The key wording is here:

 

"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.*"

 

It is converting *heat* energy to light. not electricity-to-light!!!

 

Thus, as they *lower* the forward bias V,  *electrical* efficiency INCREASES
because it is not using electrical current for operation; as Jones said,
it's the E-field which ALLOWS the HEAT-to-LIGHT conversion.  If the material
is not very conductive, one can have a large E-field with miniscule current
flow. thus, very little ELECTRICAL power use.

 

-Mark

 

From: Daniel Rocha [mailto:[email protected]] 
Sent: Tuesday, February 28, 2012 9:21 AM
To: [email protected]
Subject: Re: [Vo]:Over unity at MIT

 

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]

 

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