Opamps have transistors in their output stages to deliver these currents. A saturated output stage is one that cannot deliver any more current, even if you drive it harder. It is saturated !
Look at page 12 of the PDF data to see the output stage is two bipolar transistors, the PNP one for pull-up, and the NPN one for pull-down. A saturated transistor is not a zero resistance device. There will be some volts across it that will increase as the current increases.
Notice also that
both axes of the data graph are logarithmic.
Thus the very nearly straight sloping part of line still indicates proportionality. ie. the apparent resistance of the fully turned on transistor is nearly constant! The increasing current simply delivers increasing saturation voltage. You should appreciate that to get this measurement, the current is being controlled by some external circuit. The transistor, being saturated, no longer has any say in limiting the current.
Yet the logarithmic scale use does exaggerate the display of voltages and currents that are very small. Thus you get to see that there is a certain minimum voltage drop across the transistor that is independent of the current, and is there even when the current drops to zero. It is, however, somewhat dependent on temperature. There are three separate curves given for different temperatures. When the current is reduced to the point the voltage drop due to bulk resistance is not significant, you get to see the remaining voltage left on the collector because of the saturated state of the base-emitter junction.
Finally - be aware that the saturation voltage is
not the output voltage of the op-amp at the time. Notice there are
two data graphs. One for 'output high' and the other for 'output low'. The saturation voltage is
the difference between the (low or high) supply rail and the output terminal, when the opamp is driven so hard the output saturates.
I hope that does it, because I am better at using the things instead of thinking about their insides.
