The maximum power rating of virtually all power-semiconductors is based on it being mounted on what's called an "infinite heatsink".
In a test-lab, this is generally a 1/4" thick copper plate with water-cooling coils brazed across the bottom side.
This eliminates the heatsink as an issue, and the device's limit is then determined by the thermal-resistance between the die (chip) and the external package surface (i.e. backside of a TO-220).
Basically, they run the current up until the internal die temp is at it's rated maximum, and that's the current-rating they assign to the part.
You have to understand that power-semi ratings are a marketing game ("specmanship"), and have historically been pretty optimistic.
As far as being able to "run" a part at it's rated power, the scenario another member described is only valid for "linear" i.e. dissipative operation.
In fact, in switching mode, you can easily -control- much more power than the "maximum dissipation" rating of the device.
The example given, a 100-watt, 3-amp 200-volt part, may well be able to switch 3 amps on and off in a 200v circuit. It would then indeed be controling 600 watts...and doing it with likely much -less- than 100 watts dissipation.
As far as current-ratings go...in general, it is very difficult to use more than about half the rated current of most parts. And as mentioned already, you usually run up against a thermal limit before anything else. Especially with TO-220 packages.
With larger devices, especially large stud-packages and hockey-pucks, because of the much larger surface-area, you can safely run a lot closer to the device's ratings.
In regards to using rectifiers of a much higher PIV than the circuit requires....that practice gives up some efficiency. Whether it's important or not depends on your application and voltage level; but bear in mind that the higher the PIV rating, the higher the on-state voltage will be. If you're dealing with a 6 volt circuit, an extra 1 or 2 tenths of a volt can be several % of loss right there. In a bridge circuit, you'd have twice that additional loss; since there are -two- diodes in series with the load.
hope this helps...