How much current does an IC pin source?

Does IC pin source same amount of current regardles of its load?

Lets say a darlington pair IC could output maximum current of 500mA, source current to a normal IC, say, an inverter. Could this inverter IC take so much current?

I would be grateful if you could tell me some guideline in selecting IC, especially in current rating. Thank you.

Even if it says it can source 500mA, it all depends on the load that you have connected to it. An inverter IC will not draw 500 mA. and thie inverter will not be able to put out nearly that much current.
So the answer to your question is no.

The load will only take as much current as is required by Ohms law: I=V/R. Where R is the load resistance, in this case, the input resistance of the driven IC.

What you state as 500ma is the max current that the ic can source under full load condition.
If you require more current you will have to cascade two ICs in parallel and achieve upto 1A of current.
In short if you connect a load which requires more than 500ma of current to operate properly it will not be possoble to do so with a single chip.
But if your load requires 300ma to operate properly then your IC will provide sufficiently and exactly the required 300ma
got it?

electronist said:

What you state as 500ma is the max current that the ic can source under full load condition.
If you require more current you will have to cascade two ICs in parallel and achieve upto 1A of current.
In short if you connect a load which requires more than 500ma of current to operate properly it will not be possoble to do so with a single chip.
But if your load requires 300ma to operate properly then your IC will provide sufficiently and exactly the required 300ma
got it?

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If one does not know what the output stage looks like from an IC that can source 500mA, then shorting 2 together is not a good idea without tying together through resistors. Each output stage presumably has low impedance and the voltages will not be equal, this creates a possibility of one of them hogging all the current since the load is likely to have a greater impedance. And if you dont know details for sure, dont just short them together. Do it through at least a couple of ohms of resistance.