OK call it 1W. One 1W at 5V is 200mA. 1V drop at 200mA is only 0.2W for the regulator to dissipate. Easy for a 7805 with minimal (if any) heatsinking and will still be quite efficient - around 85% efficient in fact which is about the best you will get.

For reference a 1N4004 diode with a 200mA current, the forward voltage drop is 0.8V. So one wouldn't be quite enough and two would be way too many.
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Christian
#40104192 120Gb (no longer in my E36 M3, won't fit the E46 M3)