Infrared Detectors & The 2nd Law of Thermodynamics

[Devin-M]

But the 3.5 micrometer light from the 300k water isn’t light directly reflected from the sun or the room lights that goes away when you turn the lights off or lower the blinds. Its the black body radiation from the internal energy.

 

[collinsmark]

But the 3.5 micrometer light from the 300k water isn’t light directly reflected from the sun or the room lights that goes away when you turn the lights off or lower the blinds. Its the black body radiation from the internal energy.

Yes, but when the photodetector is submerged in the water (and the photodetector is at 300 K), it's also radiating power. The same amount of power that it's receiving.

If you want the photodetector to produce a current, it needs to receive more power than it emits.

One way to do this is to heat up the water. And if you wish, you can redirect the thermal radiation of the water (i.e., using a diffraction grating) away from the photodetector except for a small sliver of radiation around 3.5 micrometer wavelengths; That way, everything important is all around 3.5 micrometer wavelengths. And you'll need some mechanism to keep the photodetector from heating up along with the water.

But this still isn't in thermal equilibrium. There's energy leaving the system in the mechanism keeping the photodetector at a cool 300 K.

If you let the water cool down to 300 K, there isn't enough energy flux incident on the photodetector to cause it to produce current.

 

[Devin-M]

With a sensitive enough detector, the changes in the current generated in the detector from the water’s 3.5 micrometer light (in the darkness - no sun, room lights, etc) should be detectable via the inverse square law by moving the cup closer and further away, when the detector and cup of water are separated by a vacuum.

 

[collinsmark]

With a sensitive enough detector, the changes in the current generated in the detector from the water’s 3.5 micrometer light (in the darkness - no sun, room lights, etc) should be detectable via the inverse square law by moving the cup closer and further away, when the detector and cup of water are separated by a vacuum.

If all the room's walls are much cooler than 300 K, and everything in the photodetectors surroundings were much cooler than 300 K except for the water, then yes, you're right. But now you need to cool down the walls, or launch the system into deep space, or whatnot. This is not a case of thermal equilibrium.

But if you're in a closed room at 300 K, and the walls of the room are at 300 K, it doesn't matter if there's a cup of water in the room or not. The light flux is the same.

If the walls of the room, the cup of water, and the photodetector (and everything else in the room) are at 300 K, then no, the detector would not detect the cup of water, even if they were separated by a vacuum between them.

 

[Devin-M]

If even one single 3.5 micrometer photon from the cup of water bumps an electron from valence to conduction in the detector with the lights off and blinds lowered when the two are separated by vacuum I’m calling that a detection.

 

[collinsmark]

If even one 3.5 micrometer photon from the cup of water bumps an electron from valence to conduction in the detector when the two are separated by vacuum I’m calling that a detection.

But electrons are bumped from the valance band into the conduction band all the time purely from thermal reasons all the time in a PN junction. They also fall from the conduction band to the valance band all the time (emitting a photon in the process) all the time, again, for thermal reasons. The rate that all this happens is a function of the PN junction's temperature. Things radiate more energy when they're hotter. PN junctions are no different here.

If you call a single 3.5 micrometer photon reception on the photodetector a "detection," then you also must call a single 3.5 micrometer photon emission from the photodetector a "negative detection."

In thermal equilibrium, these "detections" and "negative detections" happen in equal amounts over time.

 

[Devin-M]

But electrons are bumped from the valance band into the conduction band all the time purely from thermal reasons all the time in a PN junction. They also fall from the conduction band to the valance band all the time (emitting a photon in the process) all the time, again, for thermal reasons. The rate that all this happens is a function of the PN junction's temperature. Things radiate more energy when they're hotter. PN junctions are no different here.

If you call a single 3.5 micrometer photon reception on the photodetector a "detection," then you also must call a single 3.5 micrometer photon emission from the photodetector a "negative detection."

In thermal equilibrium, these "detections" and "negative detections" happen in equal amounts over time.

Yes but the “detections” in this case are a useful form of photovoltaic electrical energy / current for powering electrical devices and the “negative detections” or “infrared emissions from the detector” happen also, yes, but do nothing useful.

 

[collinsmark]

Yes but the “detections” are a useful form of energy for powering electrical devices and the negative detections happen also, yes, but do nothing useful.

But if a device has an equal number of "detections" and "negative detections," how can you use that to harness useful power?

