Arduino Mega 2560 Rev 3: Maximum Current Draw from the 5V Out?

Hi all,

I have a fairly simple question (I think). I know that the maximum current draw from the 3v3 port is 50mA. Does anyone know the maximum current draw from the 5v port?

Thank you.

The current available from the output of the 5V regulator depends heavily on the input voltage (Vin or power jack). The recommended maximum power dissipation for the 5V regulator is 1 Watt. The regulator is not well heat sinked and can over heat with greater power dissipation and shut down. The regulator must drop the input voltage to 5V so with a 12V input the dropped voltage is 7V. To get the max current divide the 1W by the dropped voltage. 1W / 7V = 143mA. The Mega, itself, takes about 50mA so that leaves less than 100mA for the rest of the circuit (12V input). With 7.5V input there is 1W / 2.5V = 400mA available.

groundFungus:
The current available from the output of the 5V regulator depends heavily on the input voltage (Vin or power jack). The recommended maximum power dissipation for the 5V regulator is 1 Watt. The regulator is not well heat sinked and can over heat with greater power dissipation and shut down. The regulator must drop the input voltage to 5V so with a 12V input the dropped voltage is 7V. To get the max current divide the 1W by the dropped voltage. 1W / 7V = 143mA. The Mega, itself, takes about 50mA so that leaves less than 100mA for the rest of the circuit.

so with a 5V input I get infinite power ?

So the other point is in the doc:

The board can operate on an external supply of 6 to 20 volts. If supplied with less than 7V, however, the 5V pin may supply less than five volts and the board may become unstable. If using more than 12V, the voltage regulator may overheat and damage the board. The recommended range is 7 to 12 volts.

so with a 5V input I get infinite power ?

The math is always right. Except when it's not.

I assumed that the OP would know the lower limit for the input voltage, silly me.

groundFungus:
I assumed that the OP would know the lower limit for the input voltage, silly me.

You were probably assuming correctly, could not resist that one though