Infrared Detectors & The 2nd Law of Thermodynamics

[Devin-M]

You can do that, but it will accomplish nothing. You still won't get current.

So it won’t violate any laws of physics if I tried to generate electricity from the black body radiation of room temperature water with a room temperature HgCdTe PN Juction IR detector as described in the original post?

Right.

So do you disagree with Post #7?

 

[Drakkith]

No, and I don't think anything we've said contradicts Chestermiller's post. If everything is in equilibrium, detector, environment, etc, then you won't be getting any current. If not, then you might. You can add mirrors or change the size of the detector all you want, it won't change anything.

 

[Charles Link]

I'm still following this discussion, and I think @Chestermiller (post 7) needs to look more carefully at what you @Devin-M were asking. You don't get any photocurrent from a photodiode inside an enclosure when everything is at thermodynamic equilibrium. The participants in the discussion have been very patient, but don't seem to be having much luck in getting the concept to sink in that there is no mechanism for any power generation from the photodiode or solar cell when it is inside an enclosure with everything at the same temperature.

 

[Drakkith]

You don't get any photocurrent from a photodiode inside an enclosure when everything is at thermodynamic equilibrium.

Indeed, and the key here is equilibrium. Part of the difficulty here is that no one is quite sure exactly where that equilibrium is. Is it when everything is at the same temp? Should the sensor be just ever so slightly lower in temp or higher in temp given its ability to absorb a particular range of wavelengths very well? Does adding a mirror change the equilibrium? (my argument is no)

Like I said already, thermodynamics is hard. Especially when you get beyond black bodies and perfect reflectors and such.

 

[Devin-M]

I’m saying when an IR laser shines on the detector you get current in an external circuit. This means the detector is not in thermal equilibrium. The detector is not able to discriminate whether one of the IR photons it used to generate power came from the laser or was emitted by the black body spectrum of some room temperature water that happens to be nearby.

 

[Drakkith]

I’m saying when an IR laser shines on the detector you get current in an external circuit. This means the detector is not in thermal equilibrium. The detector is not able to discriminate whether on of the IR photons it used to generate power came from the laser or was emitted by the black body spectrum of some room temperature water nearby.

I don't think anyone ever doubted this.

 

[Devin-M]

Well if it’s possible one of the photons from the IR radiation from the black body spectrum of the nearby water created some power in the external circuit of the detector when the laser was on, then it should also be a possibility when the laser is turned off.

 

[Drakkith]

Well if it’s possible one of the photons from the ir radiation from the black body spectrum of the nearby water created some power in the external circuit, then it should also be a possibility when the laser is turned off.

Why?

 

[Devin-M]

Because the laser being on or off has no bearing on whether a single 3.5 micrometer photon from whatever source can produce an electron + hole pair thereby generating dc current.

 

[Charles Link]

But you don't get any photocurrent from the photodiode when the laser is off, and everything including the water is at 300 K. This is a claim you tried to make in the OP, and when we looked into it in more detail, we saw that the researchers that you had referenced did not make this claim either. They had a high temperature globar for their additional IR source, above the 300 K ambient. The 300 K ambient by itself, with water or no water, or any other material at 300 K, would not generate any photocurrent.

 

[Drakkith]

Because the laser being on or off has no bearing on whether a single 3.5 micrometer photon from whatever source can produce an electron + hole pair thereby generating dc current.

We already know the detector converts photons into electron-hole pairs, even under an equilibrium condition.

So again, why should the detector generate current while the laser is off?

 

[Devin-M]

The dc current from the laser is a quantum effect…. one photon pushes one electron into the external circuit. Why would identical photons be assumed to have different effects within the PN junction? You’re saying a 3.5 micrometer photon from one source pushes an electron into the dc load and an identical 3.5 micrometer photon from a different source doesn’t.

 

[Charles Link]

@Devin-M You need to study a book like Streetman's in detail=the section on photodiodes. It's been a number of years since I studied it in detail, but as I recall there are several things going on including electron-hole recombination, along with diffusion currents. When all of the various contributions to the junction current are summed, the part that generates the extra current seen as photocurrent is the extra photons, above the ambient, that are absorbed, generating minority carriers in the p-material (electrons) that get swept across the junction. It may be worthwhile for you to try to get a copy of Streetman's book and work through his calculations.

 

[Devin-M]

But if I have a detector and a cup of water floating through space (separated by vacuum), if even 1 IR photon from the water’s black body spectrum strikes the detector, that puts the detector out of thermal equilibrium, does it not?