In order to harness useful power, you'll need a situation that favors "detections" over "negative detections." That, you can do, but it requires bringing the system out of thermal equilibrium (for example, introducing an external light source, and/or keeping the photodetector cooler than its surroundings).

 

[Devin-M]

An ordinary solar panel reflects some of the light that hits it, and emits some energy as infrared radiation, but the useful energy comes from the visible light that bumps the electrons from valence to conduction in the PN junction in the solar panel. The problem is typical solar panels aren’t sensitive to 3.5 micrometer light but HgCdTe PN junction photodetectors are sensitive to that band which also happens to be part of the room temperature black body spectrum.

 

[collinsmark]

An ordinary solar panel reflects some of the light that hits it, and emits some energy as infrared radiation, but the useful energy comes from the visible light that bumps the electrons from valence to conduction in the PN junction in the solar panel. The problem is typical solar panels aren’t sensitive to 3.5 micrometer light but HgCdTe photodetectors are sensitive to that band which also happens to be part of the room temperature black body spectrum.

Let me try a different approach.

Consider an ideal solar panel. This solar panel is indestructible (never melts or burns, for example) and is maximally efficient at all wavelengths. And consider that this solar panel is attached to some sort of ideal energy storate device (e.g., an ideal battery or ideal capacitor that can store energy), and goes along side the solar panel, and is capable of storing an arbitrarily large amount of energy from the solar panel.

If you were to take this solar panel setup and place it solidly into the Sun's photosphere, such that there are equal amounts of light hitting it from all directions, and allow it to heat up the to same temperature of the Sun's photosphere, the solar panel would no longer produce electricity. There's plenty of sunlight -- that's not the issue -- and yet the solar panel will still not work.

That wouldn't be because the solar panel failed in any way: remember it's an ideal solar panel. No, it wouldn't produce any useful energy simply because it is at thermal equilibrium. It would be radiating the same amount of energy that it receives.

 

[Devin-M]

The real solar panels have a spectral response from 0.3 micrometers to almost 1.2 micrometers.

https://www.pveducation.org/pvcdrom/solar-cell-operation/spectral-response

 

[Devin-M]

The 300k HgCdTe photodetector goes from 2.5 to 4.5 micrometers…

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/
https://www.researchgate.net/publication/343856156_Higher_Operating_Temperature_IR_Detectors_of_the_MOCVD_Grown_HgCdTe_Heterostructures

 

[collinsmark]

And solar panels (ideal or otherwise), just like 300k HgCdTe photodetectors, cannot extract useful energy from their environment when they are in thermal equilibrium with their environment (absent any external currents or external light sources or external heat sources from outside their environment).

HgCdTe photodetectors are no different in this respect. When in thermal equilibrium with their environment, they will radiate out the same amount of energy that they receive.

 

[Devin-M]

My point is real solar panels aren’t sensitive to 300k room temperature black body photons of 3.5 micrometers but the HgCdTe photodetectors are.

 

[collinsmark]

My point is real solar panels aren’t sensitive to 300k room temperature black body photons of 3.5 micrometers but the HgCdTe photodetectors are.

But when in thermal equilibrium with their environment, they will radiate out the same amount of energy they receive. And if they are more receptive to 300 K blackbody photons, they are also likely more susceptible to emitting 300 K blackbody photons, generally speaking. What's certainly true, is that when in thermal equilibrium (with no external currents, light sources, etc.), they can't harness useful energy. This last part is true regardless of their wavlength receptivity.

 

[Devin-M]

A solar panel can’t be thermally radiating or reflecting or converting to internal energy the same power of incident electromagnetic radiation it receives since it’s also producing electric current.

 

[collinsmark]

A solar panel can’t be thermally radiating the same power of incident electromagnetic radiation it receives since it’s also producing electric current.

But if the solar panel is producing electricity, it's not in thermal equilibrium. It's kept cool by the Earth's atmosphere and by its backside, which isn't in the light.

Even if you put a solar panel in space, it's cooled by its own backside, which is facing the cold depths of space. One side is facing the hot sun, the other side facing the cold space. From that, yes, you can extract energy (because the Sun and space are at different temperatures).

Again, if you put the solar panel (the whole kit and kaboodle) squarely inside the Sun's photosphere (and allowed it to naturally heat up to the Sun's temperature), it won't work.

 

[Devin-M]

Are you saying if I point a solar panel at the sun then use a mirror to reflect light onto the back of the wafer it will stop working?

 

[collinsmark]

Are you saying if I point a solar panel at the sun then use a mirror to reflect light onto the back of the wafer it will stop working?

