The article relates it to the Peltier effect. I can get power from a thermal difference with my TEC cooler. With a heated LED I'm not sure where I should point to, to find the cold temperature.
Perhaps it's just the room? The article did say that at the highest temperature and low current they needed a lockin to measure the light from the PD, 70 pW, because it was buried in the
40 nW of black body radiation.George H.
You can find the cold beer in the dark!
An LED heated to 135C will radiate to the environment, and cool off. So will a brick.
I finished that testing, and as others predicted, LEDs fired with seven times the resistance, but for seven times as long, produce the same brightness as the pulsed base case. So I was wrong about that, and the experts here were right. That's why I ask these questions here. :-)
That means the only advantage of pulsing through the segments is that you only need one resistor instead of seven - because only one segment ever fires at a time. Not a major advantage, I think.
And the resistor has to dissipate seven times the power (though you might want to size the resistors so they don't stink if the processor locks up). No, not a big advantage, usually. Where it helps is if you want to dim the display.
No it doesn't necessarily violate the laws of thermodynamics.
The LED will be chilled by the emission of photons largely powered by the voltage drop but stealing some energy from phonons in the crystal lattice. I would only expect this to be at all relevant when the flux of photons being emitted was very very low. Thermal energy at ambient temperature emits with a peak around 10um so it can't provide very much of a boost (although the tail might). I can't see how it would possibly affect an LED with mA flowing in it but at a few pA maybe...
It is perfectly sound. The LED is emitting photons and taking some energy from the crystal lattice in the process. The cooling effect only becomes noticeably (if at all) at power levels
What does the current have to do with it? How does this differ from Maxwell's Demon? It is selecting only the more powerful electrons. It just isn't using a trap door.
-- Rick
Oh, thanks, I see it now. The other (cold) side of the heat engine is the surface that absorbs the photon. That surface has to be colder than the led, or BB radiation back to the led will balance. (Though there are a lot of messy details in how it all exactly balances.) So I predict you can't do this with a room temperature led, (unless you point it into space or something.)
George H. I would only expect this to be at all relevant when the flux of
See my previous post. My guess is the led is coupled to a temperature difference. And there is no problem for an engine to get heat out of a delta T. (Like TEC generators.)
George H.
Your previous post doesn't answer the question, that's why I asked it. What does the LED current have to do with it?
Your guess?
The only heat that can be pulled from "delta" T is related to the delta T. In this case we have no delta T. There is just one T which naturally has a spread of energies in the components. The issue of entropy is what prevents the sorting of these energies. That would be a decrease of entropy requiring the input of energy. So you can't get something for nothing which is what this seems to do.
Heat pumps and TECs work by energy being used to move heat essentially "uphill" and in the process becoming heat as well. Entropy increases because the energy that turns to heat is greater than the decreased entropy of the heat being pushed to a higher temperature.
With the LED the electrical energy is turning into light (energy which is essentially negative entropy) and the heat is turning into light. So this violates the second law of thermodynamics. The added heat of the non-idealities of the circuit are not relevant. In the ideal case heat is being turned into highly ordered energy which can't happen without an increase in entropy elsewhere.
I wrote this and thought more about it. It sure looks to me like it violates the second law of thermodynamics, but it also appears to be mechanically reasonable. I can imagine a chemical reaction which does the same thing. Chemical reactions have an activation energy which controls the rate of the reaction. The difference in energy levels tells you if heat is released or added. I can picture a reaction which requires the addition of heat but the resulting compound spontaneously decays with the release of a photon leaving the molecules in a low energy state making the reverse reaction much less likely than the forward reaction. So how does this not violate the second law of thermodynamics by cherry picking the higher energy molecules? See the ascii art of the reactions.
_ / \ _ / \____/ \ ____/ Mid \ Start \---\/\/\---> Photon \ \ \____ End
The reactants require some amount of thermal energy to rise over the first barrier. Some of this energy is returned as heat on reaching the mid point in a new molecular arrangement. So some of the heat energy is in the form of the new molecule(s). Less energy is required to reach the second barrier and all of the released energy is in the form of the photon. This energy is greater than the amount of energy difference between the starting and ending reactants, so the difference must have come from thermal energy.
The fact that some of the heat energy ends up in the photon must be the same as heat being pushed uphill by a heat pump. The amount of heat moved is limited by thermodynamics. But in this example the ratio of energy released as a photon and the thermal energy input is not limited in any way other than by kinetics. As that ratio changes the rate of reaction changes. But kinetics is not an issue in the second law of thermodynamics.
The total heat energy put into these reactions is the difference in energy of the start state and the peak of the second barrier. The energy in the photon is the difference between the second barrier and the end state. The molecular difference in energy is between the start state and the end state. The energies all add up so the first law is conserved, but heat energy ended up as a higher energy photon.
-- Rick
That doesn't hold water in that the photon is much higher energy than the BB radiation of the "cold" surface.
