I just started playing with the Duemilanove. I tried looking, but I missed the specification regarding allowable current on the 5V or Vin Power pins.
What is a safe maximum current for each of those? I need 0.3A of 5V power. That isn't much power, but the traces on the Duemilanove looked awfully thin. Can I trust the power jack on the Duemilanove to provide that much power (on either the 5V pin or the Vin pin), or should I use another power jack and power the Duemilanove through the Vin pin?
300ma can safely be drawn from the +5vdc pin, however as stated by Mike it depends on how you are powering the board. USB should be fine, 500ma limit. External power can create a heating problem in the on board +5vdc regulator if you are using say 12vdc or higher to the external power connector. 8vdc would be the optimum voltage to connect to the Arduino if you want to draw that much 5 volt current for external components.
Thanks again. I was going to use 12V, so that rules out the 5V pin for that much current.
How about the Vin pin? If I take the 12V from the Arduino power jack, run it up to the Vin pin, to a 7805 regulator, that should give plenty of 5V power. Any problems with that?
Yes, the pins cannot drive that much without some form of "interface" device. (Example: External Transistor, relay, etc)
The external transistor would draw very little current from the PIN (good) and it could be supplied power by a separate 7805 as you mentioned. This solution would isolate your Arduino pins and low current regulator from your 300ma load.
How about the Vin pin? If I take the 12V from the Arduino power jack, run it up to the Vin pin, to a 7805 regulator, that should give plenty of 5V power. Any problems with that?
The external power jack already wires to the Vin pin (the Vin pin will output whatever voltage is plugged into the external power connector), and you could take that and wire it to an external 7805 regulator. That would be a better method in my opinion then drawing higher currents from the Arduino's +5vdc pin.
However you still have to deal with heat management for the external 7805. 12v - 5v = 7 volts X .3 amps = 2.1 watts of heat dissipation for the 7805 regulator. That requires a heat-sink for the regulator for reliable operation.
Like these: Heatsink TO-220 - PRT-00121 - SparkFun Electronics
I have a 12 v external power supply connected to the arduino dc input, at the same time its connected by usb to a computer for sensor readout to a server application.
The 12 v ext. supply is there for powering some fans and a motor, which I drive from the Vin pin (by transistor on/off circuit).
From the 5 v pin, I will drive a couple of sensors intermittent (SHT15, DS18B20, led), so I guess the amps wont go up in the extremes on that pin.
Im trying to figure out what this will mean in terms of overheating problems for the built in regulator. Since the big load will go through the Vin pin, it should not be considered when calculating heat dissipation for the built-in 5 v regulator?
Since the arduino is also connected via usb, does this mean that the 5v comes from that connection and the top load still is 500ma for 5v operation?
The Duemilanove switches the power source automatically. It will always default to USB. So if its plugged in there then all the power is coming from the USB port and not you 12V power source. If the motors aren't connected to the arduino then there is no reason for the 12V power source to be. The sensors will be powered of the USB.
"Im trying to figure out what this will mean in terms of overheating problems for the built in regulator. Since the big load will go through the Vin pin, it should not be considered when calculating heat dissipation for the built-in 5 v regulator?"
Correct
"Since the arduino is also connected via usb, does this mean that the 5v comes from that connection and the top load still is 500ma for 5v operation?"
If you are suppling power via the external power connector on the Arduino then the auto voltage switch on the Arduion uses the on board +5VDC regulator to power the board, only if the USB connection is the only power source is the on-board regulator switched out and the +5vdc comes via the USB connector and has a 500ma thermo fuse to set it's current limit. The on-board +5vdc regulator is not functional at all when you only have USB connected.