Dandelion,
 You have the current source placed correctly (though I believe you meant to label your output buffer a transmitter and not a reciever) and if you are doing simulations to validate specs (Voh, etc), I would stick with the current source as a resistor will be impacted by the dynamic output resistance of your transmitter, making it difficult to control Io for spec validation purposes.

If what you are doing, assuming I understand correctly, is validating your spces (Voh, Vol, Io, etc),  the only condition with the current source is to get the polarity right.  For Vol, you are sourcing current into the buffer so your polarity as you have drawn it would be positive (+Io), while for Voh you need to be sinking current out of the buffer so your polarity should be negative (-Io).  What you are testing it for, in the Voh example, is that even when the part is sourcing Io to the load, it is still maintaining a voltage high enough to meet Voh specs, and vice versa for Vol.  (note I am assuming |Iol| = |Ioh|, which may not be true in which case make sure to match the polarity with proper value +Iol and -Ioh).

Note that this approach is how you would consider simulations for TTL or ECL type circuits.  Sometimes, in say a GTL logic family, certain aspects of these specs make no sense, such as in GTL, you would never expect to see a -Io in your schematic as all loads are pull up and so assuming the drivers voltage is never higher than the pull-up Vcc, Io will always be positive.  (polarities are specific to your schematic).  However, you can still simulate for the specs.

I am not quite sure I understand your reciever question.  Recievers do not have Voh and Vol specs per se, but they do have to trigger appropriately for such values as Vil and Vih, and how you simulate these voltages depends on your IO standard.

I hope this helps,
SRF

Dandelion,
 So here are you answers as I understand them.

I see why you called the block a reciever, no worries.  Our focus however is on the single ended output pins.  This makes sense to me now.

Regarding the spec:

For Voh, the spec is stating that if the load your output pin is driving, sinks 0.4mA, then you must atleast maintain a voltage of 3.8V all while Vid = +200mV.  That is the spec.  If I relate this to your first schematic:

Vid = +200mV  (Vin1-Vin2 ??), Vout attempts to drive the load to a high value, i.e. Vdd or 5V;  but your OUT pin is loaded down by a resistive load sinking 0.4mA ( Ioh = -0.4mA) than we want to ensure that our OUT pin, even when loaded will still drive a high value that is greater than Voh.

In otherwords,  OUT > Voh.  Your spec has a typical of 4.9V but a minimum of 3.8V.  Therefore when you tell me that your part drives the output almost to Vdd, even with the load current, that I tell you your part is meeting spec.  Remember, you are trying to drive a HIGH output value, the Voh is merely the minimum allowed voltage for a given load current.  Your part is passing spec.

For Vol is just the opposite.  Your part is trying to drive a low ouput, and when it is contending with say a pull-up resistor that sources 2mA into your part, your output is required to keep the voltage BELOW the 0.3V max Vol.  Note that typical is 0.07V.  So when you tell me your part is driving its output to 0V, even while sinking 2mA from a load, than I tell you your part is passing.

So the conclusion:  Voh and Vol are spec limits, the further you are away from them the better off your part is performing.

I am going to rewrite part of your last message to clarify:

When I validating the Voh and Vol spec using the above digarms for all the PVT corners. I found the Vout is always near at 5V (for Vid = +200mV, Io = -0.4mA)  and Vout is always near 0 (for Vid = -200mV, Io = 2mA) for all the PVT corners. I wonder why the much difference compared with the spec, after all, it is low to 3.8V in Min for Voh (my Vout >> Voh) and high to 0.3V in MAX for Vol (My Vout <<Vol)
The question here is that we used the ideal current source. so we can not get the correct output level like the actual load, Some one suggested me to use the resistor to replace the ideal current source, i.e., select the  resistor properly to get the needed current and monitor the output, then the results is near the spec now.

Do you see the difference?  You are not testing Voh or Vol as actual circuit values, they are merely specs that your Vout must meet for given conditions.  And your part is meeting spec.  You are getting the correct output levels and the current source is acting as the load, for the Io you are supplying is the loading spec.

So why then is the spec so far from actual performance?.  Because these specs have decades of history behind them and it used to be that older technologies could not drive high output and low outputs very well when they were loaded down, mostly because of high source impedances in older digital technologies.  Modern technologies have no troubling delivering a very low source impedance so as to be able to drive large currents at high voltages or sink large currents at low voltages.

You part meets spec if I understand you.  As for testing, go ahead and use resistors, but I would use a variable resistor so that you can monitor different voltages and current levels and find the margin of your part.

This was a long answer but I hope this helps.  If I am still minunderstanding what your are asking, please advise and I will attempt to take another stab at it.
-regards,
SRF