Electronic – Why does the PWM amplitude vary with duty cycle

With the resistor in series with the inductor and your scope across the LED you will get a rising voltage with rising duty cycle then the LED starts conducting at about 2 or 3 volts and you won't see any increase in voltage rise. Try putting the resistor in series with the LED and measuring across R and LED not just across the LED: -

As you can see, to raise the voltage across the LED to about 1.6 volts requires virtually zero current from the MCU pin. To raise it to 2volts requires 20mA and that will be around the limit from your device. Hey, even if it could supply 50mA the LED voltage would rise only to about 2.37 volts. See this if you want more clarification.

Also, make sure your MCU output is capable of handling the kick-back from the inductor without causing problems. Normally a fly-back diode is used or a heavier-duty push-pull stage.

Also, get rid of those meaningless scope images - we know they'll be showing a dc voltage because of the 220uF cap - just tell us what the voltage is by measuring it with a multimeter.

You can easily use a MOSFET if the Fan specifies a weak internal pull-up and a 2N7000 fits a 12V low current application.

Whether you should worry about the fan supplying more than officially specified, is another thing. If you have shorted the pin to ground for your measurement and 0.68mA came out, that's so incredibly well within the reach of a 2N7000 that you shouldn't worry about the FET too much. You're welcome to worry about the Fan itself, if that helps you in any way, but a quick test with an arbitrary PWM signal will show if it is broken, or just built different to (EDIT:) the official prescribed spec.

All in all, I say, go for it. If you are really doubtful, you can limit some FET damage by adding a resistor in the drain path, like so:

^{simulate this circuit – Schematic created using CircuitLab}

The 2N7000 has an absolute worst case drain-source resistance of 13.5Ohm (the one Fairchild makes, at least) and I expect in your use-case that it will not exceed 3Ohm even when hot.

(EDIT2:

Sunday made my maths go wobbly, changed the voltage drop and the conclusion to it in the following block:)

So at the very worst the resistor and the FET will share the risk, normally the resistor will catch the brunt of it. At 5mA maximum the resistor will only waste 0.05V, which is negligible, you can even increase it to 20 or 30 ohm and it should still be interpreted as low. In the worst kind of failure (PWM shorted to 12V in the fan) the 10Ohm will effectively limit the current through the FET to about 1A (if the FET also is at least 2 ohm then), which is an acceptable very short-term limit. 30Ohm would probably keep the FET fully safe until the resistor itself burns through.

9999 out of 10000 times the resistor is completely redundant and pointless. But it is an option if it helps you sleep better (2 cents for good sleep, I call a steal)