 

[Drakkith]

Why would identical photons be assumed to have different effects within the PN junction? You’re saying a 3.5 micrometer photon from one source pushes an electron into the dc load and an identical 3.5 micrometer photon from a different source doesn’t.

I'm saying no such thing. In fact, I've said exactly the opposite more than once. I've already explained, in detail, my understanding of the situation. Now I'm asking you why you think the detector should generate a current, and your answer, so far, completely ignores statistics, thermodynamics, and most of the operational details of the detector. This in spite of those topics being brought up time and time again in the 100+ posts in this thread. That's not a good way to learn.

But if I have a detector and a cup of water floating through space, if even 1 IR photon from the water’s black body spectrum strikes the detector, that puts the detector out of thermal equilibrium, does it not?

No, in fact the reverse is true. If the detector received zero IR photons from the water then something would be out of equilibrium in some way.

 

[Devin-M]

I would like to know whether @Chestermiller still agrees with his statement in post #7 after all the discussions that have come out since that post.

 

[Devin-M]

I would also like to know more about his Post #5

The 2nd law of thermodynamics does not say that you can't extract energy from a single temperature source and do work. It says you can't do it by a system operating in a cycle.

 

[Charles Link]

For Streetman's book see https://www.amazon.com/dp/0131587676/?tag=pfamazon01-20

Devin-M's post 119:
But if I have a detector and a cup of water floating through space, if even 1 IR photon from the water’s black body spectrum strikes the detector, that puts the detector out of thermal equilibrium, does it not?

To respond:
There are millions and millions of photons in the ambient background. They are continually getting absorbed by the photodiode while the photodiode material radiates out the same number of IR photons. It won't be precisely the same number=there is a statistical fluctuation that results in thermal noise. Thermal noise will be present in the resistor, but this fluctuation is small, and much lower than the ambient thermal background, which is not observed as photocurrent.

Edit: This is perhaps making things more complicated than necessary, but the complete details of the photodiode in an enclosure that is essentially a blackbody cavity are non-trivial. The thermal noise signal, in any case, will be of an ac nature, centered about zero, and will not be a DC level like you would get from any source, even a blackbody one, that is above the 300K ambient.

 

[collinsmark]

The active part of the PN junction is only on one side if the wafer so yes it may be receiving IR photons from all directions in the scenario but it’s more likely to generate current in the external circuit from received photons coming from preferred directions, ie the direction the sensor faces or the direction the mirror is pointing to reflect onto the sensor.

Regarding the mirror in the box, like @Drakkith said, as much as the mirror might reflect IR from some areas, it will reduce IR from other areas such that the net effect is zero. Also don't forget that the mirror itself, at 300 K, is also absorbing and radiating IR the same as everything else is.

Outside our box, a telescope can collimate regions of non-uniformity in the sky. But realize that that's all it does, really. If the entire sky, and everything else, is uniformly lit, particularly if the telescope itself is at the same temperature everything else, the telescope can't do anything useful.

If it helps, let's make the box problem even more simple. Suppose that in the 300 K box, you put two photodetectors right next to each other, facing each other face to face, PN juction to PN junction. How do you think that would effect the current in each? Would both produce current? Neither? Or would it make no difference compared to a single detector sitting alone in the 300 K box?

 

[Devin-M]

Regarding the mirror in the box, like @Drakkith said, as much as the mirror might reflect IR from some areas, it will reduce IR from other areas

But you said in the scenario the curved telescope mirror is also 300k, so not only is it emitting the same or a similar amount of black body radiation into the sensor as the 300k wall it’s “blocking,” but it’s also reflecting IR radiation from behind the sensor that would otherwise be “wasted…”

 

[collinsmark]

The 2nd law of thermodynamics does not say that you can't extract energy from a single temperature source and do work. It says you can't do it by a system operating in a cycle.

I would also like to know more about his Post #5

I'm pretty sure what was meant (using our specific examples), is that if you drop an ice-cold photodetector into a warm glass of water you can extract energy from the system until the photodetector warms up to the same temperature as the water. But this is a one-off. You can do it once, but you can't keep extracting energy without taking the photodetector out of the water and cooling it back down again (and warming up the water a little, again).

The energy that was extracted is fundamentally not enough, by itself, to achieve the same temperature separation as the original setup. If you tried to use the extracted energy later, to warm up the water and cool down the detector, it's fundamentally not possible to achive to the same level as the original setup. It's not possible to make the process cyclical, by itself. In other words, you can't use that process to form a perpetual motion machine.

 

[Devin-M]

I’m not claiming you can extract an infinite amount of blackbody radiation from a cup of water but what I am saying is that the internal thermal energy content of the world’s oceans is quite large (not infinite) and constantly replenished from the fusion power happening in the Sun.