If you illuminate the solar panel by a constant amount, equally from all directions from blackbody radiation, and you allow the panel's temperature to naturally rise uninhibited (free from any heat conduction and heat convection), which pretty much means you'll need to move it outside Earth's atmosphere, and allow its temperature to stabilize at equilibrium, then I suppose yes, that is what I am saying.

 

[Devin-M]

I would radiate the amount of incident power onto the solar panel that doesn’t generate external current, but that radiation is less than the incident power.

Why can’t an HgCdTe detector do the same? Radiate the non-current generating incident 3.5 micrometer power, less than the amount received, same as with solar.

 

[collinsmark]

I would radiate the amount of incident power onto the solar panel that doesn’t generate external current, but that radiation is less than the incident power.

Why can’t an HgCdTe detector do the same? Radiate the non-current generating incident 3.5 micrometer power, less than the amount received, same as with solar.

I don't understand.

By sticking a 300 K an HgCdTe detector into a glass of water at 300 K, the detector is receiving and emitting photons at equal rates in all directions. How do you limit the photon flux from some directions but not others?

Before you say you could put an object in-between the detector and the water, realize that as the object reaches thermal equilibrium, it too will be emitting thermal photons at the same rate as everything else.

 

[Devin-M]

It isn’t emitting as many photons as it’s receiving since some photons are producing current in the wires that exit the water.

 

[collinsmark]

It isn’t emitting as many photons as it’s receiving since some photons are producing current in the wires that exit the water.

That's only half the picture.

The photodetector is also emitting photons that effectively sap the current through the photodetector. And these current "producing" photons that arrive at the photodetector, and those current "sapping" photons that are emitted by the photodetector are happening in equal amounts when everything is in thermal equilibrium.

Consider a few examples:

• Red LEDs are more sensitive at detecting red light than other wavelengths of light. It goes without saying that red LEDs emit red light more than other wavelengths (that's why it's called a "red" LED). So if you want to use a device to detect red light, you should use a red one rather than a blue LED.
• When light passes through cold hydrogen gas (or any particular gas for that matter), the gas absorbs spectral lines. But when the gas is hot, it emits light at those same spectral lines.

Similarly, even though HgCdTe diode is more sensitive to detect photons at certain wavelengths, I'm willing to bet experiment will show it's also more sensitive to emit photons at those same wavelengths.

Any of these things, when in thermal equilibrium with their respective surroundings, cannot extract useful energy.

It's true that some materials can take a single high frequency photon and convert that into several low-frequency photons more than the reverse. But this inequality can only be exploited outside of thermal equilibrium, which isn't what we're discussing here. Once everything reaches thermal equilibrium, the HgCdTe diode will be emitting as much energy as it receives, and no net current is produced.

 

[Devin-M]

How would the fact that the HgCdTe diode emits some infrared black body light of its own (at 300k) prevent it from converting some incoming infrared 3.5 micrometer photons from an external source into current through an external electrical circuit?

 

[hutchphd]

The arguments here are essentially similar to why a diode can not spontaneously convert thermal noise into useable power. Or why a "Brownian Ratchet" fails. These are perhaps easier to understand in detail. Feynman adresses them well as I recall. Basicly they all fail because whatever mechanism is used to select the "good" processes it uses up more power than it provides on average in steady state. I would need to see the experimental apparatus in detail to discuss the graphs.
Good question!

 

[Devin-M]

Just as a solar panel diode uses the photovoltaic effect to create electrical current from incoming visible light, the HgCdTe diode uses the photovoltaic effect to create electrical current from incoming 3.5 micrometer infrared photons.

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]

How would the fact that the HgCdTe diode emits some infrared black body light of its own (at 300k) prevent it from converting some incoming infrared 3.5 micrometer photons from an external source into current through an external electrical circuit?

It's not just some infrared blackbody light that it emits. It's the same amount of blackbody light as it absorbs.

When it absorbs photons it generates emf that can be used to generate current in the external circuit. When it emits a photons it produces an emf with opposite polarity, which acts in the opposite way. Both are always happening to some degree. But when the detector is in thermal equilibrium with its surroundings, they happen at equal amounts and the effective emf, when all is said and done, is zero. No net current is produced for the external circuit.

 

[hutchphd]

Just as a solar panel diode uses the photovoltaic effect to create electrical current

The photodiode inside the sun (were it sturdy enough) would be just a diode and then the fluctuations dominate.

