Why do you think it would violate the 2nd law? I don't understand. 2012/2/28 Harry Veeder <[email protected]>
> On Tue, Feb 28, 2012 at 11:57 AM, Daniel Rocha <[email protected]> > wrote: > > Pay attention at this: > > > > " Experiments directly confirm for the first time that this behavior > > continues beyond the conventional limit of unity electrical-to-optical > power > > conversion efficiency." > > > > It is above the conventional, not that it produces energy out of nothing. > > This is just a way of saying that it exceeded expectation of light > emission > > for a LED. > > > Yes. It uses electricity to change heat into light. The abstract: > > "A heated semiconductor light-emitting diode at low forward bias > voltage V<kBT/q is shown to use electrical work to pump heat from the > lattice to the photon field. Here the rates of both radiative and > nonradiative recombination have contributions at linear order in V. As > a result the device’s wall-plug (i.e., power conversion) efficiency is > inversely proportional to its output power and diverges as V > approaches zero. Experiments directly confirm for the first time that > this behavior continues beyond the conventional limit of unity > electrical-to-optical power conversion efficiency." > > > however, wouldn't this require a violation of the second law of > thermodynamics? > > Harry > > -- Daniel Rocha - RJ [email protected]
