Arduino Ground Current from DC power jack

HI all,

I have a project where I'm providing power via the DC jack to my Arduino then accessing the DC power via the VIN pin. According to this thread I am ok to draw up to 1A from the VIN pin. I have measured 400ma out of VIN, so ok on that front.

But according to this Arduino Playground - ArduinoPinCurrentLimitations I can draw max 400ma, So I'm a bit confused, which is the correct amount of power VIN can supply if the Arduino is being powered via the DC jack?

If I take the first thread as correct and assume it's ok to draw up to 1A from the VIN pin (from the DC power jack), is it ok to ground that current on the Arduino GND pin ? if VIN can supply up to 1A then I assume that it's ok to ground up to 1A, is this correct?

All help appreciated

Look at the Arduino board. Use a magnifying glass, if you need to. Look at the skinny little traces. How thick do you think they are? 400mA at 5V is 2000 Watts. That's a lot of energy to be pushing through those skinny little paths, don't you think?

And, yet, you expect 2.5 times that much?

400mA at 5V is 2000 Watts.

2000 milliwatts = 2W. :wink:

Hi PaulS
I'm not expecting, I'm asking- I'd much rather get some concrete advice on this based on some sort of technical spec rather than what I can glean from a magnifying class, I wouldn't know what the capacity of the traces are by looking at them and I'm not sure that's the most accurate method.

Just to quote from the first thread I linked to:

"It's worth noting that I have about a thousand devices in the field drawing from 1.25A to 2A @12V through the Arduino Vin pin (to a shield), and none of them have burned out any traces or melted any pins in the Arduino =)"

Also 400ma @ 5 volts = 2W not 2000W

Thanks for your response.

DVDdoug:
2000 milliwatts = 2W. :wink:

That's what I said. OK. It's not what I typed... 8)

You confuse the power of the load, and the power dropped on the lines to the load. If the lines drop 5V from 5V, no voltage is left over for the load.

The really limiting factor for the DC jack power is the voltage regulator, which is restricted to <1W by the lack of a heat sink. The protection circuit will jump in and shut down the regulator on too high temperature or current.

DrDiettrich:
You confuse the power of the load, and the power dropped on the lines to the load. If the lines drop 5V from 5V, no voltage is left over for the load.

The really limiting factor for the DC jack power is the voltage regulator, which is restricted to <1W by the lack of a heat sink. The protection circuit will jump in and shut down the regulator on too high temperature or current.

There is no voltage regulator between the DC Jack and VIN, only a reverse protection diode D1- look at the
schematic

From the information I've managed to gather on this, it seems the limiting factor will be the trace size but is there a standard spec for these?