 

[collinsmark]

But you said in the scenario the curved telescope mirror is also 300k, so not only is it emitting the same or a similar amount of black body radiation into the sensor as the 300k wall it’s “blocking,” but it’s also reflecting IR radiation from behind the sensor that would otherwise be “wasted…”

Yes, but when in thermal equilibrium, the total number of photons leaving the reflected surface, be they reflected or thermally emitted, is the same as everywhere else in the 300 K box (the box where everything is 300 K)

This has some interesting implications for reflective surfaces (and surfaces of different colors, etc) compared to black surfaces. Reflective surfaces, since a larger fraction of the photons leaving their surface are reflected photons, means that fewer thermally radiated photons leave their surface, compared to a black body. It means that bodies coated with reflective surfaces (a white surface will also do):

• When exposed to higher-temperature, external radiation, reflective bodies will not heat up as quickly compared to a black body (which makes sense because many of the photons reaching the reflective body are reflected.
• When put in a cold environment a reflective body will maintain its higher temperature longer than a black body, because relatively fewer thermal photons are emitted. (The black body will cool down more quickly.)

But once everything is at the same temperature, the photon energy flux leaving the surface is the same as anywhere, regardless of the reflectivity, and is also the same as the flux arriving at the surface.

Back to the 300 K box problem: No, the mirror does not do "double duty." The total photon energy flux leaving the mirror is the same as everywhere else (recall that everything in the box is at the same temperature). It doesn't matter that some of those photons are reflected and some of them are thermally emitted. The total is the same as everywhere else.

 

[Devin-M]

Increasing the sensor size will increase the incident photons per second, will it not?

 

[collinsmark]

Increasing the sensor size will increase the incident photons per second, will it not?

Yes! And that's very important!

Let's talk about "flux" for a moment. The term "flux" refers to something-or-other per unit area. For example, power flux is power per unit area. Power flux might be measured in units such as W/m².

So if you double the area of the sensor, you'll double the "photon power" of the sensor.

But the real question is -- the one that's really, really important to this thread/discussion -- If you double the area of the sensor how does that affect the photon power flux leaving the detector?

What's the difference between the photon power flux of a small drop of seawater at 300 K and and an entire ocean at 300 K?

 

[Devin-M]

If you keep increasing the size of the sensor in your box eventually the 3.5 micrometer photon count per second onto the sensor from the black body radiation of the 300k-maintained walls of the box will match the wattage of the infrared beam used in the papers.

 

[collinsmark]

If you keep increasing the size of the sensor in your box eventually the 3.5 micrometer photon count per second onto the sensor from the black body radiation of the 300k-maintained walls of the box will match the wattage of the infrared beam used in the papers.

<sigh>

I'm sorry, but no. As you increase the sensor size, the photon power reaching the sensor increases, yes, but the photon power leaving the sensor also increases by the same amount. The net is still zero.

For a given temperature, the sensor size has no effect on the photon power flux leaving the sensor. And if everything in the surroundings are also at the same temperature (such as in our 300 K box example), the sensor size has no effect on the photon power flux arriving at the sensor either. What's more, when everything is at the same temperature, the photon power flux leaving the sensor is the same as that arriving.

 

[Devin-M]

The current through the external DC circuit of the 300k IR detector is directly proportional to the photon count of 3.5 micrometer photons reaching the active area of the detector and increasing the size of the sensor in your box scenario (with 300k maintained walls) will increase the photon count onto the active material in the pn junction and eventually with large enough size you get the same incident power as was used by the researchers.

https://www.speakev.com/cdn-cgi/image/format=auto,onerror=redirect,width=1920,height=1920,fit=scale-down/https://www.speakev.com/attachments/5000b9e5-011d-4ef0-af59-bedb15ff822e-jpeg.156575/

 

[collinsmark]

The current through the external DC circuit of the 300k IR detector is directly proportional to the photon count of 3.5 micrometer photons reaching the active area of the detector and increasing the size of the sensor in your box scenario (with 300k maintained walls) will increase the photon count onto the active material in the pn junction and eventually with large enough size you get the same incident power as was used by the researchers.

https://www.speakev.com/cdn-cgi/image/format=auto,onerror=redirect,width=1920,height=1920,fit=scale-down/https://www.speakev.com/attachments/5000b9e5-011d-4ef0-af59-bedb15ff822e-jpeg.156575/

Please realize that the figure you show here from the paper was not generated using a 300 K heat source. The photodetector was held at 300 K, not the source of the 3.5 micrometer photons.

