Dumb break in question...

@Millercarbon,

The output stages of all (or at least virtually all) CD players, as well as all or virtually all other devices providing line-level analog outputs, operate in class A. Which means that the transistors or tubes in the output stage will actually be conducting very slightly MORE current when no load is connected compared to when a load is connected. In the case of the typical example I cited earlier, that difference would be 0.00002 amps.

An analogy can be drawn to power amps operating in class A. As you may realize, such amps will actually dissipate less power internally, and hence operate slightly cooler, when music is being played and current is being provided to the speaker, compared to when no music is being played.

Also, regarding ...

That said, the only way to fully burn it in is to play it, and to play it into a working operating turned on amp. Because turn the amp off, now you are back to the same situation where the circuit is open and nothing is happening.

... Many and perhaps most power amps (and preamps, if that is what the CD player would be driving) will present an input impedance when turned off that is not much different than when turned on. So in those cases the CD player won’t "know" the difference.

Regards,
-- Al

If you were to connect the outputs of the CD player to a typical input impedance of say 47K, and if the CD player is putting out an average of say 1 volt while playing a disc, the player would be supplying an average of about 0.00002 amperes (i.e., 20 microamperes) into that load impedance. Which is so little that it would seem very unlikely to make any difference, compared to breaking the player in without a load.

IMO. Enjoy the new player!
-- Al