# Can someone explain the science behind MIT's 230% efficient LEDs?

I was reading Gizmodo the other day and I didn't quite understand the Physics behind this. Could anybody shed some light on how this effect actually works?

-
Good question. Though I don't know enough to make an answer, it seems to me that what is going on is that the electricity is essentially catalyzing light production in the LED. The light production itself is merely the result of an endoenergetic event and that energy can come from many sources. –  AdamRedwine Mar 7 '12 at 20:01
This is really interesting. It sounds crazy at first but they basically invented a phonon to photon heat pump and I am curious what the limits of these processes are. –  Alexander Mar 7 '12 at 20:41
The paper is here (behind paywall) link.aps.org/doi/10.1103 and the synopsis at Physics is here /PhysRevLett.108.097403 physics.aps.org/synopsis-for/10.1103/PhysRevLett.108.097403 . The synopsis confirms its a heat pump. –  Frédéric Grosshans Mar 7 '12 at 21:38
The only real answer here is more incompetent science journalism. –  Colin K Mar 8 '12 at 4:03
@ColinK : It is not incompetent science journalism. The abstract of the Physical Review Letter itself and the answer below show that the efficiency is indeed above 1. –  Frédéric Grosshans Mar 13 '12 at 14:26

As others have said, a heat pump with an efficiency greater than unity doesn't have any problem with the laws of thermodynamics. There are still a few problems with this which mean that it isn't as good as it seems.

The efficiency of a heat pump depends on the temperature difference - you can get higher efficiencies when you're just increasing the temperature a bit. Now the effective temperature of white light is the temperature of the sun's surface, so there's no way that this can be a light emitting diode. In fact it emits at a wavelength of about 2.5 microns, which is in the infrared.

However, that still corresponds to a temperature of about 1200 K. The device is heated to 135 degrees C, i.e about 400 K. Thats still too much of a temperature difference for a heat pump to have 230% efficiency, so there's clearly something else going on.

Were they adding in the normal black body radiation from the device? No, they were subtracting it, but the interesting thing is that it is much larger than the emission from the device. In the paper they say that the black body emission in the appropriate wavelength range is about 40 nW, while the emission from the device is 69pW.

This explains how they can have a heat pump with such high efficiency. If you think of the effective temperature for the emission, well 40 nW is 135 Celsius, so 40.069 nW will be a bit more than 135 Celsius. So there's no problem in having a heat pump with 230% efficiency.

-
I think that this is the most coherent and easy to understand answer –  Kian Mar 13 '12 at 11:57

I can only hypothesize that the extra energy is due to cooling of environment. As there is some input of energy, that would not contradict the second law of thermodynamics.

-
That seems right -- quoting from the abstract, "The extra energy comes from lattice vibrations, so the device should be cooled slightly, as occurs in thermoelectric coolers." What they actually claim is that their device outputs more light than the electrical energy it consumes. Why this is not in violation of the second law of thermodynamics -- I still don't understand. –  Anonymous Mar 7 '12 at 21:17
One of the wordings of the second law: "No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature." (en.wikipedia.org/wiki/Second_law_of_thermodynamics ). However, if there is also some input of external energy, such a process is possible. –  akhmeteli Mar 7 '12 at 21:51
I skimmed their paper, one thing that I noticed is that they heat their LED well above ambient temperature. I'm curious if it could still work if it was in equilibrium or if blackbody radiation would disrupt this effect. –  user2963 Mar 7 '12 at 22:28
@Anonymous : The first line of the paper is clear about the second law : "The presence of entropy in incoherent electromagnetic radiation permits semiconductor light-emitting diodes (LEDs) to emit more optical power than they consume in electrical power, with the remainder drawn from lattice heat." –  Frédéric Grosshans Mar 9 '12 at 11:53

2012-03-08.17:37 EST:

Kian Mayne: Good question! Since everyone knows energy is conserved, the title of the Gizmodo article summary sounds instantly self-contradictory. It certainly makes you want to figure out what the authors really meant, since unless they completely misunderstood the original paper, they clearly did not mean that you get 2.3 times as much energy out as light as you put in as electrical energy.

A careful reading of the summary articles points out near the end that what they are really talking about is higher overall efficiency achieved by converting not just electricity into photons, but also "waste heat" into light also. The article defines "waste heat" as "... vibrations in the bulb's atomic lattice ...", which is certainly true since that pretty much is the definition of heat. They also mention ambient temperature affecting the process, but the summary itself doesn't give enough information to assess that part of the claim.

So, what's almost certainly going on is this: Sending electricity through their device is generating both light and heat, with more of the energy going into heat than into light. That's not surprising, since all light systems generate heat also. The clever part of what this group has done -- and it really is clever, with some nice potential if it can be scaled up to regular scale LEDs -- is that they've found a way to recapture a large chunk of that heat energy and convert that into light also.

So, if you do the percentages in the more obvious fashion of looking at joules of energy in the form of light over joules of electrical power sent into the device while it is at equilibrium operating temperature (to avoid that ambient heat confusion), you will get a percentage that is less than 100%, as you would expect given conservation of energy.

