## Energy Problem Solved By LEDs!!!!

I really don’t have much serious comment about this, yet, because I can only get the synopsis of the paper, and the press article surrounding this.  But, it appears that they are claiming a perpetual motion machine in the making.  Not only that, but an air conditioner as well!!!

### Synopsis: Optical Device is More Than 100% Efficient

A light-emitting diode (LED) that emits more light energy than it consumes in electrical energy has been unveiled by researchers in the US.

I’ll give excepts from both……..

The device – which has a conventional efficiency of greater than 200% – behaves as a kind of optical heat pump that converts lattice vibrations into infrared photons, cooling its surroundings in the process. The possibility of such a device was first predicted in 1957, but a practical version had proved impossible to create until now.

…….there was no barrier in principle to an LED being more than 100% efficient, in which case it would actually cool its surroundings.

##### Obeys the second law

…….Lead researcher Parthiban Santhanam of the Massachusetts Institute of Technology explains that the process is perfectly consistent with the second law of thermodynamics…..”The most counterintuitive aspect of this result is that we don’t typically think of light as being a form of heat. Usually we ignore the entropy and think of light as work,” he explains. “If the photons didn’t have entropy (i.e. if they were a form of work, rather than heat), this would break the second law. Instead, the entropy shows up in the outgoing photons, so the second law is satisfied.

The researchers chose a light-emitting diode with a small band gap, and applied such small voltages that it acted like a normal resistor. With each halving of the voltage, they reduced the electrical power by a factor of 4, even though the number of electrons, and thus the light power emitted, dropped by only a factor of 2. Decreasing the input power to 30 picowatts, the team detected nearly 70 picowatts of emitted light. The extra energy comes from lattice vibrations, so the device should be cooled slightly, as occurs in thermoelectric coolers.

These initial results provide too little light for most applications. However, heating the light emitters increases their output power and efficiency, meaning they are like thermodynamic heat engines, except they come with the fast electrical control of modern semiconductor devices.

Now, this science babel is going to take me a while to decipher.  But, halving the voltage, in the world I play in, halves the energy as well.  Volts * Amps = Watts.  So, I don’t know what they’re talking about.  But, they probably don’t, either.  Can any of my sciency friends articulate how halving the voltage doesn’t equate to halving the electrical power?  I keep bumping into Ohm’s law. ….. or did they falsify that as well as making a perpetual motion/air conditioner machine?

In pretend world, this is great news!  We no longer need to have power plants!  Just get a bunch of LEDs, fire an electric charge with a hand crank into them and put some photo cells on the receiving of some fiber optic cable and then have the charge go back into the LEDs!!!!  Done!  It would prolly ‘splode within a few minutes because of the energy overload.  But, on a hot summer day, think of how much cooler things could get with a 200% efficiency!

They provide too little light, so just make them on a larger scale!!!  We can go from picowatts to megawatts!!!!

Sonnofa……..!!!  Color me skeptical.

This entry was posted in Energy. Bookmark the permalink.

### 28 Responses to Energy Problem Solved By LEDs!!!!

1. kelly liddle says:

Now we can have a perpetual energy machine so long as the temperature does not reach 0 celcius http://newenergyandfuel.com/http:/newenergyandfuel/com/2010/10/08/maximum-solar-cell-efficiency-with-frog-foam/ I am sure we can work it out.

• suyts says:

lol, hell yeh!!! All we have to do is turn on those special LEDs and that’ll suck the hot right out of the air!!!! Now we can have double perpetual energy!!!!

Remember it was our idea right here!!!!

2. miked1947 says:

I would say it is PFM. P for pure. M for magic and I do not think they know what the F they are talking about.

3. mr166 says:

If you double the voltage across a resistor the current thru it also doubles thus the power quadruples.

Just how does one measure picowatts of power? The “Gain” looks to be measurement error to me.

• suyts says:

“Just how does one measure picowatts of power? The “Gain” looks to be measurement error to me.”

Looks that way to me, too. Or they haven’t calculated the medium properly… or something of that nature.

“If you double the voltage across a resistor the current thru it also doubles thus the power quadruples.”

Wouldn’t that depend upon the resistor? And what does the potential(volt) tell us without mentioning the current(amps)?

• suyts says:

Electrical resistance is equal to the drop in voltage across the terminals of the resistor divided by the current(amps) being applied to the resistor. The current through a resistor is in direct proportion to the voltage across the resistor’s terminals. Thus, the ratio of the voltage applied across a resistor’s terminals to the intensity of current through the circuit is called resistance. This relation is represented by Ohm’s law: Or I= V/R

So, if amps = 2(I) and V = 8 and r = 4…. to make 2 = 8/4 so doubling V to 16, then 16/4=4 …. … that’s only doubling, not quadrupling…. unless they’re using some sort of potentiometer, but that’s not really discovering anything magical.

