MCF4213DE.pdf
Table 28. DC Electrical Specifications
Input low voltage VIL (min) VSS – 0.3 (max) 0.35 × VDD
Weak internal pull Up device current, tested at VIL Max. IAPU 10 –130 μA
What that means is that if you have external circuitry that is meant to pull the pin down to a valid low level (which is "below VIL Max") then it should be designed to expect to sink between 10 and 130uA at "VIL Max". Your design figure is thus 130uA.
On the other hand, if your external circuitry has to guarantee the pin will be at a valid high level (which is "above VIH min") then it must draw less than how much at VIH? There is no written specification for this, as designing for a maximum of 10uA may still pull the pin down to "VIL Max" and not "VIH Min". So for this condition you have to guess and add a large "fudge factor".
> the pull-ups tend to be about 20k.
(3.3V - (0.35 * 0.33V)) / 10uA = 214.5k
(3.3V - (0.35 * 0.33V)) / 130uA = 16.5k
But they are within specifications anywhere from 16.5k to 214.5k, so you should make sure your design works properly over that full range.
Actually the "pullups" probably aren't resistors and you shouldn't design assuming they are. Google finds the following forum answer for pullup questions on a PIC chiip:
http://old.nabble.com/Re%3A--pic--Weak-pullups-p4693014.html
They are not resistors. They are small p-channel MOSFETs typically
operating in the pinch-off region when conducting, so they are
(relatively) constant currrent sources for small values of output
voltage relative to ground. As the voltage on the output approaches
Vdd-Vth, the current will drop in a nonlinear fashion. The gate of
the MOSFET is switched to ground to turn the weak pullup on, and to
Vdd to turn it off (see port pin schematics in your datasheet).
Small FETs are easier/cheaper/smaller to make than actual resistors. The only specification is the given current range at a specific voltage, and after that you're pretty much on your own, and should add external pullups everywhere it matters.
Tom
