Mega 2560 Ampere Tolerance?

Hello all:

I’m quite new to electrical engineering, but not so new to software engineering. I just got an Arduino Mega 2560 last week and have been toying with it. I’ve come up with an idea for a new, but still very simple project that requires it be powered by the 2.1mm barrel plug when it’s complete. I found a cable laying around that should work, but I wan’t to make sure. The plug I found is 12v and center-pin-positive, which I understand to be good, but what I’m worried about is the amperage. This cable produces 1.0 ampere. Is that too much? What is my arduino’s amperage tolerance and recommended rage?

I’ve spent three days searching around for this and have found no useful information. Any help would be appreciated.

Thank you,
KD0BPV

kd0bpv:
This cable produces 1.0 ampere. Is that too much? What is my arduino's amperage tolerance and recommended rage?

This is a very (very) common misconception. A source provides certain amount of current. Devices draw a certain amount of current. Your source does not produce 1.0A. It is capable of providing 1.0A.

If your devices only require 50mA they will only draw 50mA from the source.

Ah. Thank you very much for clearing that up. I also have one more quick question for now: I've heard that the Arduinos have a resistor built into pin 13 for the on-board LED, and as such, if I bypass the on-board LED with my own, I don't need the resistor. I know it's true that there is a resister there, but if I plug a wire into pin 13, isn't that connection then made BEFORE the resistor, thus bypassing it as well as the LED?

kd0bpv:
I've heard that the Arduinos have a resistor built into pin 13 for the on-board LED

True.

kd0bpv:
if I plug a wire into pin 13, isn't that connection then made BEFORE the resistor, thus bypassing it as well as the LED?

Correct.

The only affect this has on your use of the I/O pin is that the LED and resistor will draw about 2mA of current in addition to whatever you are doing. Which means you can pretty much ignore it.

The only affect this has on your use of the I/O pin

Well there is another, if you use that pin as an input, the internal pull up resistor is not small enough to pull it up, so you will need an external pull up resistor.