So, the title of the Gizmodo summary is technically accurate, but it is also confusing because it refers to relationship that exists within the overall energy budget of the electrical current driving it. For example -- and I'm making these figures up! -- if the device converted 30% of its input electrical power directly into photons via traditional hole-electron recombination, and 39% of input power to photons via the new heat-capture, the overall budget would still lose 31% of the original input power, but would nonetheless radiate 2.3 times as many photons (230%) as would a 30% traditional LED for the same electrical power input.

So why might the be important if it is scaled up? Two reasons, really. The first of course is flat-out efficiency. A 130% additional output for the same energy inputs is a very, very impressive gain if it can be scaled up!

But the second is every bit as important, and possibly even more important in the broader scheme of things: Such a bulb would run far, far cooler than a regular LED. Most people don't realize that at this point, the main reason why LED lights have not yet displaced other forms of lighting is that despite how cool they are in comparison both to incandescent and florescent lights, heat is still a huge factor in creating large, bright LED lights, since the LEDs are solid state devices with some definite limits on how much heat they can handle before the lose efficiency or begin to incur physical damage.

So, this is good stuff. Scale up is always the issue in such cases, but simply showing that something like this is even possible is the real start. This group seems to have done just that.

-
If an article receives an 'Editor's suggestion' from PRL it is seems that this is not only due to some 'superb press corp' from MIT. Harshly critizing someone else's publication without even reading it is not constructive in any way. –  Alexander Mar 8 '12 at 19:31
Alexander, you are correct, and I apologize. Please see my complete rewrite above. –  Terry Bollinger Mar 8 '12 at 23:31
Reading the PRL paper, it's not what you describe : the 230% efficiency is from the plug power: So sending electricity in the device does not generate heat, ti transfers it from the device to the light. –  Frédéric Grosshans Mar 9 '12 at 11:53
Sort of a heat pump then, hmm? Interesting. Heat pumps of course still require electrical energy to run, but are more efficient. –  Terry Bollinger Mar 10 '12 at 0:51
Terry: It really seems to be an LED heat pump with a 230% efficiency from the input electric power. Unfortunately their effect only showed these properties at an 'ambient' temperature level above 100°C, but it is just a prove of concept so far. Thank you for your rewrite, removing the -1. –  Alexander Mar 11 '12 at 14:52

I just read the paper today and here's my take, with several points still not really worked out, any help would be appreciated.

The last paragraph of the paper is extremely illuminating, I'm quoting it here (it should be fine right?):

In conclusion, a new regime for electroluminescent cooling of a semiconductor diode was experimentally demonstrated. In this regime application of a forward bias voltage V less than the thermal voltage $k_BT/q$ im- poses a small deviation from thermodynamic equilibrium on the device. In response, the rates of both radiative and nonradiative recombination in the device’s active region have contributions at linear order in V, so that the external quantum efficiency EQE is voltage independent. As a result, the LED’s optical output power scales linearly with voltage while the input power scales quadratically, resulting in arbitrarily efficient photon generation accom- panied by net electroluminescent cooling of the solid at low bias. Experimental evidence was presented for small band gap emitters at high temperature.

So, the clever part of this approach is operating at the low-bias region (where the thermal voltage is a lot greater than the forward-bias). In this regime then, the steady-state carrier concentration is raised and this quantity ($\delta n$ and $\delta p$) are then proportional to the forward-bias (from Taylor series of the $e^{qV/k_BT}$ relationship).

Both radiative (producing light) and non-radiative (producing heat) recombination processes are proportional to $$np=n_0p_0e^{qV/k_BT}$$ This means the rate of recombination is also proportional to the voltage applied.

I'm not exactly convinced that $\eta_{EQE}$, the external quantum efficiency is subsequently voltage independent. But mathematically it probably comes from: $$\eta_{EQE}=\frac{recombination_{radiative}}{recombination_{total}}$$, voltage drops out and the result is indepedent from V.

Finally, the above unity efficiency comes from how optical power and input power scales differently according to the equation they used: $$P_{cool}=I(\hbar\omega/q)\eta_{EQE}-I^2R$$, where the first term is the total optical power, the second term is the total electrical power from the plug.

I think the equation makes more sense if we arrange the first term into: $$\frac{I}{q}(\hbar\omega)\eta_{EQE}$$, which says the total optical power is the emitted energy from $\eta_{EQE}$ of the total electron-hole recombinations.

So following this train of thought, the paper makes sense mathematically. A few questions remain for me though:

1) This relationship reveals that there is an optimal forward-bias at which the heat-pumping is optimal. Yet in the abstract, it states "the device's wall-plug efficiency diverges as V approaches 0". Am I missing something here?

2) The whole train of math makes sense. The physical picture I'm getting is that, in low-bias, excess carriers are created and a fixed fraction of them will combine radiatively. And the total light emitted somehow exceed the total input electrical power, thus leaving the $(1-\eta_{EQE})$ carriers...have no idea...

Glancing through Ram's other papers (http://www.rle.mit.edu/sclaser/Publications/Bias-dependent.pdf), seems like this would be explained pretty easily with a knowledge of optical cooling. Can anyone explain physically how low forward-bias FORCES extracting lattice energy?

-

Again, another hypothesis: LED's work on semiconductors. If the outside environment is hotter, we get excitation of majority carriers, which will supplement the light. It may be possible that the majority carriers are not excited enough when there is no electricity. Could be a laser-like stable-metastable effect.

-

## protected by Qmechanic♦Dec 19 '12 at 12:53

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.