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I have read this article, Which says that conversion efficiency of the LED have exceeded 100%. The results are published in Physical Review Letters

In their experiments, the researchers reduced the LED’s input power to just 30 picowatts and measured an output of 69 picowatts of light - an efficiency of 230%.

How is this possible, is it not the violation of conservation of energy or I am missing something.

Edit: I missed that the LED worked as TE cooler and absorb heat from atmosphere to convert it to light. But still it is overwhelming observation for me.

Is it really possible that a device use thermal energy to produce light.

hsinghal
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    Yes, you are missing that the LED was converting heat from the environment. It's mentioned in the article. – CuriousOne Jun 09 '16 at 14:40
  • Note that although it's called an LED, it's actually emitting in the infrared (around 2.4 microns) and it's operating at a temperature of 135 degrees Celsius. Thus the thermodynamics seems fairly reasonable. – quantropy Jun 14 '16 at 20:20
  • Actually I'm not so sure about the thermodynamics - I haven't read the article in detail, but I'd guess that what's being measured is an increase over the existing blackbody radiation from the device, in which case you have to be very careful to include everything in thermodynamic calculations – quantropy Jun 14 '16 at 20:28

4 Answers4

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The device is apparently working as a heat pump, for which I give a brief theoretical analysis here.

In the example given, the $P_h=69{\rm pW}$ light output comprises the $W=30{\rm pW}$ input by the researchers together with $P_c=39{\rm pW}$ of heat that was formerly in the chip.

We can model the process by ideal heat pumping as follows. Heat drawn from the chip will lead to a drop in the chip's entropy of $\Delta S_c = -\frac{P_c}{T_c}$, and the light and output to the ambient World increases the entropy of the latter by $\Delta S_w = \frac{W + P_c}{T_h}$, where $T_h$ is the effective temperature of the light (measuring the latter's degree of thermalization together with its optical grasp). Since the light ends up in the environment, its effective temperature is ambient or greater.

The total entropy change of the World is then

$$\frac{W}{T_h} + P_c\,\left(\frac{1}{T_h} - \frac{1}{T_c}\right)$$

We know that $T_h>T_c$ because the effective "exhaust" temperature is at least ambient and $T_c$ must wind up less than this, because heat is being pumped out. So the second term with the brackets is negative: this means we must supply enough work $W$ to at least make the quantity positive. So the device can very plausibly (and probably does) work as claimed and comply with both the first and second laws of thermodynamics.

Community
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Selene Routley
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  • @Mark Fischler and WetSavanna aka Rod Vance, I am not so convinced about the working of heat pumps, I have earlier posted the question here. I will now try harder to understand the working of heat pumps. Thanks for your response, I will ponder over your answer and try to grasp the gist of it. – hsinghal Jun 09 '16 at 17:16
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    Is this the sort of thing that would only work well in the dark? My (possibly wrong) intuition is that the free energy comes from channeling the flow of heat from an excited bath of phonons to an empty (cold?) reservoir of photons. A well-lit environment will have $T_h$ be large, necessitating larger $W$, ... and asymptotically bringing the efficiency to 100% I hope? –  Jun 09 '16 at 21:02
  • @ChrisWhite I agree, but I can't for the life of me begin to imagine how you would calculate effective $T_h$ rigorously: there's so much to it! My main point is that there don't have to be thermodynamic surprises here. – Selene Routley Jun 10 '16 at 01:53
  • @hsinghal Also, look up "co-efficient of performance" for a heat pump: the efficiency, measured by the amount of heat pumped divided by the energy input, is always greater than unity. – Selene Routley Jun 10 '16 at 01:54
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    Thanks I finally understood, after reading the working of heat engines. In order to transfer the energy from sink to source, the gas inside heat engine must be expanded isothermally when in contact with sink, in this way the energy from the sink is transferred in the gas. Do you think any such process is occurring here. – hsinghal Jun 10 '16 at 18:38
  • @hsinghal That's a correct summary of heat engines. But I mean things a little more abstractly. If you take a quantity of heat out of a cold body, you have lowered that body's entropy. The ideal process would put that heat into the hotter environment - balancing energy, but with a nett decrease in entropy: the same quantity of heat put into a hot body always raises the latter's entropy less than the same quantity of heat into a colder body. So even the most ideal process must somehow make up for that entropy deficit: the only way is that if further some low entropy energy (work) winds .... – Selene Routley Jun 11 '16 at 08:26
  • @hsinghal .... up as heat in the hotter body. So we know, even without reasoning how the process works, that it must require an input of zero entropy energy $\Delta E = \Delta Q (T_{cold}^{-1}-T_{hot}^{-1})$ to make up the deficit. I'm not making any assumptions on what form the $\Delta Q$ takes. – Selene Routley Jun 11 '16 at 08:29
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Given that the device is extracting heat (vibrational energy) from the lattice that the LED is embedded in, the conservation of energy issue can be understood. The question you really should be worried about is, "does this violate the second law of thermodynamis?" Is the overall free energy increasing?

You can use thermoelectric coooling, for example, exploiting the the Peltier effect to make a device which cools one side and transfers the heat to the opposite side of a plate. When you place this in an environment of initially uniform temperature, the free energy apparently gained (you could after all exploit the new temerature gradient by a Carnot engine to extract work) is less than or equal to the free energy supplied by the current driving the cooler. You can certainly use this to heat the "other side" of a plate by more than you would by just using the current in a resistor.

Similarly, the LED device is converting heat on one side to heat-converted-to-light on the other and is exploiting the free energy associated with the current to do this (so that the entropy is not decreasing.

Mark Fischler
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  • I was actually thinking about violation of second law of thermodynamics. Also I am not so convinced about heat pumps and their workings, but thanks for your answer I will try to understand it. – hsinghal Jun 09 '16 at 17:20
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When you reading the plot in the paper, when efficiency is above 100%, the temperature is 135 degC (275 F) and the light output power is low. This makes me think, when you heat an steel bar to hot red, it emits light without electricity. You can pretend to inject electricity into the bar, and your efficiency is infinite. Energy is from hot environment. Electric power may play a role as in transistor amplifier circuit.

user115350
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  • At room temperature the Thermal emission is peaked in far infrared regime, on the other hand LED emit quasi monoenergetic light hence it is not possible that these two can be possibly combined and result in increase in efficiency. To me heat pump explanation and Peltier cooler explanation looks more plausible. – hsinghal Jun 10 '16 at 02:54
  • @hsinghal I think what he meant was reductio ad absurdum argument how easy is to claim efficiency over 100% – Agent_L Jun 10 '16 at 11:42
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The efficiency is being miscalculated because they left out the energy being used to heat up the device to 135 degrees! This is similar to claiming that an amplifier that takes a 1 mW input signal and "converts" it into 1 W output signal, as having an efficiency of 1000%! They are leaving out the energy being supplied to the amplifier "external" to the signal!

Guill
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