I'm now going to come clean and give you the answer to the resizable box problem: The answer is there is nothing you can do increase the photon power flux reaching the sensor, given the restrictions of the problem. It can't be done. The power flux is dependent upon temperature, and that's it.

And you can make the box as big as you like. Put an entire ocean in the box (let's ignore the corresponding vapor pressure for the moment). One might say, "Good golly, an entire ocean, there must be a lot of photons from that!" But all those photons are spread out over a huge area. The photon power flux does not change, so long as everything is kept at 300 K, not matter how big something is. It's always the same, tiny, tiny number.

And at 300 K, the photon power flux of photons in the region of 3.5 micrometers is tiny. Really, really, tiny tiny. Furthermore, when everything is at the same temperature (glass of water + detector), there is no conceivable way to collimate these tiny, tiny amount of photons onto the detector. The power flux leaving the detector is the same as the power flux arriving at the detector, and there isn't any way to change that so long as everything involved is at the same temperature.

The figure you quoted from the paper was made with 3.5 micrometer photon power flux that I'm guessing is billions or trillions (I don't know, some huge, huge number) of times greater than that fixed, small, tiny, tiny photon power flux that could be achieved from objects at 300 K. In order to make such a power flux, a source much hotter than 300 K is needed (and yes, an IR laser counts as a much hotter source. An incandecent bulb and a diffraction grating will also suffice).

And finally, there's no point in claiming that one doesn't need billions or trillions of times greater, and a smaller number generated at 300 K will suffice. Because if the photon power flux is generated at 300 K, that's the same power flux leaving the detector when it is at 300 K. The net power flux is zero. No power can be extracted from that.

 

[Devin-M]

I thought my understanding was the mechanism of IR absorption in the detector (corresponding to an electron jumping an energy level from valence to conduction, producing DC current) is entirely different from the mechanism of IR emission from the detector (which I thought doesn’t correspond to an electron falling an energy level… I thought that’s only the case for most visible light, but I thought the IR emissions of the detector itself are from random molecular motions, not from an electron falling from a higher energy level to a lower one in an atom.)

 

[collinsmark]

It's both. The mechanism is both molecular motion and electrons changing excited states. And that's true for both emission and absorption.

An electron falling to its ground state can often be too high of energy for infrared for many atoms. But an electron transitioning from one highly excited state to a slightly lower excited state can certainly produce an IR photon.

 

[Devin-M]

When 3.5 micrometer IR photons knock electrons from valence to conduction, generating current in the external circuit of the detector, does the energy from those photons always get re-emitted from the detector itself or does some of that energy turn into heat over in the external DC circuit that has current running through it?

 

[Devin-M]

Suppose the detector is initially at 300 k.

A 1kW infrared 3.5 micrometer laser turns on & is shining on the detector (the detector is large enough so this won’t damage it or even heat it rapidly).

The external DC circuit attached to the detector consists of a copper wire leading from one terminal of the detector, through a block of ice which then coils through a well insulated 320k tank of water (a wire section in the tank is initially 320k), then back through the ice and back to the opposite terminal of the detector. The wire in the ice is initially the same temperature as the ice.

There’s DC current through the wire in the 320k water tank the moment the laser turns on so it immediately begins heating up that water via current through resistance in the wire, even though the detector is still only somewhere between 300k and 301k and the wires in between the detector and the tank are still freezing.

The energy traveled at the speed of light, from the laser to the 300k - 301k detector, which immediately resulted in the heating of a tank of 320k water which had been separated from the detector by freezing wires which passed through a block of ice.

Also incidentally a few of the 3.5 micrometer photons that hit the detector after the laser turned on were from a different source— black body IR radiation from the molecular motions of some 300k water near the detector.

 

[Drakkith]

No, no, no, no. There's no reason to change to a much more complicated setup when you still don't understand the first one(s). Just use a detector at equilibrium in a perfect blackbody box. No mirrors, no ice blocks, no water, nothing else. Not even a laser.

 

[jartsa]

So there is some electric device in a tank filled with warm water. And said device produces a small amount of high entropy electricity.

This high entropy electric power can heat a low entropy place. But not a equally high entropy place.

The reason being that a heating element is also an electric device that produces a small amount of high entropy electricity if it is warm. Always the warmer electric device generates more high entropy electric power than the cooler one.As for a laser device and warm water and an "infrared solar panel" generating electricity, the warm water that generates a flux of photons in the same direction as the laser increases the electric power of the panel, while the water that generates a flux of photons in the opposite direction as the laser decreases the electric power of the panel.