4. Billy Liar says:

I think if you had about 1.3 trillion of these LEDs in your ceiling you could just about light the room.

I hope they don’t cost much.

5. Marc77 says:

A thermal pomp can also output more than 100% in heat. I think an air conditioner often outputs more than 200% the power it consumes as it cools the room. It’s probably because a differential of temperatures stores a very small amount of energy. So the cooling of the air conditioner has to be match with an equal amount of warming plus the inefficiency of the system is turned into additional heat. I think a heat pump is guarantied to increase this amount of warming for has long as the outside is warmer than the cool pole of the pump.

• suyts says:

Marc, you’re going to have to show me the numbers on that one.

I’m not sure what you’re referring to, are you talking about a heat pump used to warm homes? If that’s the case, then you’re talking about external sources of energy. Which is a different subject, altogether.

But, no, as far as I know, we’ve never been able to create a machine the outputs more energy than was input. If we had, then there’s no point in the argument over coal or wind or solar…… we can just deploy these magical machines and harness the excess energy they’ve created.

For a review, for everyone, even before the 2nd law of thermodynamics, the basic physics is that the total energy in a closed or isolated system is constant, no matter what happens. Another law stated that the mass in an isolated system is constant. Einstein expanded the thought with E=mc2 (in other words that mass was a manifestation of energy) the law was said to refer to the conservation of mass-energy. The total of both mass and energy is retained, just moved around.

Further, there is the 3rd law of thermodynamics….. “states that it is impossible to create a thermodynamic process which is perfectly efficient.”….. but maybe 200% efficient?

6. HankH says:

The abstract mentions the resistance of the LED and emitted power. The correct formula to use to factor voltage [V], resistance [R], and power [P] is P = V^2/R.

As a thought experiment, lets use some simple numbers. Assume voltage [V] = 2V and resistance = 100 ohms [R]. P = (2 * 2)/100 = .04W. Now halve the voltage. P = 1/100 = .01W. So you can see that with resistance remaining the same, halving the voltage results in the power being dropped by a factor of 4.

• suyts says:

Ok Hank, I’m trying to understand this, how are we expressing power? Because your formula seems to contradict the formula I used. Where I (amps) = Volts/Ohms ….. additionally, then volts*amp= watts

Ohm’s law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Introducing the constant of proportionality, the resistance, one arrives at the usual mathematical equation that describes this relationship…… I=V/R

Again, in the area I work in, when you’ve got volts and ohms, then amperes is the only thing left to talk about. And the relationship is directly proportional, not logarithmic.

• HankH says:

We can make this work. Using ohms law and calculating for current (I = V/R), lets use the same values as before. We know V = 2 and and R = 100. Thus, I = 2/100 = 0.02A. Now plug that into your earlier formula P=I*V. P = .02*2 = 0.04W. Now halve the voltage from 2V to 1V and solve for current [I] again: I = 1/100 = .01A. Again, plug our calculated current into the formula P = I*V: P = 0.01*1 = 0.01W. As you can see we’re getting the exact same results as if we were to use P=E2/R only we can save a step and not have to solve for current.

• HankH says:

It appears the HTML SUP tag doesn’t work on wordpress. That should be Power = E squared over R, not P = E sub 2 over R.

• suyts says:

Thanks Hank, I see what you’re saying …. sometimes I get too focused and only see what I’m focusing on.

I’m still going to have to mull over this wonderful LED energy creating machine.

7. Bruce of Newcastle says:

Ah, they’ve combined the LED with thermoelectric refrigeration. Very cool. Sort of the opposite of an Edison light bulb, which is really a combined light source and heater.

Unfortunately there’s a problem that in winter you don’t want your room cooled down, so all the efficiency benefit is lost as soon as you turn on the heater. And no one is going to have separate lights for summer and winter. I suspect though the cooling effect is so small that it wouldn’t be noticeable.

Still, I like it.

LED lighting is improving almost at Moore’s Law rates. My Kindle has an LED light which runs off the internal battery. It consumes so little power that you can read for hours without doing much to the battery level. Some greenery I think is stupid (base load solar) and horrible (bird munchers), but LED lighting will be brilliant if they can get the price down to squiggly bulb levels.

• suyts says:

Right….. too bad the U.S. forced conversion from incandescent to those mercury filled ones before the LED tech had a chance to mature.

Bruce, you have any thoughts on this energy gain?

