Is there an input current limit in the digital pins on DUE?

Hi,

Following this page there is a max current sink in some digital pins of up to 9mA, this means if one of those digital pins are selected as an input and receive 3.3V 100mA it will only let 9mA go by without damaging the board?

F1_:
this means if one of those digital pins are selected as an input and receive 3.3V 100mA it will only let 9mA go by without damaging the board?

No, it does not mean that. The Due will not restrict the current to 9mA. What the spec means is that you must restrict the current yourself, using a resistor external to the Due. If you do not restrict the current, and the current exceeds 9mA, then the pin will be damaged.

Edit: my mistake, as Whandall points out below, if the pin is configured as input, it has high impedance, and will not draw very much current. External resistors are not needed in this case.

Doug

Okay, so I will apply resistors or something else to reduce the current, thank you.

F1_:
there is a max current sink in some digital pins of up to 9mA, this means if one of those digital pins are selected as an input and receive 3.3V 100mA it will only let 9mA go by without damaging the board?

If configured as input the pins do not draw any significant current,
you do not need resistors for inputs to delimit the current.

Thanks!

The value of current limiting resistor for maximum sink current as Output Port:

Given for PA0 (as per data sheets):
VDD = 3.3V, VOL = 0.4V, and IOL = 9 mA

R = (3.3 - 0.4)/9*10-3 =~ 320 ohm.

Q How much current would PA0-pin of DUE sink when driving a typical TTL load?

IOL = (3.3 - 0.7-0.4)/4*10-3 = 550 uA.