
16V: it was dissipating 5.5Watts.
That's a lot of power for a regulator with only a PCB as a heat sink. However, it all depends on the actual current being drawn.
Where was the 3.3V regulator getting its input power?
At first without all the other peripherals it was working ok after an upload.
and all the perifferals were working without the sdcard reader.
after putting them together, the arduino began to reboot every time it wanted to write to the card..
i could see that because of my diagnostic messages to the serial port ..
tested it again without all the perifferals, and it worked fine .. so because of that my conclusion was the current ...
and as for the power supply wich was 16v unconnected usualy drops a few volts (old coiltransformer)..
but since my arduino died, i could not measure the actual voltage.
).