• Bruce says:

Well reading the article there is no energy gain. All that is happening is that background heat is being converted to light (well, IR in this case), taking away both energy and entropy (so delta G is negative). You have to have a voltage to make it work, after that it takes half the energy from the current and the other half from the external surroundings. Pretty similar to frequency multiplication of light, except here its a photon plus a heat generated phonon.

The author is pretty clear that it is not the most efficient way to cool your beer, but I see nothing wrong with the physics. I’m no physicist apart from a couple years at uni many years ago but I have had plenty experience with thermodynamics.

This lab rig was heated too, so in actual efficiency it was terrible because they only factored efficiency from the current not the heating. And the efficiency of electron use was terrible (0.1%), but the efficiency of those electrons used by the light emission process that was 200%. Heroic spin, but hey that’s usual for a first attempt. If someone ever invents antigravity I’d forgive worse spin than this.

But assuming they can make it practical there’s no reason why it couldn’t work at room temperature. Thermoelectric coolers are commercially available, its just they’re not very effective. So if you have an overclocked PC you probably still need to use water cooling despite the advantages that thermoelectric would have.

In fact you could say all this actually is is a thermoelectric cooler – they get cold on one side and hot on the other – and radiate IR photons from that side. The only difference is its quantised directional IR not black body radiation in this case.

• suyts says:

lol, thanks Bruce. I can appreciate the dance. I’m good with the coolers that don’t work……sorry, I wasn’t specific. I was asking about the 200% efficiency claim.

Clearly, I’m not a physicist, either. But, the claims are, IMHO, outrageous. And, maybe that’s the way it happened, but, that’s not what was stated.

What I’m trying to get at,…….. is it okay to tout 200% efficiency with caveats and people are suppose to understand the statements aren’t really real? Or, do people really believe that 200% efficiency is obtained and that all that is required is us catching up with the thoughts of the authors?

Yes, these are rhetorical questions.

8. pjie2 says:

1) Current = Volts / Resistance

2) Power = Volts x Current

Substituting (1) into (2) gives you Power = Volts^2 / Resistance

Therefore, if you hold the resistance constant, you expect the power output to vary as the square of the voltage. This is really quite simple algebra.

9. HankH says:

I’ve been thinking about this for a few days. From what I read and understand, the general workings of the paper make sense if not just a little overhyped and poorly explained.

My understanding of how it works is the current fed into the LED raises the excitation level of the atoms in the semiconductor to enter a state of quantum harmonic oscillation (QHO or referred to as lattice vibration in the article). The vibration occurs at the propagational velocity of sound through the N substrate or P junction semiconductor solid.

When in a state of QHO, the LED semiconductor can absorb energy as a specific quanta of phonons from the heat source and a specific quanta of phonons from electricity applied to the LED. Both energies combine additively. Then using what might be best thought of as piezoelectric conversion more electrons are knocked off their atoms to jump from the conduction band of the PN junction to the valence band across the band gap. When electrons jump the band gap, light photons are produced and emitted in the IR frequency range. The amount of photons emitted are greater than what would be expected for just the energy (electricity) supplied to the LED leads because of the energy added by the phonons from the heat source.

Here’s the kicker. While the article states that the LED is 200% efficient I think it is a bit of a misnomer. There are direct conversion losses involved in QHO, refractive losses in the photoemission conversion, and resistive losses in the semiconductor and power system. The LED cannot generate twice as much energy as put into it. Practically speaking, the LED can put out twice as much light while cooling a beer but it will not produce more electricity than pumped into it. The conversion of photons back to electricity is somewhere in the 13% – 23% efficiency range. So if I’m figuring right, a broad array of these LEDs, if configured to generate electricity would output less than 46% of the energy pumped into them. It’s a loosing proposition.

• suyts says:

Thanks Hank, I appreciate the well thought comment here.

Yes, I’ve been going back to this as well, though not at the depth you have. And, yes, the 200% is a silly statement.

So, does this mean we’ll still need power sources other than a LED array? Still, this could have some good functionality….. we could eliminate the need for those push switches on our refrigerators!!!

• HankH says:

Yes, I’m afraid we’ll still need power sources other than a LED array. I agree, they could be further developed to have some commercial value. I don’t see them playing much of a role in lighting given that the light they emit is in the IR spectrum. Perhaps they may make more efficient Peltier effect junction coolers, which are only 8%-10% efficient at present.

I should make one minor clarification to my previous comment. The LED array would put out no more than 46% of the electricity required to run them. I’m ignoring the heat energy contribution (as the authors are doing in their boasting of 200% efficiency) and assuming it is coming from an endless source like Trenberth’s missing heat. ;-P

• suyts says:

lol, no doubt………. I’m wondering if they tested these things in different temperatures. And, I’m wondering why they couldn’t scale them up a bit.