B Infrared Detectors & The 2nd Law of Thermodynamics

[Charles Link]

Even with a strong source such as a globar, when a monochromator is used at moderately high resolution to only have a small spectral increment $\Delta \lambda$ incident from the globar as each reading is taken, the incident power on the photodiode is still rather small. With the lock-in amplifier with an optical chopper you can then get a very accurate measurement.

 

[Devin-M]

If I put the incident power source in a box separated from the 300k operating temperature detector by a vacuum, and I expose the detector to 3.5 micrometer light from a laser or from a 300k cup of water what is the difference?

 

[collinsmark]

If I put the incident power source in a box separated from the 300k operating temperature detector by a vacuum, and I expose the detector to 3.5 micrometer light from a laser or from a 300k cup of water what is the difference?

It's not just about the wavelengths of photons hitting or leaving the photodetector (although, yes, that's part of it), it's also about the quantity.

In the case of the 300 K cup of water with with the photodetector submerged in the water, also at 300 K, the photodetector receives some level of power flux incident on it, but it also radiates an equal amount, such that the net power flux reaching the photodetector is zero.

In the case of the 3.5 micrometer laser (which can also be applied to the case with the broadband light source + monochromater), the power flux is much greater. The amount of photons incident on the photodetector is much greater. But the photodetector, held at 300 K, doesn't naturally radiate any more than it did in the glass of water example. A small fraction of that extra power gets converted to photodetector current, a lot of it is reflected (or passes through it; we are talking deep IR here), and the rest of it goes into the mechanism keeping the photodetector cool (the mechanism that keeps the photodetector and its immediate surroundings from heating up).

 

[Devin-M]

In the case of the 300 K cup of water with with the photodetector submerged in the water, also at 300 K, the photodetector receives some level of power flux incident on it, but it also radiates an equal amount, such that the net power flux reaching the photodetector is zero.

If I understand the PN junction correctly, the material absorbs the infrared photon but it does not re-emit an infrared photon. Instead that photon jumps an electron from the valence band to the conduction band in the material, and since it's a diode it only has one way it can go.

 

[collinsmark]

If I understand the PN junction correctly, the material absorbs the infrared photon but it does not re-emit an infrared photon. Instead that photon jumps an electron from the valence band to the conduction band in the material, and since it's a diode it only has one way it can go.

In terms of overall energy flux (regardless of particular wavelengths), it works both ways.

Photodetectors can emit light, contrary to their typical operation.
LEDs can detect light, contrary to their typical operation.



When everything is in thermal equilibrium (no external light sources, no external current through the device), the device operates right in-between. It is neither a detector nor source. It doesn't receive any more energy than it emits.

 

[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!