But maybe that is the point. There can be some heat energy captured by the photon just as a heat pump can move heat uphill. The entropy does not decrease because energy is used to push the heat uphill and retrieving the energy will require restoring the entropy by letting the heat run back downhill. "Some" heat can be included with the photon, in the end no different from BB radiation?
It has been over 40 years since I studied this. Thermo doesn't show up much in digital electronics.
-- Rick
Woah, Read my response to Martin, Delta T is T(diode) - T(absorber) the temperature of the surface that absorbs the photon, led light.
If the absorbing surface was as hot as the diode, the led would be flooded with BB (black body) radiation.
I'm skipping the violation of the 2nd law,
George H.
There is just one T which
That doesn't make sense. The fact that the LED emitted a photon doesn't raise it's temperature. The absorber would have to be *much* hotter to emit significant quantities of the same energy of photon.
I believe you are skipping serious thought about it.
-- Rick
The LED is at 135 C. (emitting a photon makes it slightly colder.) The BB radiation (according to the article) from the 135 C led was
40 nW, There would be something similar going back the other way if the surroundings were also at 135 C. The emitted power due to e-h recombo was only 70 pW... factors of 1000.Yeah whatever, if the authors of the article (who I assume are smart guys.) thought they had violated the 2nd law they would be screaming it from the tree tops. That would be a very big deal physics-wise.
George H.
What was the wavelength of the LED light generated from the current? What portion was the thermal contribution? Compare that to the BB radiation. Talking about nW and pW is not the issue. The issue is entropy. Energy is the 1st law of thermo.
I'm trying to have a reasonable discussion of it. Just waving your arms and saying "you can't violate the 2nd law" doesn't explain anything.
-- Rick
Scotopic visual acuity is worse but at the onset of scotopic vision and with appropriate correction I estimate that the available resolution is just about enough to read 5mm high lettering at 300mm.
In strong light you should be able to manage more like 1mm at 300mm. You should see something between 10-30 pixels across the moon naked eye.
I spend a fair amount of time reading charts in the pitch dark using as little light as possible.
You won't be able to read the small print on your insurance policy but if you cant read for example the main number on your credit card then you really ought to go and see an optician for a retinal scan.
Red has the advantage of preserving scotopic vision and modern red LEDs are way better for this than the old cheap glass filters.
Retinal test? Your dark adapted vision should not be as poor as you seem to be implying here.
Scotopic vision is an order of magnitude worse resolution than optimal strong light vision and then degrades gracefully with decreasing light levels in a Bayesian sort of way until you hit the noise floor. Even then outdoors the (clear) night sky is bright even in the wilderness.
In a true darkroom or a cave with no lights on you eventually get to the point where you see patterns in the noise floor after a couple of hours in absolute total darkness.
-- Regards, Martin Brown
You are absolutely right. The 2nd law can be very hard to pin down when you get into the details. Look how long it took to figure out Maxwells demon.. Szilard got a handle on it and then the information guys (information as entropy) nailed it down.
In this case it looks a lot like a Peltier device. You can get energy from a temperature difference. I was just trying to identify the temperature difference. In this case, if you want to argue with my identification as photon absorber as the lower temperature that would be fine. I could be totally wrong. Maybe one has to look at the entropy of the photon number too.
But I will say I tend to treat the 1st and 2nd law's as sacrosanct. And treat any apparent disagreement a sign that we are missing something. (To me that seems like the best way to further our understanding of nature.)
George H.
+1.
It isn't just LEDs that have this...BJTs do too. Collector current has to pass through the base, whose potential can easily be higher than the collector's (even if the transistor isn't quite saturated yet). The built-in potential has to be taken into account.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Optometrists always write me prescriptions that I can't use. I now make my own and order glasses from Zenni for $22 a pair.
I haven't figured out the sign convention for astigmatism. I'm using the 105 degree value that I inherited from a prescription, and it's clearly off by about 15 degrees, which means I should probably use 90. Or maybe 120? The Zennis are so cheap I should just try both.
I had cataract surgery and had one eye set to 10" FL and one to 20. Works great for reading and computing. I only need glasses for driving and skiing.
But I'm talking about dark-adapted light levels where you can barely tell whether an LED right in your face is lit 10cm away. Different kettle of fish.
I found white LED light fine for preserving scoptic vision too. I assume this is because I was already well below photopic illumination levels.
Can you see all a fox's details at night in the dark woods? Or do you see a shadowy figure, moving? That's what I see.
My photopic corrected acuity is 20/10, in U.S. notation. My retinas are fine. If you're into it you can map all sorts of localized bumps and blobs on your corneas with homespun techniques. That's what limits my resolution, not retinas.
I think we're talking about different light levels. I'm talking about levels far below starlight.
That closely describes the conditions. It was a closed room, inside a darkened building, at night, dark-adapted, trying to pick illumination sources out of the noise floor.
Duplicate that, then try taking notes in the dark.
Cheers, James Arthur
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