 

[Devin-M]

It's not just some infrared blackbody light that it emits. It's the same amount of blackbody light as it absorbs.

Yes some of the absorbed IR light can be re-emitted from the detector itself and the rest of the energy it absorbs can be emitted as heat generated by the current through the resistance in the external wires coming out of the detector diode.

 

[collinsmark]

Yes some of the absorbed IR light can be re-emitted from the detector itself and the rest of the energy it absorbs can be emitted as heat generated by the current through the resistance in the external wires coming out of the detector diode.

That would require the photodetector to emit less energy into its surroundings than it absorbs; the difference going into the electric circuit.

And if the energy incident on the detector has a thermal spectrum, and the detector naturally emits radiation according to its thermal spectrum, it means that the detector must be colder than its surroundings. But we've established everything is at the same temperature.

Do you see the contradiction here?

 

[Devin-M]

I’m saying shine an IR 3.5 micrometer monochrome laser of whatever power you wish on the 300k HgCdTe detector and you get current responsivity in amps per watt through an external circuit in photovoltaic mode with 0v applied bias volts.

Now shine some 3.5 micrometer photons from 300k water onto it through the vacuum and aside from the number of photons, it can’t tell the difference between the individual 3.5 micrometer photons whether laser generated or 300k black body generated.

 

[collinsmark]

I’m saying shine an IR 3.5 micrometer monochrome laser of whatever power you wish on the 300k HgCdTe detector and you get current responsivity in amps per watt through an external circuit in photovoltaic mode with 0v applied bias volts.

The details of what happens depends upon the intensity of the laser (or whatever source). But if the intensity is above minuscule, the HgCdTe detector will start to produce current, but it will also heat up. And it will keep increasing in temperature and its efficiency will decrease.

If the detector is bathed in 3.5 micrometer light in all directions, and its temperature is not inhibited (no conduction, no convection, etc.), it will eventually stop producing current. Assuming it doesn't burn up, it will eventually become hot enough such that it is thermally radiating as much energy as it is receiving from its surroundings (including the 3.5 micrometer laser source) and no more current will be generated.

Now shine some 3.5 micrometer photons from 300k water onto it through the vacuum

In that case the photodetector will get no hotter than 300 K. There's simply not enough energy flux to increase its temperature more than that (i.e., not enough photons).

it can’t tell the difference between the individual 3.5 micrometer photons whether laser generated or 300k black body generated.

Good god. The number of photons makes a huge difference.

Consider a microwave oven. Microwave ovens utilize photons of around 12.2 cm. For reference, blackbody who's peak wavelength is 12.2 cm really cold compared to the 3.5 micrometer we were previously discussing.

Your popcorn can certainly tell the difference regarding the number of photons.

 

[Devin-M]

This is how many 3.5 micrometer photons the initially 300k water in the dark is emitting into the vacuum…
https://www.speakev.com/cdn-cgi/image/format=auto,onerror=redirect,width=1920,height=1920,fit=scale-down/https://www.speakev.com/attachments/aea0d9fd-14b5-4308-a81c-f5fb6ef06d83-png.156577/

 

[collinsmark]

This is how many 3.5 micrometer photons the initially 300k water in the dark is emitting into the vacuum…
https://www.speakev.com/cdn-cgi/image/format=auto,onerror=redirect,width=1920,height=1920,fit=scale-down/https://www.speakev.com/attachments/aea0d9fd-14b5-4308-a81c-f5fb6ef06d83-png.156577/

I'm not sure I understand what I should take from that.

Getting back to my previous example, deep space (far from any stars or galaxies) has a thermal signature from the cosmic microwave background corresponding to a black body of around a handful of Kelvin. It's really cold. And yet microwaves from a microwave oven of roughly the same ballpark wavelength regularly heats my hot dogs. They are not the same. The microwave oven has a much, much larger photon flux. And that makes a difference.

So, how do you increase 3.5 micrometer photon flux above what you would get by surrounding the area with 300 K water? There are several ways, but one way is to use a hotter source, and then filter off the other wavelengths. Sure, the hot source doesn't have a blackbody peak at 3.5 micrometers, it's much hotter. None-the-less, its flux around the 3.5 micrometer region is still plenty more than surrounding the target with 300 K water.

 

[Devin-M]

If all I care about is whether a single individual incoming IR photon pushes one particular electron and electron hole out towards the external circuit from the PN junction diode, why would it be of consequence whether that photon came from the IR laser or the surface of some 300k water in the dark?